id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-7700 | Expectation of Negative Binomial Distribution/Type 1 | Let $X$ be a discrete random variable with the type 1 negative binomial distribution with parameters $n$ and $p$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac n p$ | From Probability Generating Function of Negative Binomial Distribution (Type 1), we have:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 1)|type 1 negative binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = \dfrac n p$ | From [[Probability Generating Function of Negative Binomial Distribution (Type 1)]], we have:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
where $q = 1 - p$.
From [[Expectation of Discrete Random Variable from PGF]], we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X... | Expectation of Negative Binomial Distribution/Type 1 | https://proofwiki.org/wiki/Expectation_of_Negative_Binomial_Distribution/Type_1 | https://proofwiki.org/wiki/Expectation_of_Negative_Binomial_Distribution/Type_1 | [
"Expectation",
"Negative Binomial Distribution (Type 1)"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 1",
"Definition:Expectation"
] | [
"Probability Generating Function of Negative Binomial Distribution/Type 1",
"Expectation of Discrete Random Variable from PGF",
"First Derivative of PGF of Negative Binomial Distribution/Type 1"
] |
proofwiki-7701 | Derivatives of PGF of Negative Binomial Distribution/Type 1 | Let $X$ be a discrete random variable with the type $1$ negative binomial distribution with parameters $n$ and $p$.
Then the derivatives of the PGF of $X$ {{WRT|Differentiation}} $s$ are:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = ...$ | The Probability Generating Function of Negative Binomial Distribution (Type 1) is:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
We have that for a given type $1$ negative binomial distribution , $n, p$ and $q$ are constant.
{{ProofWanted}}
Category:Derivatives of PGF of Negative Binomial Distribution
Category... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 1)|type $1$ negative binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Higher Derivative|derivatives]] of the [[Definition:Probability Generating Function|PG... | The [[Probability Generating Function of Negative Binomial Distribution (Type 1)]] is:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
We have that for a given [[Definition:Negative Binomial Distribution (Type 1)|type $1$ negative binomial distribution ]], $n, p$ and $q$ are constant.
{{ProofWanted}}
[[Cate... | Derivatives of PGF of Negative Binomial Distribution/Type 1 | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Negative_Binomial_Distribution/Type_1 | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Negative_Binomial_Distribution/Type_1 | [
"Derivatives of PGF of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 1)",
"Derivatives of PGFs"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 1",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Probability Generating Function"
] | [
"Probability Generating Function of Negative Binomial Distribution/Type 1",
"Definition:Negative Binomial Distribution/Type 1",
"Category:Derivatives of PGF of Negative Binomial Distribution",
"Category:Negative Binomial Distribution (Type 1)",
"Category:Derivatives of PGFs"
] |
proofwiki-7702 | First Derivative of PGF of Negative Binomial Distribution/Type 1 | Let $X$ be a discrete random variable with the type 1 negative binomial distribution (second form) with parameters $n$ and $p$.
Then the first derivative of the PGF of $X$ {{WRT|Differentiation}} $s$ is:
:$\dfrac \d {\d s} \map {\Pi_X} s = n p \paren {\dfrac {\paren {p s}^{n - 1} } {\paren {1 - q s}^{n + 1} } }$ | The Probability Generating Function of Negative Binomial Distribution (Type 1) is:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
We have that for a given type 1 negative binomial distribution, $n, p$ and $q$ are constant.
Thus we have:
{{begin-eqn}}
{{eqn | l = \frac \d {\d s} \map {\Pi_X} s
| r = \frac ... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 1)|type 1 negative binomial distribution (second form) with parameters $n$ and $p$]].
Then the [[Definition:Derivative|first derivative]] of the [[Definition:Probability Generating F... | The [[Probability Generating Function of Negative Binomial Distribution (Type 1)]] is:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
We have that for a given [[Definition:Negative Binomial Distribution (Type 1)|type 1 negative binomial distribution]], $n, p$ and $q$ are constant.
Thus we have:
{{begin-eqn... | First Derivative of PGF of Negative Binomial Distribution/Type 1 | https://proofwiki.org/wiki/First_Derivative_of_PGF_of_Negative_Binomial_Distribution/Type_1 | https://proofwiki.org/wiki/First_Derivative_of_PGF_of_Negative_Binomial_Distribution/Type_1 | [
"First Derivative of PGF of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 1)",
"Derivatives of PGFs"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 1",
"Definition:Derivative",
"Definition:Probability Generating Function"
] | [
"Probability Generating Function of Negative Binomial Distribution/Type 1",
"Definition:Negative Binomial Distribution/Type 1",
"Power Rule for Derivatives",
"Derivative of Composite Function",
"Quotient Rule for Derivatives",
"Derivative of Composite Function",
"Power Rule for Derivatives",
"Category... |
proofwiki-7703 | Nth Derivative of Reciprocal of Mth Power/Corollary | The $n$th derivative of $\dfrac 1 x$ {{WRT|Differentiation}} $x$ is:
:$\dfrac {\d^n} {\d x^n} \dfrac 1 x = \dfrac {\paren {-1}^n n!} {x^{n + 1} }$
where $n!$ denotes $n$ factorial. | Follows directly from Nth Derivative of Reciprocal of Mth Power by putting $m = 1$.
{{Qed}}
Category:Derivatives
Category:Reciprocals
5esfog7mo3vmt4mb8i0vdfikz5koxty | The [[Definition:Higher Derivative|$n$th derivative]] of $\dfrac 1 x$ {{WRT|Differentiation}} $x$ is:
:$\dfrac {\d^n} {\d x^n} \dfrac 1 x = \dfrac {\paren {-1}^n n!} {x^{n + 1} }$
where $n!$ denotes [[Definition:Factorial|$n$ factorial]]. | Follows directly from [[Nth Derivative of Reciprocal of Mth Power]] by putting $m = 1$.
{{Qed}}
[[Category:Derivatives]]
[[Category:Reciprocals]]
5esfog7mo3vmt4mb8i0vdfikz5koxty | Nth Derivative of Reciprocal of Mth Power/Corollary | https://proofwiki.org/wiki/Nth_Derivative_of_Reciprocal_of_Mth_Power/Corollary | https://proofwiki.org/wiki/Nth_Derivative_of_Reciprocal_of_Mth_Power/Corollary | [
"Derivatives",
"Reciprocals"
] | [
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Factorial"
] | [
"Nth Derivative of Reciprocal of Mth Power",
"Category:Derivatives",
"Category:Reciprocals"
] |
proofwiki-7704 | First Derivative of PGF of Negative Binomial Distribution/Type 2 | Let $X$ be a discrete random variable with the type 2 negative binomial distribution with parameters $n$ and $p$.
Then the first derivative of the PGF of $X$ {{WRT|Differentiation}} $s$ is:
:$\dfrac \d {\d s} \map {\Pi_X} s = \dfrac {n p} q \paren {\dfrac q {1 - p s} }^{n + 1}$ | The Probability Generating Function of Negative Binomial Distribution (Type 2) is:
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^n$
We have that for a given type 2 negative binomial distribution, $n, p$ and $q$ are constant.
Thus we have:
{{begin-eqn}}
{{eqn | l = \frac \d {\d s} \map {\Pi_X} s
| r = \map {\fra... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 2)|type 2 negative binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Derivative|first derivative]] of the [[Definition:Probability Generating Function|PGF]] ... | The [[Probability Generating Function of Negative Binomial Distribution (Type 2)]] is:
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^n$
We have that for a given [[Definition:Negative Binomial Distribution (Type 2)|type 2 negative binomial distribution]], $n, p$ and $q$ are constant.
Thus we have:
{{begin-eqn}}
{... | First Derivative of PGF of Negative Binomial Distribution/Type 2 | https://proofwiki.org/wiki/First_Derivative_of_PGF_of_Negative_Binomial_Distribution/Type_2 | https://proofwiki.org/wiki/First_Derivative_of_PGF_of_Negative_Binomial_Distribution/Type_2 | [
"First Derivative of PGF of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 2)",
"Derivatives of PGFs"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 2",
"Definition:Derivative",
"Definition:Probability Generating Function"
] | [
"Probability Generating Function of Negative Binomial Distribution/Type 2",
"Definition:Negative Binomial Distribution/Type 2",
"Power Rule for Derivatives",
"Derivative of Composite Function",
"Power Rule for Derivatives",
"Derivative of Composite Function",
"Category:First Derivative of PGF of Negativ... |
proofwiki-7705 | Expectation of Negative Binomial Distribution/Type 2 | Let $X$ be a discrete random variable with the type $2$ negative binomial distribution with parameters $r$ and $p$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac {r p} q$
where $q = 1 - p$. | From Probability Generating Function of Negative Binomial Distribution (Type 2):
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^r$
From Expectation of Discrete Random Variable from PGF:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \map {\frac \d {\d s} } {\paren {\dfrac... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 2)|type $2$ negative binomial distribution with parameters $r$ and $p$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = \dfrac {r p} q$
where $... | From [[Probability Generating Function of Negative Binomial Distribution (Type 2)]]:
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^r$
From [[Expectation of Discrete Random Variable from PGF]]:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \map {\frac \d {\d s} } {\... | Expectation of Negative Binomial Distribution/Type 2 | https://proofwiki.org/wiki/Expectation_of_Negative_Binomial_Distribution/Type_2 | https://proofwiki.org/wiki/Expectation_of_Negative_Binomial_Distribution/Type_2 | [
"Expectation of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 2)",
"Expectation"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 2",
"Definition:Expectation"
] | [
"Probability Generating Function of Negative Binomial Distribution/Type 2",
"Expectation of Discrete Random Variable from PGF",
"First Derivative of PGF of Negative Binomial Distribution/Type 2"
] |
proofwiki-7706 | Closure Condition for Hausdorff Space | Let $\struct {S, \tau}$ be a topological space.
Then $\struct {S, \tau}$ is a Hausdorff space {{iff}}:
:For all $x, y \in S$ such that $x \ne y$, there exists an open set $U$ such that $x \in U$ and $y \notin U^-$, where $U^-$ is the closure of $U$. | === Necessary Condition ===
Let $\struct {S, \tau}$ be a Hausdorff space.
Let $x, y \in S$ with $x \ne y$.
Then by the definition of Hausdorff space there exist open sets $U, V \subseteq S$ such that:
:$x \in U$
:$y \in V$
:$U \cap V = \O$
By Empty Intersection iff Subset of Complement, $U \subseteq S \setminus V$.
By ... | Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then $\struct {S, \tau}$ is a [[Definition:Hausdorff Space|Hausdorff space]] {{iff}}:
:For all $x, y \in S$ such that $x \ne y$, there exists an [[Definition:Open Set (Topology)|open set]] $U$ such that $x \in U$ and $y \notin U^-$, whe... | === Necessary Condition ===
Let $\struct {S, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]].
Let $x, y \in S$ with $x \ne y$.
Then by the definition of [[Definition:Hausdorff Space|Hausdorff space]] there exist [[Definition:Open Set (Topology)|open sets]] $U, V \subseteq S$ such that:
:$x \in U$
:$y \in ... | Closure Condition for Hausdorff Space | https://proofwiki.org/wiki/Closure_Condition_for_Hausdorff_Space | https://proofwiki.org/wiki/Closure_Condition_for_Hausdorff_Space | [
"Hausdorff Spaces"
] | [
"Definition:Topological Space",
"Definition:T2 Space",
"Definition:Open Set/Topology",
"Definition:Closure (Topology)"
] | [
"Definition:T2 Space",
"Definition:T2 Space",
"Definition:Open Set/Topology",
"Empty Intersection iff Subset of Complement",
"Definition:Closed Set/Topology",
"Definition:Closure (Topology)",
"Definition:Relative Complement",
"Definition:Subset",
"Definition:Open Set/Topology",
"Definition:T2 Spac... |
proofwiki-7707 | Second Derivative of PGF of Negative Binomial Distribution/Type 1 | Let $X$ be a discrete random variable with the type $1$ negative binomial distribution with parameters $n$ and $p$.
Then the second derivative of the PGF of $X$ {{WRT|Differentiation}} $s$ is:
:$\dfrac {\d^2} {\d s^2} \map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^{n + 2} \paren {\dfrac {n \paren {n - 1} + 2 n q s} ... | The Probability Generating Function of Negative Binomial Distribution (Type 1) is:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
We have that for a given negative binomial distribution, $n, p$ and $q$ are constant.
From First Derivative of PGF of Negative Binomial Distribution (Type 1):
{{begin-eqn}}
{{eqn | l... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 1)|type $1$ negative binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Second Derivative|second derivative]] of the [[Definition:Probability Generating Funct... | The [[Probability Generating Function of Negative Binomial Distribution (Type 1)]] is:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
We have that for a given [[Definition:Negative Binomial Distribution (Type 1)|negative binomial distribution]], $n, p$ and $q$ are constant.
From [[First Derivative of PGF of... | Second Derivative of PGF of Negative Binomial Distribution/Type 1 | https://proofwiki.org/wiki/Second_Derivative_of_PGF_of_Negative_Binomial_Distribution/Type_1 | https://proofwiki.org/wiki/Second_Derivative_of_PGF_of_Negative_Binomial_Distribution/Type_1 | [
"Second Derivative of PGF of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 1)",
"Derivatives of PGFs"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 1",
"Definition:Derivative/Higher Derivatives/Second Derivative",
"Definition:Probability Generating Function"
] | [
"Probability Generating Function of Negative Binomial Distribution/Type 1",
"Definition:Negative Binomial Distribution/Type 1",
"First Derivative of PGF of Negative Binomial Distribution/Type 1",
"Product Rule for Derivatives",
"First Derivative of PGF of Negative Binomial Distribution/Type 1",
"Power Rul... |
proofwiki-7708 | Derivatives of PGF of Negative Binomial Distribution/Type 2 | Let $X$ be a discrete random variable with the type $2$ negative binomial distribution with parameters $n$ and $p$.
Then the derivatives of the PGF of $X$ {{WRT|Differentiation}} $s$ are:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \dfrac {n^{\overline k} p^k} {q^k} \paren {\dfrac q {1 - p s} }^{n + k}$
where:
:$n^{\over... | Proof by induction:
The Probability Generating Function of Negative Binomial Distribution (Type 2) is:
:$\ds \map {\Pi_X} s = \paren {\frac q {1 - p s} }^n$
For all $k \in \N_{> 0}$, let $\map P k$ be the proposition:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \dfrac {n^{\overline k} p^k} {q^k} \paren {\dfrac q {1 - p s... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 2)|type $2$ negative binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Higher Derivative|derivatives]] of the [[Definition:Probability Generating Function|PG... | Proof by [[Principle of Mathematical Induction|induction]]:
The [[Probability Generating Function of Negative Binomial Distribution (Type 2)]] is:
:$\ds \map {\Pi_X} s = \paren {\frac q {1 - p s} }^n$
For all $k \in \N_{> 0}$, let $\map P k$ be the [[Definition:Proposition|proposition]]:
:$\dfrac {\d^k} {\d s^k} \... | Derivatives of PGF of Negative Binomial Distribution/Type 2 | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Negative_Binomial_Distribution/Type_2 | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Negative_Binomial_Distribution/Type_2 | [
"Derivatives of PGF of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 2)",
"Derivatives of PGFs"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 2",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Probability Generating Function",
"Definition:Rising Factorial"
] | [
"Principle of Mathematical Induction",
"Probability Generating Function of Negative Binomial Distribution/Type 2",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-7709 | Second Derivative of PGF of Negative Binomial Distribution/Type 2 | Let $X$ be a discrete random variable with the type $2$ negative binomial distribution with parameters $n$ and $p$.
Then the second derivative of the PGF of $X$ {{WRT|Differentiation}} $s$ is:
:$\dfrac {\d^2} {\d s^2} \map {\Pi_X} s = \dfrac {n \paren {n + 1} p^2} {q^2} \paren {\dfrac q {1 - p s} }^{n + 2}$
where $q = ... | The Probability Generating Function of Negative Binomial Distribution (Type 2) is:
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^n$
From Derivatives of PGF of Negative Binomial Distribution: Type 2:
:$(1): \quad \dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \dfrac {n^{\overline k} p^k} {q^k} \paren {\dfrac q {1 - p s} }^{n... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 2)|type $2$ negative binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Second Derivative|second derivative]] of the [[Definition:Probability Generating Funct... | The [[Probability Generating Function of Negative Binomial Distribution (Type 2)]] is:
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^n$
From [[Derivatives of PGF of Negative Binomial Distribution/Type 2|Derivatives of PGF of Negative Binomial Distribution: Type 2]]:
:$(1): \quad \dfrac {\d^k} {\d s^k} \map {\Pi_X}... | Second Derivative of PGF of Negative Binomial Distribution/Type 2 | https://proofwiki.org/wiki/Second_Derivative_of_PGF_of_Negative_Binomial_Distribution/Type_2 | https://proofwiki.org/wiki/Second_Derivative_of_PGF_of_Negative_Binomial_Distribution/Type_2 | [
"Second Derivative of PGF of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 2)",
"Derivatives of PGFs"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 2",
"Definition:Derivative/Higher Derivatives/Second Derivative",
"Definition:Probability Generating Function"
] | [
"Probability Generating Function of Negative Binomial Distribution/Type 2",
"Derivatives of PGF of Negative Binomial Distribution/Type 2",
"Definition:Rising Factorial",
"Category:Second Derivative of PGF of Negative Binomial Distribution",
"Category:Negative Binomial Distribution (Type 2)",
"Category:Der... |
proofwiki-7710 | Variance of Negative Binomial Distribution/Type 1 | Let $X$ be a discrete random variable with the type $1$ negative binomial distribution with parameters $r$ and $p$.
Then the variance of $X$ is given by:
:$\var X = \dfrac {r q} {p^2}$
where $q = 1 - p$. | From Variance of Discrete Random Variable from PGF:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Negative Binomial Distribution (Type 1):
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
From Expectation of Negative Binomial Distrib... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 1)|type $1$ negative binomial distribution with parameters $r$ and $p$]].
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = \dfrac... | From [[Variance of Discrete Random Variable from PGF]]:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the [[Definition:Expectation|expectation]] of $X$.
From the [[Probability Generating Function of Negative Binomial Distribution (Type 1)]]:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
From [... | Variance of Negative Binomial Distribution/Type 1 | https://proofwiki.org/wiki/Variance_of_Negative_Binomial_Distribution/Type_1 | https://proofwiki.org/wiki/Variance_of_Negative_Binomial_Distribution/Type_1 | [
"Variance of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 1)",
"Variance"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 1",
"Definition:Variance/Discrete"
] | [
"Variance of Discrete Random Variable from PGF",
"Definition:Expectation",
"Probability Generating Function of Negative Binomial Distribution/Type 1",
"Expectation of Negative Binomial Distribution/Type 1",
"Second Derivative of PGF of Negative Binomial Distribution/Type 1"
] |
proofwiki-7711 | PGF of Sum of Independent Discrete Random Variables/General Result | Let:
:$Z = X_1 + X_2 + \cdots + X_n$
where each of $X_1, X_2, \ldots, X_n$ are independent discrete random variables with PGFs $\map {\Pi_{X_1} } s, \map {\Pi_{X_2} } s, \ldots, \map {\Pi_{X_n} } s$.
Then:
:$\ds \map {\Pi_Z} s = \prod_{j \mathop = 1}^n \map {\Pi_{X_j} } s$ | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \map {\Pi_Z} s = \prod_{j \mathop = 1}^m \map {\Pi_{X_j} } s$
whenever $m \le n$
$\map P 1$ is true, as this just says $\map {\Pi_{X_1} } s = \map {\Pi_{X_1} } s$. | Let:
:$Z = X_1 + X_2 + \cdots + X_n$
where each of $X_1, X_2, \ldots, X_n$ are [[Definition:Independent Random Variables|independent]] [[Definition:Discrete Random Variable|discrete random variables]] with [[Definition:Probability Generating Function|PGFs]] $\map {\Pi_{X_1} } s, \map {\Pi_{X_2} } s, \ldots, \map {\Pi_{... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \map {\Pi_Z} s = \prod_{j \mathop = 1}^m \map {\Pi_{X_j} } s$
whenever $m \le n$
$\map P 1$ is true, as this just says $\map {\Pi_{X_1} } s = \map {\Pi_{X_1} } s$. | PGF of Sum of Independent Discrete Random Variables/General Result | https://proofwiki.org/wiki/PGF_of_Sum_of_Independent_Discrete_Random_Variables/General_Result | https://proofwiki.org/wiki/PGF_of_Sum_of_Independent_Discrete_Random_Variables/General_Result | [
"Probability Generating Functions"
] | [
"Definition:Independent Random Variables",
"Definition:Random Variable/Discrete",
"Definition:Probability Generating Function"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-7712 | Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods/Lemma | Let $\struct {S, \tau}$ be a Hausdorff space.
Let $C$ be a compact subspace of $S$.
Let $x \in S \setminus C$.
Then there exist open sets $U$ and $V$ such that $x \in U$, $C \subseteq V$, and $U \cap V = \O$. | Let $\FF$ be the set of all ordered pairs $\tuple {A, B}$ such that $A$ and $B$ are open, $x \in A$, and $A \cap B = \O$.
As a set of ordered pairs, $\FF$ constitutes a relation on $\tau$:
:$\FF \subseteq \tau \times \tau$
By the definition of Hausdorff space, for each $y \in C$ there exists an element $\tuple {A, B} \... | Let $\struct {S, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]].
Let $C$ be a [[Definition:Compact Topological Space|compact subspace]] of $S$.
Let $x \in S \setminus C$.
Then there exist [[Definition:Open Set (Topology)|open sets]] $U$ and $V$ such that $x \in U$, $C \subseteq V$, and $U \cap V = \O$. | Let $\FF$ be the [[Definition:Set|set]] of all [[Definition:Ordered Pair|ordered pairs]] $\tuple {A, B}$ such that $A$ and $B$ are [[Definition:Open Set (Topology)|open]], $x \in A$, and $A \cap B = \O$.
As a [[Definition:Set|set]] of [[Definition:Ordered Pair|ordered pairs]], $\FF$ constitutes a [[Definition:Endorela... | Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods/Lemma | https://proofwiki.org/wiki/Disjoint_Compact_Sets_in_Hausdorff_Space_have_Disjoint_Neighborhoods/Lemma | https://proofwiki.org/wiki/Disjoint_Compact_Sets_in_Hausdorff_Space_have_Disjoint_Neighborhoods/Lemma | [
"Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods"
] | [
"Definition:T2 Space",
"Definition:Compact Topological Space",
"Definition:Open Set/Topology"
] | [
"Definition:Set",
"Definition:Ordered Pair",
"Definition:Open Set/Topology",
"Definition:Set",
"Definition:Ordered Pair",
"Definition:Endorelation",
"Definition:T2 Space",
"Definition:Element",
"Definition:Image (Set Theory)/Relation/Relation",
"Definition:Compact Topological Space",
"Definition... |
proofwiki-7713 | Sum of Independent Binomial Random Variables | Let $X$ and $Y$ be discrete random variables with a binomial distribution:
:$X \sim \Binomial m p$
and
:$Y \sim \Binomial n p$
Let $X$ and $Y$ be independent.
Then their sum $Z = X + Y$ is distributed as:
:$Z \sim \Binomial {m + n} p$ | From Probability Generating Function of Poisson Distribution, we have that the probability generating functions of $X$ and $Y$ are given by:
:$\map {\Pi_X} s = \paren {q + p s}^m$
:$\map {\Pi_Y} s = \paren {q + p s}^n$
respectively.
Now because of their independence, we have:
{{begin-eqn}}
{{eqn | l = \map {\Pi_{X + Y}... | Let $X$ and $Y$ be [[Definition:Discrete Random Variable|discrete random variables]] with a [[Definition:Binomial Distribution|binomial distribution]]:
:$X \sim \Binomial m p$
and
:$Y \sim \Binomial n p$
Let $X$ and $Y$ be [[Definition:Independent Random Variables|independent]].
Then their sum $Z = X + Y$ is distri... | From [[Probability Generating Function of Poisson Distribution]], we have that the [[Definition:Probability Generating Function|probability generating functions]] of $X$ and $Y$ are given by:
:$\map {\Pi_X} s = \paren {q + p s}^m$
:$\map {\Pi_Y} s = \paren {q + p s}^n$
respectively.
Now because of their [[Definition:I... | Sum of Independent Binomial Random Variables | https://proofwiki.org/wiki/Sum_of_Independent_Binomial_Random_Variables | https://proofwiki.org/wiki/Sum_of_Independent_Binomial_Random_Variables | [
"Binomial Distribution",
"Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Independent Random Variables"
] | [
"Probability Generating Function of Poisson Distribution",
"Definition:Probability Generating Function",
"Definition:Independent Random Variables",
"PGF of Sum of Independent Discrete Random Variables",
"Exponent Combination Laws/Product of Powers",
"Definition:Probability Generating Function",
"Definit... |
proofwiki-7714 | Field has Algebraic Closure | Every field has an algebraic closure. | Let $F$ be a field.
Let $\FF$ be the collection of all extensions of $F$.
{{explain|Is $\FF$ a set or a proper class? It is not obvious that this collection is "small enough" to be a set, and I was unable to find a proof of this on ProofWiki.<br/>(On the other hand, does it matter which it is? // Yes, because Zorn's l... | Every [[Definition:Field (Abstract Algebra)|field]] has an [[Definition:Algebraic Closure|algebraic closure]]. | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\FF$ be the collection of all [[Definition:Field Extension|extensions]] of $F$.
{{explain|Is $\FF$ a [[Definition:Set|set]] or a [[Definition:Proper Class|proper class]]? It is not obvious that this collection is "small enough" to be a set, and I was un... | Field has Algebraic Closure | https://proofwiki.org/wiki/Field_has_Algebraic_Closure | https://proofwiki.org/wiki/Field_has_Algebraic_Closure | [
"Field Theory"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Algebraic Closure"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Extension",
"Definition:Set",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Ordering",
"Definition:Field Extension",
"Definition:Chain (Order Theory)",
"Definition:Field (Abstract Algebra)",
"Set is Subset of Union",
"Definitio... |
proofwiki-7715 | Point Finite Set of Open Sets in Separable Space is Countable | Let $\struct {X, \tau}$ be a separable space.
Let $\FF$ be a point finite set of open sets of $X$.
Then $\FF$ is countable. | Since $\struct {X, \tau}$ is separable, $X$ has a countable everywhere dense subset $S$.
{{WLOG}}, assume that $\O \notin \FF$.
By the definition of point finite, $\set {V \in \FF: x \in V}$ is finite for each $x \in S$.
From Open Set Characterization of Denseness, each element of $\FF$ contains an element of $S$.
From... | Let $\struct {X, \tau}$ be a [[Definition:Separable Space|separable space]].
Let $\FF$ be a [[Definition:Point Finite|point finite]] [[Definition:Set of Sets|set]] of [[Definition:Open Set (Topology)|open sets]] of $X$.
Then $\FF$ is [[Definition:Countable Set|countable]]. | Since $\struct {X, \tau}$ is [[Definition:Separable Space|separable]], $X$ has a [[Definition:Countable Set|countable]] [[Definition:Everywhere Dense|everywhere dense subset]] $S$.
{{WLOG}}, assume that $\O \notin \FF$.
By the definition of [[Definition:Point Finite|point finite]], $\set {V \in \FF: x \in V}$ is [[De... | Point Finite Set of Open Sets in Separable Space is Countable | https://proofwiki.org/wiki/Point_Finite_Set_of_Open_Sets_in_Separable_Space_is_Countable | https://proofwiki.org/wiki/Point_Finite_Set_of_Open_Sets_in_Separable_Space_is_Countable | [
"Topology"
] | [
"Definition:Separable Space",
"Definition:Point Finite",
"Definition:Set of Sets",
"Definition:Open Set/Topology",
"Definition:Countable Set"
] | [
"Definition:Separable Space",
"Definition:Countable Set",
"Definition:Everywhere Dense",
"Definition:Point Finite",
"Definition:Finite Set",
"Open Set Characterization of Denseness",
"Definition:Element",
"Definition:Element",
"Union of Set of Sets when a Set Intersects All",
"Countable Union of F... |
proofwiki-7716 | Countable Union of Finite Sets is Countable | The following statements are equivalent in $\mathrm{ZF}^-$:
:The Axiom of Countable Choice for Finite Sets holds.
:The union of any countable set of finite sets is countable. | === Axiom of Countable Choice for Finite Sets implies Countable Union Condition for Finite Sets ===
Suppose that the Axiom of Countable Choice for Finite Sets holds.
Let $\FF$ be a countable set of non-empty finite sets.
Then $\FF$ is either finite or countably infinite.
If $\FF$ is finite, then $\bigcup \FF$ is finite... | The following statements are equivalent in $\mathrm{ZF}^-$:
:The [[Axiom:Axiom of Countable Choice for Finite Sets|Axiom of Countable Choice for Finite Sets]] holds.
:The [[Definition:Set Union|union]] of any [[Definition:Countable Set|countable set]] of [[Definition:Finite Set|finite sets]] is [[Definition:Countable... | === Axiom of Countable Choice for Finite Sets implies Countable Union Condition for Finite Sets ===
Suppose that the [[Axiom:Axiom of Countable Choice for Finite Sets|Axiom of Countable Choice for Finite Sets]] holds.
Let $\FF$ be a [[Definition:Countable Set|countable set]] of non-empty [[Definition:Finite Set|finit... | Countable Union of Finite Sets is Countable | https://proofwiki.org/wiki/Countable_Union_of_Finite_Sets_is_Countable | https://proofwiki.org/wiki/Countable_Union_of_Finite_Sets_is_Countable | [
"Set Union"
] | [
"Axiom:Axiom of Countable Choice for Finite Sets",
"Definition:Set Union",
"Definition:Countable Set",
"Definition:Finite Set",
"Definition:Countable Set"
] | [
"Axiom:Axiom of Countable Choice for Finite Sets",
"Definition:Countable Set",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Countably Infinite/Set",
"Definition:Finite Set",
"Definition:Finite Set",
"Finite Union of Finite Sets is Finite",
"Definition:Countable Set",
"Definition:C... |
proofwiki-7717 | Reverse Triangle Inequality/Normed Vector Space | Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then:
:$\forall x, y \in X: \norm {x - y} \ge \size {\norm x - \norm y}$ | Let $d$ denote the metric induced by $\norm {\, \cdot \,}$, that is:
:$\map d {x, y} = \norm {x - y}$
From Metric Induced by Norm is Metric we have that $d$ is indeed a metric.
Then, from the Reverse Triangle Inequality as applied to metric spaces:
:$\forall x, y, z \in X: \size {\norm {x - z} - \norm {y - z} } \le \n... | Let $\struct {X, \norm {\, \cdot \,} }$ be a [[Definition:Normed Vector Space|normed vector space]].
Then:
:$\forall x, y \in X: \norm {x - y} \ge \size {\norm x - \norm y}$ | Let $d$ denote the [[Definition:Metric Induced by Norm|metric induced by $\norm {\, \cdot \,}$]], that is:
:$\map d {x, y} = \norm {x - y}$
From [[Metric Induced by Norm is Metric]] we have that $d$ is indeed a [[Definition:Metric|metric]].
Then, from the [[Reverse Triangle Inequality]] as applied to [[Definition:Me... | Reverse Triangle Inequality/Normed Vector Space | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Normed_Vector_Space | https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Normed_Vector_Space | [
"Triangle Inequality"
] | [
"Definition:Normed Vector Space"
] | [
"Definition:Metric Induced by Norm",
"Metric Induced by Norm is Metric",
"Definition:Metric Space/Metric",
"Reverse Triangle Inequality",
"Definition:Metric Space"
] |
proofwiki-7718 | Equivalence of Forms of Axiom of Countable Choice | The following forms of the Axiom of Countable Choice are equivalent in $\mathrm{ZF}^-$:
{{explain|$\mathrm{ZF}^-$}} | === Form 1 implies Form 2 ===
Suppose that Axiom of Countable Choice: Form 1 holds.
Let $S$ be a countable set of non-empty sets.
Then $S$ is either finite or countably infinite.
If $S$ is finite, then it has a choice function by the Principle of Finite Choice.
Suppose instead that $S$ is countably infinite.
Then there... | The following forms of the [[Axiom:Axiom of Countable Choice|Axiom of Countable Choice]] are equivalent in $\mathrm{ZF}^-$:
{{explain|$\mathrm{ZF}^-$}} | === Form 1 implies Form 2 ===
Suppose that [[Axiom:Axiom of Countable Choice/Form 1|Axiom of Countable Choice: Form 1]] holds.
Let $S$ be a [[Definition:Countable Set|countable set]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|sets]].
Then $S$ is either [[Definition:Finite Set|finite]] or [[Definition... | Equivalence of Forms of Axiom of Countable Choice | https://proofwiki.org/wiki/Equivalence_of_Forms_of_Axiom_of_Countable_Choice | https://proofwiki.org/wiki/Equivalence_of_Forms_of_Axiom_of_Countable_Choice | [
"Axiom of Countable Choice"
] | [
"Axiom:Axiom of Countable Choice"
] | [
"Axiom:Axiom of Countable Choice/Form 1",
"Definition:Countable Set",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Finite Set",
"Definition:Countably Infinite/Set",
"Definition:Finite Set",
"Definition:Choice Function",
"Principle of Finite Choice",
"Definition:Countably Infinite/Set"... |
proofwiki-7719 | GCD from Generator of Ideal | Let $m, n \in \Z$, with either $m \ne 0$ or $n \ne 0$.
Let $I = \ideal {m, n}$ be the ideal generated by $m$ and $n$.
Let $d$ be a non-negative generator for the principal ideal $I$.
Then:
:$\gcd \set {m, n} = d$
where $\gcd \set {m, n}$ denotes the greatest common divisor of $m$ and $n$. | First we show that such an element $d$ exists.
By Ring of Integers is Principal Ideal Domain there exists a generator $e$ of $I$.
If $e < 0$, then since the units of $\Z$ are $\set {\pm 1}$, we have by definition that $-e$ is an associate of $e$.
Therefore by $(3)$ of Principal Ideals in Integral Domain $-e > 0$ is als... | Let $m, n \in \Z$, with either $m \ne 0$ or $n \ne 0$.
Let $I = \ideal {m, n}$ be the [[Definition:Ideal of Ring|ideal]] [[Definition:Generator of Ideal|generated]] by $m$ and $n$.
Let $d$ be a non-[[Definition:Negative|negative]] generator for the [[Definition:Principal Ideal of Ring|principal ideal]] $I$.
Then:
:$... | First we show that such an element $d$ exists.
By [[Ring of Integers is Principal Ideal Domain]] there exists a generator $e$ of $I$.
If $e < 0$, then since the [[Units of Ring of Integers|units of $\Z$]] are $\set {\pm 1}$, we have by definition that $-e$ is an [[Definition:Associate of Integer|associate]] of $e$.
... | GCD from Generator of Ideal | https://proofwiki.org/wiki/GCD_from_Generator_of_Ideal | https://proofwiki.org/wiki/GCD_from_Generator_of_Ideal | [
"Greatest Common Divisor"
] | [
"Definition:Ideal of Ring",
"Definition:Generator of Ideal of Ring",
"Definition:Negative",
"Definition:Principal Ideal of Ring",
"Definition:Greatest Common Divisor/Integers"
] | [
"Ring of Integers is Principal Ideal Domain",
"Invertible Integers under Multiplication",
"Definition:Associate/Integers",
"Principal Ideals in Integral Domain",
"Definition:Generator of Ideal of Ring",
"Bézout's Identity",
"Definition:Smallest Element",
"Definition:Positive/Integer",
"Definition:In... |
proofwiki-7720 | Finite Union of Finite Sets is Finite | Let $S$ be a finite set of finite sets.
Then the union of $S$ is finite. | The proof proceeds by induction.
Let $S$ be a finite set with cardinality $n$.
If $n = 0$ then $S = \O$, so $\bigcup S = \O$, which is finite.
Suppose that an arbitrary finite set with cardinality $n$ of finite sets has a finite union.
Let $S$ have cardinality $n^+$.
Then there is a bijection $f: n^+ \to S$.
Then:
:$\d... | Let $S$ be a [[Definition:Finite Set|finite set]] of [[Definition:Finite Set|finite sets]].
Then the [[Definition:Finite Union|union]] of $S$ is [[Definition:Finite Set|finite]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
Let $S$ be a [[Definition:Finite Set|finite set]] with [[Definition:Cardinality|cardinality]] $n$.
If $n = 0$ then $S = \O$, so $\bigcup S = \O$, which is [[Definition:Finite Set|finite]].
Suppose that an arbitrary [[Definition:Finite Set|finit... | Finite Union of Finite Sets is Finite/Proof 1 | https://proofwiki.org/wiki/Finite_Union_of_Finite_Sets_is_Finite | https://proofwiki.org/wiki/Finite_Union_of_Finite_Sets_is_Finite/Proof_1 | [
"Set Union",
"Finite Sets",
"Finite Union of Finite Sets is Finite"
] | [
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Set Union/Finite Union",
"Definition:Finite Set"
] | [
"Principle of Mathematical Induction",
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Finite Set",
"Definition:Set Union/Finite Union",
"Definition:Cardinality",
"Definition:Bijection",
"Union of Finite S... |
proofwiki-7721 | Finite Union of Finite Sets is Finite | Let $S$ be a finite set of finite sets.
Then the union of $S$ is finite. | Let $S = \set {A_1, \ldots, A_n}$ such that $A_k$ is finite $\forall k = 1, \ldots, n$.
Set:
: $m = \max \set {\card {A_1}, \ldots, \card {A_n} }$
Then:
: $\ds \card {\bigcup_{k \mathop = 1}^n A_k} \le \sum_{k \mathop = 1}^n \card {A_k} \le \sum_{k \mathop = 1}^n m = n m$
{{explain|needs to invoke a link to result that... | Let $S$ be a [[Definition:Finite Set|finite set]] of [[Definition:Finite Set|finite sets]].
Then the [[Definition:Finite Union|union]] of $S$ is [[Definition:Finite Set|finite]]. | Let $S = \set {A_1, \ldots, A_n}$ such that $A_k$ is [[Definition:Finite Set|finite]] $\forall k = 1, \ldots, n$.
Set:
: $m = \max \set {\card {A_1}, \ldots, \card {A_n} }$
Then:
: $\ds \card {\bigcup_{k \mathop = 1}^n A_k} \le \sum_{k \mathop = 1}^n \card {A_k} \le \sum_{k \mathop = 1}^n m = n m$
{{explain|needs to... | Finite Union of Finite Sets is Finite/Proof 2 | https://proofwiki.org/wiki/Finite_Union_of_Finite_Sets_is_Finite | https://proofwiki.org/wiki/Finite_Union_of_Finite_Sets_is_Finite/Proof_2 | [
"Set Union",
"Finite Sets",
"Finite Union of Finite Sets is Finite"
] | [
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Set Union/Finite Union",
"Definition:Finite Set"
] | [
"Definition:Finite Set"
] |
proofwiki-7722 | Open Set Characterization of Denseness | Let $\struct {X, \tau}$ be a topological space.
Let $S \subseteq X$.
Then $S$ is (everywhere) dense in $X$ {{iff}} every non-empty ($\tau$-)open set contains an element of $S$. | === Necessary Condition ===
Let $S$ be everywhere dense in $X$.
Let $U$ be open and non-empty.
Then $U$ has an element $x$.
Since $S$ is everywhere dense in $X$, $x \in S^-$, the closure of $S$.
By Equivalence of Definitions of Adherent Point of Set, every open neighborhood of $x$ contains an element of $S$.
Thus in pa... | Let $\struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $S \subseteq X$.
Then $S$ is [[Definition:Everywhere Dense|(everywhere) dense]] in $X$ {{iff}} every [[Definition:Non-Empty Set|non-empty]] ($\tau$-)[[Definition:Open Set (Topology)|open]] set contains an [[Definition:Element|elemen... | === Necessary Condition ===
Let $S$ be [[Definition:Everywhere Dense|everywhere dense]] in $X$.
Let $U$ be [[Definition:Open Set (Topology)|open]] and [[Definition:Non-Empty Set|non-empty]].
Then $U$ has an [[Definition:Element|element]] $x$.
Since $S$ is [[Definition:Everywhere Dense|everywhere dense]] in $X$, $x ... | Open Set Characterization of Denseness | https://proofwiki.org/wiki/Open_Set_Characterization_of_Denseness | https://proofwiki.org/wiki/Open_Set_Characterization_of_Denseness | [
"Denseness"
] | [
"Definition:Topological Space",
"Definition:Everywhere Dense",
"Definition:Non-Empty Set",
"Definition:Open Set/Topology",
"Definition:Element"
] | [
"Definition:Everywhere Dense",
"Definition:Open Set/Topology",
"Definition:Non-Empty Set",
"Definition:Element",
"Definition:Everywhere Dense",
"Definition:Closure (Topology)",
"Equivalence of Definitions of Adherent Point of Set",
"Definition:Open Neighborhood/Point",
"Definition:Element",
"Defin... |
proofwiki-7723 | Union of Set of Sets when a Set Intersects All | Let $F$ be a set of sets.
Let $S$ be a set or class.
Suppose that:
:$\forall A \in F: A \cap S \ne \O$
Then:
:$\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$ | Suppose that $B \in F$.
Then $B \cap S$ has an element $x_B$.
Thus:
:$B \in \set {A \in F: x_B \in A}$
By the definition of union:
:$\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
Suppose instead that:
:$\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
Then by the definition of union, there ex... | Let $F$ be a [[Definition:Set of Sets|set of sets]].
Let $S$ be a [[Definition:Set|set]] or [[Definition:Class (Class Theory)|class]].
Suppose that:
:$\forall A \in F: A \cap S \ne \O$
Then:
:$\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$ | Suppose that $B \in F$.
Then $B \cap S$ has an element $x_B$.
Thus:
:$B \in \set {A \in F: x_B \in A}$
By the definition of [[Definition:Union of Family|union]]:
:$\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
Suppose instead that:
:$\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
The... | Union of Set of Sets when a Set Intersects All | https://proofwiki.org/wiki/Union_of_Set_of_Sets_when_a_Set_Intersects_All | https://proofwiki.org/wiki/Union_of_Set_of_Sets_when_a_Set_Intersects_All | [
"Set Union"
] | [
"Definition:Set of Sets",
"Definition:Set",
"Definition:Class (Class Theory)"
] | [
"Definition:Set Union/Family of Sets",
"Definition:Set Union/Family of Sets",
"Definition:Set Equality",
"Category:Set Union"
] |
proofwiki-7724 | Order of Group of Units of Integers Modulo n | Let $n \in \Z_{\ge 0}$ be an integer.
Let $\struct {\Z / n \Z, +, \cdot}$ be the ring of integers modulo $n$.
Let $U = \struct {\paren {\Z / n \Z}^\times, \cdot}$ denote the group of units of this ring.
Then:
:$\order U = \map \phi n$
where $\phi$ denotes the Euler $\phi$-function. | By Reduced Residue System under Multiplication forms Abelian Group, $U$ is equal to the set of integers modulo $n$ which are coprime to $n$.
It follows by Cardinality of Reduced Residue System:
:$\order U = \map \phi n$
{{qed}} | Let $n \in \Z_{\ge 0}$ be an [[Definition:Integer|integer]].
Let $\struct {\Z / n \Z, +, \cdot}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $n$]].
Let $U = \struct {\paren {\Z / n \Z}^\times, \cdot}$ denote the [[Definition:Group of Units of Ring|group of units]] of this ring.
Then:
:$\or... | By [[Reduced Residue System under Multiplication forms Abelian Group]], $U$ is equal to the [[Definition:Integers Modulo m|set of integers modulo $n$]] which are [[Definition:Coprime Integers|coprime]] to $n$.
It follows by [[Cardinality of Reduced Residue System]]:
:$\order U = \map \phi n$
{{qed}} | Order of Group of Units of Integers Modulo n | https://proofwiki.org/wiki/Order_of_Group_of_Units_of_Integers_Modulo_n | https://proofwiki.org/wiki/Order_of_Group_of_Units_of_Integers_Modulo_n | [
"Ring of Integers Modulo m",
"Euler Phi Function"
] | [
"Definition:Integer",
"Definition:Ring of Integers Modulo m",
"Definition:Group of Units/Ring",
"Definition:Euler Phi Function"
] | [
"Reduced Residue System under Multiplication forms Abelian Group",
"Definition:Integers Modulo m",
"Definition:Coprime/Integers",
"Cardinality of Reduced Residue System"
] |
proofwiki-7725 | Cyclicity Condition for Units of Ring of Integers Modulo n | Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $\struct {\Z / n \Z, +, \times}$ be the ring of integers modulo $n$.
Let $U = \struct {\paren {\Z / n \Z}^\times, \times}$ denote the group of units of $\struct {\Z / n \Z, +, \times}$.
Then $U$ is cyclic {{iff}} either:
:$n = p^\alpha$
or:
:$n = 2 p^\alpha$
where $p \g... | {{tidy}}
{{MissingLinks}} | Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Let $\struct {\Z / n \Z, +, \times}$ be the [[Definition:Ring (Abstract Algebra)|ring]] of [[Definition:Integers Modulo m|integers modulo $n$]].
Let $U = \struct {\paren {\Z / n \Z}^\times, \times}$ denote the [[Definition:Group of Units of... | {{tidy}}
{{MissingLinks}} | Cyclicity Condition for Units of Ring of Integers Modulo n | https://proofwiki.org/wiki/Cyclicity_Condition_for_Units_of_Ring_of_Integers_Modulo_n | https://proofwiki.org/wiki/Cyclicity_Condition_for_Units_of_Ring_of_Integers_Modulo_n | [
"Ring of Integers Modulo m",
"Proofs by Induction"
] | [
"Definition:Positive/Integer",
"Definition:Ring (Abstract Algebra)",
"Definition:Integers Modulo m",
"Definition:Group of Units/Ring",
"Definition:Cyclic Group",
"Definition:Prime Number"
] | [] |
proofwiki-7726 | Chinese Remainder Theorem/Corollary | Let $n_1, n_2, \ldots, n_r$ be pairwise coprime positive integers.
Let $\ds N = \prod_{i \mathop = 1}^r n_i$.
For an integer $k$, let $\Z / k \Z$ denote the ring of integers modulo $k$.
Then we have a ring isomorphism:
:$\Z / N \Z \simeq \Z / n_1 \Z \times \cdots \times \Z / n_r \Z$ | Define a mapping:
:$\phi: \Z / N \Z \to \Z / n_1 \Z \times \cdots \times \Z / n_r \Z$
by:
:$\map \phi {d \pmod N} = \paren {d \pmod {n_1}, \ldots, d \pmod {n_r} }$
Then, by Mappings Between Residue Classes, $\phi$ is well-defined.
By the definition of multiplication and addition in $\Z / k \Z$, $k \in \Z$ we have:
:$\p... | Let $n_1, n_2, \ldots, n_r$ be [[Definition:Pairwise Coprime Integers|pairwise coprime]] [[Definition:Positive Integer|positive integers]].
Let $\ds N = \prod_{i \mathop = 1}^r n_i$.
For an integer $k$, let $\Z / k \Z$ denote the [[Definition:Ring of Integers Modulo m|ring of integers modulo $k$]].
Then we have a [[... | Define a mapping:
:$\phi: \Z / N \Z \to \Z / n_1 \Z \times \cdots \times \Z / n_r \Z$
by:
:$\map \phi {d \pmod N} = \paren {d \pmod {n_1}, \ldots, d \pmod {n_r} }$
Then, by [[Mappings Between Residue Classes]], $\phi$ is [[Definition:Well-Defined Mapping|well-defined]].
By the definition of multiplication and additi... | Chinese Remainder Theorem/Corollary | https://proofwiki.org/wiki/Chinese_Remainder_Theorem/Corollary | https://proofwiki.org/wiki/Chinese_Remainder_Theorem/Corollary | [
"Chinese Remainder Theorem",
"Commutative Algebra"
] | [
"Definition:Pairwise Coprime/Integers",
"Definition:Positive/Integer",
"Definition:Ring of Integers Modulo m",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism"
] | [
"Mappings Between Residue Classes",
"Definition:Well-Defined/Mapping",
"Definition:Ring Homomorphism",
"Chinese Remainder Theorem",
"Definition:Surjection",
"Definition:Injection"
] |
proofwiki-7727 | Hilbert's Basis Theorem/Corollary | Let $A$ be a Noetherian ring.
Let $n \ge 1$ be an integer.
Let $A \sqbrk {x_1, \ldots, x_n}$ be the ring of polynomial forms over $A$ in the indeterminates $x_1, \ldots, x_n$.
Then $A \sqbrk {x_1, \ldots, x_n}$ is also a Noetherian ring. | We proceed by induction over $n \ge 1$. | Let $A$ be a [[Definition:Noetherian Ring|Noetherian ring]].
Let $n \ge 1$ be an [[Definition:Integer|integer]].
Let $A \sqbrk {x_1, \ldots, x_n}$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms over $A$ in the indeterminates $x_1, \ldots, x_n$]].
Then $A \sqbrk {x_1, \ldots, x_n}$ is also a [... | We proceed by [[Principle of Mathematical Induction|induction]] over $n \ge 1$. | Hilbert's Basis Theorem/Corollary | https://proofwiki.org/wiki/Hilbert's_Basis_Theorem/Corollary | https://proofwiki.org/wiki/Hilbert's_Basis_Theorem/Corollary | [
"Commutative Algebra"
] | [
"Definition:Noetherian Ring",
"Definition:Integer",
"Definition:Ring of Polynomial Forms",
"Definition:Noetherian Ring"
] | [
"Principle of Mathematical Induction"
] |
proofwiki-7728 | Minkowski's Theorem | Let $L$ be a lattice in $\R^n$.
Let $d$ be the covolume of $L$.
Let $\mu$ be a translation invariant measure on $\R^n$
Let $S$ be a convex subset of $\R^n$ that is symmetric about the origin, i.e. such that:
:$\forall p \in S : -p \in S$
Let the volume of $S$ be greater than $2^n d$.
Then $S$ contains a non-zero point... | Let $D$ be any fundamental parallelepiped.
Then by definition:
:$\ds \R^n = \coprod \limits_{\vec x \mathop \in L} \paren {D + \vec x}$
where:
: $A + \vec x := \set {\vec a + \vec x : \vec a \in A}$
By Intersection with Subset is Subset:
:$\dfrac 1 2 S \cap \R^n = \dfrac 1 2 S \iff \dfrac 1 2 S \subseteq \R^n$
Hence by... | Let $L$ be a [[Definition:Integral Lattice|lattice]] in $\R^n$.
Let $d$ be the [[Definition:Covolume of Lattice|covolume]] of $L$.
Let $\mu$ be a [[Definition:Translation Invariant Measure|translation invariant measure]] on $\R^n$
Let $S$ be a [[Definition:Convex Set (Vector Space)|convex]] [[Definition:Subset|subse... | Let $D$ be any [[Definition:Fundamental Parallelepiped|fundamental parallelepiped]].
Then by definition:
:$\ds \R^n = \coprod \limits_{\vec x \mathop \in L} \paren {D + \vec x}$
where:
: $A + \vec x := \set {\vec a + \vec x : \vec a \in A}$
By [[Intersection with Subset is Subset]]:
:$\dfrac 1 2 S \cap \R^n = \dfr... | Minkowski's Theorem | https://proofwiki.org/wiki/Minkowski's_Theorem | https://proofwiki.org/wiki/Minkowski's_Theorem | [
"Number Theory",
"Geometry of Numbers",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Integral Lattice",
"Definition:Covolume of Lattice",
"Definition:Translation Invariant Measure",
"Definition:Convex Set (Vector Space)",
"Definition:Subset",
"Definition:Coordinate System/Origin",
"Definition:Volume",
"Definition:Zero Vector"
] | [
"Definition:Fundamental Domain (Lattice)",
"Intersection with Subset is Subset",
"Intersection Distributes over Union",
"Definition:Set Intersection",
"Definition:Translation Invariant Measure",
"Definition:Set",
"Definition:Pairwise Disjoint",
"Dilation of Lebesgue-Measurable Set is Lebesgue-Measurab... |
proofwiki-7729 | Component of Finite Union in Ultrafilter | Let $S$ be a non-empty set.
Let $\UU$ be a ultrafilter on $S$.
Let $\set {Y_1, \dots, Y_n}$ be a pairwise disjoint set of subsets of $S$ such that $S = Y_1 \cup \cdots \cup Y_n$.
Then there is a unique $k \in \set {1, \dots, n}$ such that $Y_k \in \UU$. | Assume that none of the $Y_k$ are empty.
If not, then any empty ones can simply be removed. | Let $S$ be a [[Definition:Non-Empty Set|non-empty set]].
Let $\UU$ be a [[Definition:Ultrafilter on Set|ultrafilter]] on $S$.
Let $\set {Y_1, \dots, Y_n}$ be a [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$ such that $S = Y_1 \cup \cdots \cup Y_n$.
... | Assume that none of the $Y_k$ are [[Definition:Empty Set|empty]].
If not, then any [[Definition:Empty Set|empty]] ones can simply be removed. | Component of Finite Union in Ultrafilter | https://proofwiki.org/wiki/Component_of_Finite_Union_in_Ultrafilter | https://proofwiki.org/wiki/Component_of_Finite_Union_in_Ultrafilter | [
"Filter Theory"
] | [
"Definition:Non-Empty Set",
"Definition:Ultrafilter on Set",
"Definition:Pairwise Disjoint",
"Definition:Set",
"Definition:Subset"
] | [
"Definition:Empty Set",
"Definition:Empty Set"
] |
proofwiki-7730 | Restricted Tukey's Theorem/Weak Form | Let $X$ be a set.
Let $\AA$ be a non-empty set of subsets of $X$.
Let $'$ be a unary operation on $X$.
Let $\AA$ have finite character.
For all $A \in \AA$ and all $x \in X$, let either:
:$A \cup \set x \in \AA$
or:
:$A \cup \set {x'} \in \AA$
Then there exists a $B \in \AA$ such that for all $x \in X$, either $x \in B... | {{ProofWanted}}
{{Namedfor|John Wilder Tukey|cat = Tukey}} | Let $X$ be a [[Definition:Set|set]].
Let $\AA$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Subset|subsets]] of $X$.
Let $'$ be a [[Definition:Unary Operation|unary operation]] on $X$.
Let $\AA$ have finite character.
For all $A \in \AA$ and all $x \in X$, let either:
:$A \cup \set x \in \AA$
or:... | {{ProofWanted}}
{{Namedfor|John Wilder Tukey|cat = Tukey}} | Restricted Tukey's Theorem/Weak Form | https://proofwiki.org/wiki/Restricted_Tukey's_Theorem/Weak_Form | https://proofwiki.org/wiki/Restricted_Tukey's_Theorem/Weak_Form | [
"Tukey's Lemma"
] | [
"Definition:Set",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Operation/Unary Operation"
] | [] |
proofwiki-7731 | Restricted Tukey's Theorem/Strong Form | Let $X$ be a set.
Let $\AA$ be a non-empty set of subsets of $X$.
Let $'$ be a unary operation on $X$.
Let $\AA$ have finite character.
For all $A \in \AA$ and all $x \in X$, let either:
: $A \cup \set x \in \AA$
or:
:$A \cup \set {x'} \in \AA$
Then for each $A \in \AA$ there exists a $C \in \AA$ such that:
:$A \subset... | Let $A \in \AA$.
Let:
:$\BB = \set {B: B \subseteq X \text{ and } A \cup B \in \AA}$
It is to be shown that $\BB$ has finite character:
First suppose that $B \in \BB$ and $F$ is a Definition:Finite Subset subset of $B$.
Then since $B \in \BB$, $B \subseteq X$ and $A \cup B \in \AA$.
We wish to show that $F \in \BB$.
Si... | Let $X$ be a [[Definition:Set|set]].
Let $\AA$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Subset|subsets]] of $X$.
Let $'$ be a [[Definition:Unary Operation|unary operation]] on $X$.
Let $\AA$ have [[Definition:Finite Character|finite character]].
For all $A \in \AA$ and all $x \in X$, let eit... | Let $A \in \AA$.
Let:
:$\BB = \set {B: B \subseteq X \text{ and } A \cup B \in \AA}$
It is to be shown that $\BB$ has [[Definition:Finite Character|finite character]]:
First suppose that $B \in \BB$ and $F$ is a [[Definition:Finite Subset subset]] of $B$.
Then since $B \in \BB$, $B \subseteq X$ and $A \cup B \in \... | Restricted Tukey's Theorem/Strong Form | https://proofwiki.org/wiki/Restricted_Tukey's_Theorem/Strong_Form | https://proofwiki.org/wiki/Restricted_Tukey's_Theorem/Strong_Form | [
"Tukey's Lemma"
] | [
"Definition:Set",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Operation/Unary Operation",
"Definition:Finite Character"
] | [
"Definition:Finite Character",
"Definition:Finite Subset subset",
"Restricted Tukey's Theorem/Weak Form"
] |
proofwiki-7732 | Boolean Prime Ideal Theorem/Extension Lemma | Let $\struct {B, \vee, \wedge, \le}$ be a Boolean lattice.
Let $F \subseteq B$ be a filter.
Let $J \subseteq B$ have the finite join property with respect to $F$.
Let $z \in B$.
Then either $J \cup \set z$ or $J \cup \set {\neg z}$ also has the finite join property with respect to $F$. | {{AimForCont}} that neither $J \vee z$ nor $J \vee \neg z$ has the finite join property with respect to $F$.
Then there are $x_1, \dots, x_n, y_1, \dots, y_m \in J$ such that $x_1 \vee \dots \vee x_n \vee z \in F$ and $y_1 \vee \dots \vee y_m \vee \neg z \in F$.
Let $q = x_1 \vee \dots \vee x_n \vee y_1 \vee \dots \vee... | Let $\struct {B, \vee, \wedge, \le}$ be a [[Definition:Boolean Lattice|Boolean lattice]].
Let $F \subseteq B$ be a [[Definition:Filter|filter]].
Let $J \subseteq B$ have the [[Definition:Finite Join Property|finite join property]] with respect to $F$.
Let $z \in B$.
Then either $J \cup \set z$ or $J \cup \set {\ne... | {{AimForCont}} that neither $J \vee z$ nor $J \vee \neg z$ has the [[Definition:Finite Join Property|finite join property]] with respect to $F$.
Then there are $x_1, \dots, x_n, y_1, \dots, y_m \in J$ such that $x_1 \vee \dots \vee x_n \vee z \in F$ and $y_1 \vee \dots \vee y_m \vee \neg z \in F$.
Let $q = x_1 \vee \... | Boolean Prime Ideal Theorem/Extension Lemma | https://proofwiki.org/wiki/Boolean_Prime_Ideal_Theorem/Extension_Lemma | https://proofwiki.org/wiki/Boolean_Prime_Ideal_Theorem/Extension_Lemma | [
"Lattice Theory"
] | [
"Definition:Boolean Lattice",
"Definition:Filter",
"Definition:Finite Join Property",
"Definition:Finite Join Property"
] | [
"Definition:Finite Join Property",
"Join Succeeds Operands",
"Axiom:Filter Axioms",
"Filter is Closed under Meet",
"Axiom:Distributive Lattice Axioms",
"Meet Absorbs Join",
"Meet is Commutative",
"Axiom:Distributive Lattice Axioms",
"Join Absorbs Meet",
"Definition:Finite Join Property",
"Catego... |
proofwiki-7733 | Finite Character for Sets of Mappings | Let $S$ and $T$ be sets.
Let $\FF$ be a set of mappings from subsets of $S$ to $T$.
That is, let $\FF$ be a set of partial mappings from $S$ to $T$.
Then the following are equivalent:
{{begin-axiom}}
{{axiom | n = 1
| t = $\FF$ has finite character in the sense of Definition:Finite Character/Mappings.
}}
{{axio... | === $(1)$ implies $(2)$ ===
Let $\FF$ have finite character in the sense of Definition:Finite Character/Mappings.
That is, suppose that for each partial mapping $f \subseteq S \times T$:
:$f \in \FF$ {{iff}} for each finite subset $K$ of the domain of $f$, the restriction of $f$ to $K$ is in $\FF$.
Let $q \subseteq S \... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\FF$ be a [[Definition:Set|set]] of [[Definition:Mapping|mappings]] from [[Definition:Subset|subsets]] of $S$ to $T$.
That is, let $\FF$ be a [[Definition:Set|set]] of [[Definition:Partial Mapping|partial mappings]] from $S$ to $T$.
Then the following are equivalent:... | === $(1)$ implies $(2)$ ===
Let $\FF$ have finite character in the sense of [[Definition:Finite Character/Mappings]].
That is, suppose that for each [[Definition:Partial Mapping|partial mapping]] $f \subseteq S \times T$:
:$f \in \FF$ {{iff}} for each [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $K$... | Finite Character for Sets of Mappings | https://proofwiki.org/wiki/Finite_Character_for_Sets_of_Mappings | https://proofwiki.org/wiki/Finite_Character_for_Sets_of_Mappings | [
"Set Theory"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Set",
"Definition:Many-to-One Relation",
"Definition:Finite Character/Mappings",
"Definition:Set",
"Definition:Subset",
"Definition:Finite Character"
] | [
"Definition:Finite Character/Mappings",
"Definition:Many-to-One Relation",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Restriction",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Many-to-One Relation",
"Definition:Domain (Set Theo... |
proofwiki-7734 | Cowen-Engeler Lemma | Let $X$ and $Y$ be sets.
Let $M$ be a set of mappings from subsets of $X$ to $Y$.
Thus each element of $M$ is a partial mapping from $X$ to $Y$.
Define a mapping:
:$\Phi: X \to \powerset Y$
thus:
:$\map \Phi x = \set {\map f x: f \in M \text{ and } x \in \Dom f}$
That is:
:$\map \Phi x = \set {y \in Y: \exists f \in M... | Let $\map {\operatorname {Fin} } X$ be the set of finite subsets of $X$.
For each $S \in \map {\operatorname {Fin} } X$, let:
:$\Gamma_S = \set {f \in M: S \subseteq \Dom f}$
By $(2)$, $\Gamma_S$ is non-empty for each $S \in \map {\operatorname {Fin} } X$.
{{tidy|presentation is ugly and difficult to follow}}
For each ... | Let $X$ and $Y$ be [[Definition:Set|sets]].
Let $M$ be a [[Definition:Set|set]] of [[Definition:Mapping|mappings]] from [[Definition:Subset|subsets]] of $X$ to $Y$.
Thus each [[Definition:Element|element]] of $M$ is a [[Definition:Partial Mapping|partial mapping]] from $X$ to $Y$.
Define a [[Definition:Mapping|mapp... | Let $\map {\operatorname {Fin} } X$ be the [[Definition:Set|set]] of [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $X$.
For each $S \in \map {\operatorname {Fin} } X$, let:
:$\Gamma_S = \set {f \in M: S \subseteq \Dom f}$
By $(2)$, $\Gamma_S$ is [[Definition:Non-Empty Set|non-empty]] for each $S \... | Cowen-Engeler Lemma | https://proofwiki.org/wiki/Cowen-Engeler_Lemma | https://proofwiki.org/wiki/Cowen-Engeler_Lemma | [
"Set Theory"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Element",
"Definition:Many-to-One Relation",
"Definition:Mapping",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Element",
"Definition:Domain (Set Theory)/Mapping",
... | [
"Definition:Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Non-Empty Set",
"Definition:Finite Intersection Property",
"Definition:Ultrafilter on Set",
"Definition:Subset",
"Definition:Mapping",
"Definition:Set Partition",
"Component of Finite Union in Ultrafilter",
"Definition:F... |
proofwiki-7735 | Ultrafilter Lemma/Corollary | Let $S$ be a non-empty set.
Let $\AA$ be a set of subsets of $S$.
Suppose that $\AA$ has the finite intersection property.
Then there is an ultrafilter $\UU$ on $S$ such that $\AA \subseteq \UU$. | Let $\II$ be the set of intersections of non-empty finite subsets of $\AA$.
Let $\FF = \set {T \in \powerset S: \exists B \in \II: B \subseteq T}$.
Note that $\AA \subseteq \II \subseteq \FF$.
$\FF$ is a filter on $S$:
Because $\AA$ has the finite intersection property:
:$\O \notin \II$
Because each element of $\FF$ is... | Let $S$ be a non-empty [[Definition:Set|set]].
Let $\AA$ be a set of [[Definition:Subset|subsets]] of $S$.
Suppose that $\AA$ has the [[Definition:Finite Intersection Property|finite intersection property]].
Then there is an [[Definition:Ultrafilter|ultrafilter]] $\UU$ on $S$ such that $\AA \subseteq \UU$. | Let $\II$ be the set of [[Definition:Set Intersection|intersections]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $\AA$.
Let $\FF = \set {T \in \powerset S: \exists B \in \II: B \subseteq T}$.
Note that $\AA \subseteq \II \subseteq \FF$.
$\FF$ is a [[D... | Ultrafilter Lemma/Corollary | https://proofwiki.org/wiki/Ultrafilter_Lemma/Corollary | https://proofwiki.org/wiki/Ultrafilter_Lemma/Corollary | [
"Ultrafilter Lemma"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Finite Intersection Property",
"Definition:Ultrafilter on Set"
] | [
"Definition:Set Intersection",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Filter on Set",
"Definition:Finite Intersection Property",
"Definition:Element",
"Definition:Subset/Superset",
"Definition:Element",
"Definition:Filter on Set",
"Ultrafilter Lemma... |
proofwiki-7736 | Order-Extension Principle/Strict | Let $S$ be a set.
Let $\prec$ be a strict ordering on $S$.
Then there exists a strict total ordering $<$ on $S$ such that:
:$\forall a, b \in S: a \prec b \implies a < b$ | Let $\AA$ be the set of relations $A$ on $S$ with the property that the transitive closure $A^+$ of $A$ is a strict ordering of $S$.
For each $\tuple {x, y} \in S \times S$, let $\tuple {x, y}' = \tuple {y, x}$.
Let $A \in \AA$.
Let $\tuple {x, y} \in S \times S$.
Let $\tuple {x, y} \in A^+$.
Then:
:$\paren {A \cup \se... | Let $S$ be a [[Definition:Set|set]].
Let $\prec$ be a [[Definition:Strict Ordering|strict ordering]] on $S$.
Then there exists a [[Definition:Strict Total Ordering|strict total ordering]] $<$ on $S$ such that:
:$\forall a, b \in S: a \prec b \implies a < b$ | Let $\AA$ be the [[Definition:Set|set]] of [[Definition:Endorelation|relations]] $A$ on $S$ with the property that the [[Definition:Transitive Closure of Relation|transitive closure]] $A^+$ of $A$ is a [[Definition:Strict Ordering|strict ordering]] of $S$.
For each $\tuple {x, y} \in S \times S$, let $\tuple {x, y}' =... | Order-Extension Principle/Strict/Proof 1 | https://proofwiki.org/wiki/Order-Extension_Principle/Strict | https://proofwiki.org/wiki/Order-Extension_Principle/Strict/Proof_1 | [
"Set Theory",
"Order Theory"
] | [
"Definition:Set",
"Definition:Strict Ordering",
"Definition:Strict Total Ordering"
] | [
"Definition:Set",
"Definition:Endorelation",
"Definition:Transitive Closure of Relation",
"Definition:Strict Ordering",
"Definition:Non-Comparable Elements",
"Strict Ordering can be Expanded to Compare Additional Pair",
"Definition:Finite Subset",
"Definition:Strict Ordering",
"Definition:Asymmetric... |
proofwiki-7737 | Order-Extension Principle/Strict | Let $S$ be a set.
Let $\prec$ be a strict ordering on $S$.
Then there exists a strict total ordering $<$ on $S$ such that:
:$\forall a, b \in S: a \prec b \implies a < b$ | For the purposes of this proof, a relation $<_U$ on a subset $U$ of $S$ will be considered '''compatible''' with $\prec$ {{iff}}:
:$\forall a, b \in U: a \prec b \implies a < b$
Let $M$ be the set of partial mappings $f$ from $S \times S$ to $\left\{ {0, 1}\right\}$ such that for all $x, y, z \in S$:
:$(a): \quad \left... | Let $S$ be a [[Definition:Set|set]].
Let $\prec$ be a [[Definition:Strict Ordering|strict ordering]] on $S$.
Then there exists a [[Definition:Strict Total Ordering|strict total ordering]] $<$ on $S$ such that:
:$\forall a, b \in S: a \prec b \implies a < b$ | For the purposes of this proof, a [[Definition:Endorelation|relation]] $<_U$ on a [[Definition:Subset|subset]] $U$ of $S$ will be considered '''compatible''' with $\prec$ {{iff}}:
:$\forall a, b \in U: a \prec b \implies a < b$
Let $M$ be the [[Definition:Set|set]] of [[Definition:Partial Mapping|partial mappings]] $... | Order-Extension Principle/Strict/Proof 2 | https://proofwiki.org/wiki/Order-Extension_Principle/Strict | https://proofwiki.org/wiki/Order-Extension_Principle/Strict/Proof_2 | [
"Set Theory",
"Order Theory"
] | [
"Definition:Set",
"Definition:Strict Ordering",
"Definition:Strict Total Ordering"
] | [
"Definition:Endorelation",
"Definition:Subset",
"Definition:Set",
"Definition:Many-to-One Relation",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Characteristic Function (Set Theory)/Relation",
"Definition:Strict Total Ordering",
"Cowen-Engeler Lemma",
"Definition:Finite Set",
"Definition... |
proofwiki-7738 | Order-Extension Principle/Strict/Finite Set | Let $T$ be a finite set.
Let $\prec$ be a strict ordering on $T$.
Then there exists a strict total ordering $<$ on $T$ such that:
:$\forall a, b \in T: \paren {a \prec b \implies a < b}$ | {{ProofWanted|Use Strict Ordering can be Expanded to Compare Additional Pair.}}
Category:Set Theory
Category:Order Theory
3rvr5p803rau0xia8t260ztfwhhah99 | Let $T$ be a [[Definition:Finite Set|finite set]].
Let $\prec$ be a [[Definition:Strict Ordering|strict ordering]] on $T$.
Then there exists a [[Definition:Strict Total Ordering|strict total ordering]] $<$ on $T$ such that:
:$\forall a, b \in T: \paren {a \prec b \implies a < b}$ | {{ProofWanted|Use [[Strict Ordering can be Expanded to Compare Additional Pair]].}}
[[Category:Set Theory]]
[[Category:Order Theory]]
3rvr5p803rau0xia8t260ztfwhhah99 | Order-Extension Principle/Strict/Finite Set | https://proofwiki.org/wiki/Order-Extension_Principle/Strict/Finite_Set | https://proofwiki.org/wiki/Order-Extension_Principle/Strict/Finite_Set | [
"Set Theory",
"Order Theory"
] | [
"Definition:Finite Set",
"Definition:Strict Ordering",
"Definition:Strict Total Ordering"
] | [
"Strict Ordering can be Expanded to Compare Additional Pair",
"Category:Set Theory",
"Category:Order Theory"
] |
proofwiki-7739 | Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma | Let $\LL$ be a language of predicate logic.
Let $T$ be a finitely satisfiable $\LL$-theory.
Let $\phi$ be an $\LL$-sentence.
Then either:
:$T \cup \set \phi$
or:
:$T \cup \set {\neg \phi}$
is finitely satisfiable. | {{WLOG}}, suppose that $T \not\models \phi$ and $T \not\models \neg \phi$.
{{AimForCont}} that $T \cup \set \phi$ is not finitely satisfiable.
Then by definition there must be a finite subset $K$ of $T \cup \set \phi$ which is not satisfiable.
Since $T$ is finitely satisfiable, $\phi \in K$.
Therefore $\Delta = K \setm... | Let $\LL$ be a [[Definition:Language of Predicate Logic|language of predicate logic]].
Let $T$ be a [[Definition:Finitely Satisfiable|finitely satisfiable]] $\LL$-[[Definition:Theory|theory]].
Let $\phi$ be an $\LL$-[[Definition:Sentence|sentence]].
Then either:
:$T \cup \set \phi$
or:
:$T \cup \set {\neg \phi}$
is... | {{WLOG}}, suppose that $T \not\models \phi$ and $T \not\models \neg \phi$.
{{AimForCont}} that $T \cup \set \phi$ is not [[Definition:Finitely Satisfiable|finitely satisfiable]].
Then by definition there must be a [[Definition:Finite Subset|finite subset]] $K$ of $T \cup \set \phi$ which is not [[Definition:Satisfiab... | Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma | https://proofwiki.org/wiki/Finitely_Satisfiable_Theory_has_Maximal_Finitely_Satisfiable_Extension/Lemma | https://proofwiki.org/wiki/Finitely_Satisfiable_Theory_has_Maximal_Finitely_Satisfiable_Extension/Lemma | [
"Model Theory for Predicate Logic"
] | [
"Definition:Language of Predicate Logic",
"Definition:Finitely Satisfiable",
"Definition:Theory",
"Definition:Classes of WFFs/Sentence",
"Definition:Finitely Satisfiable"
] | [
"Definition:Finitely Satisfiable",
"Definition:Finite Subset",
"Definition:Satisfiable",
"Definition:Finitely Satisfiable",
"Definition:Finite Subset",
"Definition:Structure for Predicate Logic",
"Definition:Model (Logic)/Set of Logical Formulas",
"Definition:Structure for Predicate Logic/Formal Seman... |
proofwiki-7740 | Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Proof 2 | Let $T$ be a finitely satisfiable $\LL$-theory.
There is a finitely satisfiable $\LL$-theory $T'$ which contains $T$ as a subset such that for all $\LL$-sentences $\phi$, either $\phi \in T'$ or $\neg\phi \in T'$.
{{explain|Does this actually mean the same as the statement of the theorem in Finitely Satisfiable Theory ... | Let $\AA$ be the set of finitely satisfiable extensions of $T$.
By the lemma, for each element $S$ of $\AA$ and each $\LL$-sentence $\phi$, either $S \cup \set \phi \in \AA$ or $S \cup \set {\neg \phi} \in \AA$.
$\AA$ has finite character, by the following argument:
Let $S \in \AA$.
Let $F$ be a finite subset of $S$.
T... | Let $T$ be a finitely satisfiable $\LL$-theory.
There is a finitely satisfiable $\LL$-theory $T'$ which contains $T$ as a subset such that for all $\LL$-sentences $\phi$, either $\phi \in T'$ or $\neg\phi \in T'$.
{{explain|Does this actually mean the same as the statement of the theorem in [[Finitely Satisfiable The... | Let $\AA$ be the [[Definition:Set|set]] of finitely satisfiable extensions of $T$.
By the [[Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma|lemma]], for each element $S$ of $\AA$ and each $\LL$-sentence $\phi$, either $S \cup \set \phi \in \AA$ or $S \cup \set {\neg \phi} \in \AA$.
$\AA$ ... | Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Proof 2 | https://proofwiki.org/wiki/Finitely_Satisfiable_Theory_has_Maximal_Finitely_Satisfiable_Extension/Proof_2 | https://proofwiki.org/wiki/Finitely_Satisfiable_Theory_has_Maximal_Finitely_Satisfiable_Extension/Proof_2 | [
"Model Theory for Predicate Logic"
] | [
"Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension"
] | [
"Definition:Set",
"Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma",
"Definition:Finite Character",
"Definition:Finite Subset",
"Definition:Finite Subset",
"Definition:Finite Subset",
"Restricted Tukey's Theorem"
] |
proofwiki-7741 | Reflexive Closure of Strict Total Ordering is Total Ordering | Let $S$ be a set.
Let $\prec$ be a strict total ordering of $S$.
Let $\preceq$ be the reflexive closure of $\prec$.
Then $\preceq$ is an total ordering of $S$. | By the definition of strict total ordering, $\prec$ is a strict ordering which connects $S$.
By Reflexive Closure of Strict Ordering is Ordering, $\preceq$ is a ordering.
Since $\prec$ connects $S$, for each $a, b \in S$, either $a = b$, $a \prec b$ or $b \prec a$.
If $a = b$, then $a \preceq b$.
If $a \prec b$, then $... | Let $S$ be a [[Definition:set|set]].
Let $\prec$ be a [[Definition:Strict Total Ordering|strict total ordering]] of $S$.
Let $\preceq$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\prec$.
Then $\preceq$ is an [[Definition:Total Ordering|total ordering]] of $S$. | By the definition of [[Definition:Strict Total Ordering|strict total ordering]], $\prec$ is a [[Definition:Strict Ordering|strict ordering]] which [[Definition:Connected Relation|connects]] $S$.
By [[Reflexive Closure of Strict Ordering is Ordering]], $\preceq$ is a [[Definition:ordering|ordering]].
Since $\prec$ [[D... | Reflexive Closure of Strict Total Ordering is Total Ordering | https://proofwiki.org/wiki/Reflexive_Closure_of_Strict_Total_Ordering_is_Total_Ordering | https://proofwiki.org/wiki/Reflexive_Closure_of_Strict_Total_Ordering_is_Total_Ordering | [
"Strict Orderings",
"Total Orderings",
"Reflexive Closures"
] | [
"Definition:set",
"Definition:Strict Total Ordering",
"Definition:Reflexive Closure",
"Definition:Total Ordering"
] | [
"Definition:Strict Total Ordering",
"Definition:Strict Ordering",
"Definition:Connected Relation",
"Reflexive Closure of Strict Ordering is Ordering",
"Definition:ordering",
"Definition:Connected Relation",
"Definition:Total Ordering",
"Category:Strict Orderings",
"Category:Total Orderings",
"Cate... |
proofwiki-7742 | Galois Field of Order q Exists iff q is Prime Power | Let $q \ge 0$ be a positive integer.
Then there exists a Galois field of order $q$ {{iff}} $q$ is a prime power. | {{tidy}}
{{MissingLinks}} | Let $q \ge 0$ be a [[Definition:Positive Integer|positive integer]].
Then there exists a [[Definition:Galois Field|Galois field]] of [[Definition:Order of Structure|order]] $q$ {{iff}} $q$ is a [[Definition:Prime Power|prime power]]. | {{tidy}}
{{MissingLinks}} | Galois Field of Order q Exists iff q is Prime Power | https://proofwiki.org/wiki/Galois_Field_of_Order_q_Exists_iff_q_is_Prime_Power | https://proofwiki.org/wiki/Galois_Field_of_Order_q_Exists_iff_q_is_Prime_Power | [
"Galois Fields"
] | [
"Definition:Positive/Integer",
"Definition:Galois Field",
"Definition:Order of Structure",
"Definition:Prime Power"
] | [] |
proofwiki-7743 | Leaf of Rooted Tree is on One Branch | Let $T$ be a rooted tree with root node $r_T$.
Let $t$ be a leaf node of $T$.
Then there exists a unique branch $\Gamma$ of $T$ such that $t \in \Gamma$. | Let $t$ be on the branch $\Gamma$.
By definition of branch, $\Gamma$ is a path from the root node $r_T$ to $t$.
By Path in Tree is Unique, such $\Gamma$ is unique.
{{qed}} | Let $T$ be a [[Definition:Rooted Tree|rooted tree]] with [[Definition:Root Node|root node]] $r_T$.
Let $t$ be a [[Definition:Leaf Node|leaf node]] of $T$.
Then there exists a [[Definition:Unique|unique]] [[Definition:Branch of Tree|branch]] $\Gamma$ of $T$ such that $t \in \Gamma$. | Let $t$ be on the branch $\Gamma$.
By definition of [[Definition:Branch of Tree|branch]], $\Gamma$ is a [[Definition:Path (Graph Theory)|path]] from the [[Definition:Root Node|root node]] $r_T$ to $t$.
By [[Path in Tree is Unique]], such $\Gamma$ is unique.
{{qed}} | Leaf of Rooted Tree is on One Branch | https://proofwiki.org/wiki/Leaf_of_Rooted_Tree_is_on_One_Branch | https://proofwiki.org/wiki/Leaf_of_Rooted_Tree_is_on_One_Branch | [
"Rooted Trees"
] | [
"Definition:Rooted Tree",
"Definition:Rooted Tree/Root Node",
"Definition:Tree (Graph Theory)/Leaf Node",
"Definition:Unique",
"Definition:Rooted Tree/Branch"
] | [
"Definition:Rooted Tree/Branch",
"Definition:Path (Graph Theory)",
"Definition:Rooted Tree/Root Node",
"Path in Tree is Unique"
] |
proofwiki-7744 | Branch of Finite Tree is Finite | Let $T$ be a finite rooted tree with root node $r_T$.
Let $\Gamma$ be a branch of $T$.
Then $\Gamma$ is a finite branch. | Let $\Gamma$ be a branch of a finite rooted tree $T$.
{{AimForCont}} $\Gamma$ were an infinite branch of $T$.
By definition $\Gamma$ contains an infinite number of nodes.
{{explain|Technically, from the definition currently posted, it doesn't - that still needs to be demonstrated.}}
From Subset of Finite Set is Finite,... | Let $T$ be a [[Definition:Finite Tree|finite]] [[Definition:Rooted Tree|rooted tree]] with [[Definition:Root Node|root node]] $r_T$.
Let $\Gamma$ be a [[Definition:Branch (Graph Theory)|branch]] of $T$.
Then $\Gamma$ is a [[Definition:Finite Branch|finite branch]]. | Let $\Gamma$ be a [[Definition:Branch (Graph Theory)|branch]] of a [[Definition:Finite Tree|finite]] [[Definition:Rooted Tree|rooted tree]] $T$.
{{AimForCont}} $\Gamma$ were an [[Definition:Infinite Branch|infinite branch]] of $T$.
By definition $\Gamma$ contains an [[Definition:Infinite Set|infinite number]] of [[De... | Branch of Finite Tree is Finite | https://proofwiki.org/wiki/Branch_of_Finite_Tree_is_Finite | https://proofwiki.org/wiki/Branch_of_Finite_Tree_is_Finite | [
"Rooted Trees"
] | [
"Definition:Tree (Graph Theory)/Finite",
"Definition:Rooted Tree",
"Definition:Rooted Tree/Root Node",
"Definition:Rooted Tree/Branch",
"Definition:Rooted Tree/Branch/Finite"
] | [
"Definition:Rooted Tree/Branch",
"Definition:Tree (Graph Theory)/Finite",
"Definition:Rooted Tree",
"Definition:Rooted Tree/Branch/Infinite",
"Definition:Infinite Set",
"Definition:Tree (Graph Theory)/Node",
"Subset of Finite Set is Finite",
"Definition:Finite Set",
"Definition:Tree (Graph Theory)/N... |
proofwiki-7745 | Same Dimensional Vector Spaces are Isomorphic | Let $K$ be a division ring.
Let $V$, $W$ be finite dimensional $K$-vector spaces.
Suppose that $\dim_K V = \dim_K W$.
Then:
:$V \cong W$
That is, $V$ and $W$ are isomorphic. | Let $\mathbb V$, $\mathbb W$ be bases for $V$, $W$ respectively.
We have {{hypothesis}} that:
:$\dim_K V = \dim_K W$
Thus by the definition of dimension:
:$\mathbb V \sim \mathbb W$
Therefore we can choose a bijection $\phi: \mathbb V \leftrightarrow \mathbb W$.
Define the mapping $\lambda: V \to W$ by:
:$\ds \map \lam... | Let $K$ be a [[Definition:Division Ring|division ring]].
Let $V$, $W$ be [[Definition:Finite Dimensional Vector Space|finite dimensional]] $K$-[[Definition:Vector Space|vector spaces]].
Suppose that $\dim_K V = \dim_K W$.
Then:
:$V \cong W$
That is, $V$ and $W$ are [[Definition:Vector Space Isomorphism|isomorphic]... | Let $\mathbb V$, $\mathbb W$ be [[Definition:Basis (Linear Algebra)|bases]] for $V$, $W$ respectively.
We have {{hypothesis}} that:
:$\dim_K V = \dim_K W$
Thus by the definition of [[Definition:Dimension of Vector Space|dimension]]:
:$\mathbb V \sim \mathbb W$
Therefore we can choose a [[Definition:Bijection|biject... | Same Dimensional Vector Spaces are Isomorphic | https://proofwiki.org/wiki/Same_Dimensional_Vector_Spaces_are_Isomorphic | https://proofwiki.org/wiki/Same_Dimensional_Vector_Spaces_are_Isomorphic | [
"Linear Algebra",
"Dimension of Vector Space"
] | [
"Definition:Division Ring",
"Definition:Dimension of Vector Space/Finite",
"Definition:Vector Space",
"Definition:Isomorphism (Abstract Algebra)/R-Algebraic Structure Isomorphism/Vector Space Isomorphism"
] | [
"Definition:Basis (Linear Algebra)",
"Definition:Dimension of Vector Space",
"Definition:Bijection",
"Definition:Mapping",
"Unique Linear Transformation Between Vector Spaces",
"Definition:Kronecker Delta",
"Definition:Algebraic Dual",
"Vector Scaled by Zero is Zero Vector",
"Definition:Term of Expr... |
proofwiki-7746 | Five Lemma | Let $A$ be a commutative ring with unity.
Let:
::<nowiki>$\begin{xy}\xymatrix@L+2mu@+1em{
M_1 \ar[r]^*{\alpha_1}
\ar[d]^*{\phi_1}
&
M_2 \ar[r]^*{\alpha_2}
\ar[d]^*{\phi_2}
&
M_3 \ar[r]^*{\alpha_3}
\ar[d]^*{\phi_3}
&
M_4 \ar[r]^*{\alpha_4}
\ar[d]^*{\phi_4}
&
M_5 \ar[d]^*{\phi_5}
\\
N_1 \ar[r]_*... | First suppose that $\phi_2$ and $\phi_4$ are surjective and $\phi_5$ is injective.
Let $n_3 \in N_3$ be any element.
We want to find $x \in M_3$ such that $\map {\phi_3} x = n_3$.
Let $n_4 = \map {\beta_3} {n_3} \in N_4$.
Since $\phi_4$ is surjective, there exists $m_4 \in M_4$ such that $\map {\phi_4} {m_4} = n_4$.
Si... | Let $A$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let:
::<nowiki>$\begin{xy}\xymatrix@L+2mu@+1em{
M_1 \ar[r]^*{\alpha_1}
\ar[d]^*{\phi_1}
&
M_2 \ar[r]^*{\alpha_2}
\ar[d]^*{\phi_2}
&
M_3 \ar[r]^*{\alpha_3}
\ar[d]^*{\phi_3}
&
M_4 \ar[r]^*{\alpha_4}
\ar[d]^*{\ph... | First suppose that $\phi_2$ and $\phi_4$ are [[Definition:Surjection|surjective]] and $\phi_5$ is [[Definition:Injection|injective]].
Let $n_3 \in N_3$ be any [[Definition:Element|element]].
We want to find $x \in M_3$ such that $\map {\phi_3} x = n_3$.
Let $n_4 = \map {\beta_3} {n_3} \in N_4$.
Since $\phi_4$ is [... | Five Lemma | https://proofwiki.org/wiki/Five_Lemma | https://proofwiki.org/wiki/Five_Lemma | [
"Homological Algebra",
"Named Theorems"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Commutative Diagram",
"Definition:Module over Ring",
"Definition:Exact Sequence of Modules",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Injection"
] | [
"Definition:Surjection",
"Definition:Injection",
"Definition:Element",
"Definition:Surjection",
"Definition:Injection",
"Definition:Ring Homomorphism",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Injecti... |
proofwiki-7747 | Snake Lemma | Let $A$ be a commutative ring with unity.
Let:
::<nowiki>$\begin{xy}\xymatrix@L+2mu@+1em{
&
M_1 \ar[r]_*{\alpha_1}
\ar[d]^*{\phi_1}
&
M_2 \ar[r]_*{\alpha_2}
\ar[d]^*{\phi_2}
&
M_3 \ar[d]^*{\phi_3}
\ar[r]
&
0
\\
0 \ar[r]
&
N_1 \ar[r]_*{\beta_1}
&
N_2 \ar[r]_*{\beta_2}
&
N_3
&
}\end{xy}$</nowiki>
... | {{ProofWanted}}
Category:Homological Algebra
Category:Named Theorems
gv5hyctbr94a8kkgou9pchdutitewfy | Let $A$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let:
::<nowiki>$\begin{xy}\xymatrix@L+2mu@+1em{
&
M_1 \ar[r]_*{\alpha_1}
\ar[d]^*{\phi_1}
&
M_2 \ar[r]_*{\alpha_2}
\ar[d]^*{\phi_2}
&
M_3 \ar[d]^*{\phi_3}
\ar[r]
&
0
\\
0 \ar[r]
&
N_1 \ar[r]_*{\beta_1}
&
N_2 \a... | {{ProofWanted}}
[[Category:Homological Algebra]]
[[Category:Named Theorems]]
gv5hyctbr94a8kkgou9pchdutitewfy | Snake Lemma | https://proofwiki.org/wiki/Snake_Lemma | https://proofwiki.org/wiki/Snake_Lemma | [
"Homological Algebra",
"Named Theorems"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Commutative Diagram",
"Definition:Module over Ring",
"Definition:Exact Sequence of Modules",
"Definition:Kernel",
"Definition:Cokernel",
"Definition:Inclusion Mapping",
"Definition:Quotient Epimorphism",
"Definition:Restriction/Mapping",
"Defi... | [
"Category:Homological Algebra",
"Category:Named Theorems"
] |
proofwiki-7748 | Hall's Marriage Theorem/Finite Set | Let $\SS = \family {S_k}_{k \mathop \in I}$ be a finite indexed family of finite sets.
For each $F \subseteq I$, let $\ds Y_F = \bigcup_{k \mathop \in F} S_k$.
Let $Y = Y_I$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\SS$ satisfies the marriage condition: {{:Axiom:Marriage Condition}} }}
{{item|(2):|There exists an injec... | === $(2)$ implies $(1)$ ===
Let:
:$\exists P \subseteq I: \card P > \card {Y_P}$
Then:
:$\card P \not \le \card {Y_P}$
By contrapositive of Injection implies Cardinal Inequality, there can be no injection from $P$ to $Y_P$.
Thus there can be no injection from $I$ to $Y$ satisfying the requirements.
{{qed|lemma}} | Let $\SS = \family {S_k}_{k \mathop \in I}$ be a finite [[Definition:Indexed Family|indexed family]] of finite sets.
For each $F \subseteq I$, let $\ds Y_F = \bigcup_{k \mathop \in F} S_k$.
Let $Y = Y_I$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\SS$ satisfies the [[Axiom:Marriage Condition|marriage condition]]: {{:... | === $(2)$ implies $(1)$ ===
Let:
:$\exists P \subseteq I: \card P > \card {Y_P}$
Then:
:$\card P \not \le \card {Y_P}$
By [[Definition:Contrapositive Statement|contrapositive]] of [[Injection implies Cardinal Inequality]], there can be no [[Definition:Injection|injection]] from $P$ to $Y_P$.
Thus there can be no [[... | Hall's Marriage Theorem/Finite Set | https://proofwiki.org/wiki/Hall's_Marriage_Theorem/Finite_Set | https://proofwiki.org/wiki/Hall's_Marriage_Theorem/Finite_Set | [
"Hall's Marriage Theorem"
] | [
"Definition:Indexing Set/Family",
"Axiom:Marriage Condition",
"Definition:Injection"
] | [
"Definition:Contrapositive Statement",
"Injection implies Cardinal Inequality",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection"
] |
proofwiki-7749 | Connecting Homomorphism is Functorial | Let $A$ be a ring with unity.
Let:
:<nowiki>$\begin{xy}\xymatrix{
&&& M_1 \ar@{->}[rr]
\ar@{->}[dl]^{f_1}
\ar@{->}[dd]^{\phi_1}|!{[d];[d]}\hole
&& M_2 %
\ar@{->}[rr]
\ar@{->}[dl]^{f_2}
\ar@{->}[dd]^{\phi_2}|!{[d];[d]}\hole
&& M_3 \ar@{->}[dl]^{f_3}
\ar@{->}[dd]^{\phi_3}|!{[d];[d]}\hole
... | {{proof wanted}}
Category:Homological Algebra
88br8tds8012egp9su74479ll2ycs7b | Let $A$ be a [[Definition:Ring with Unity|ring with unity]].
Let:
:<nowiki>$\begin{xy}\xymatrix{
&&& M_1 \ar@{->}[rr]
\ar@{->}[dl]^{f_1}
\ar@{->}[dd]^{\phi_1}|!{[d];[d]}\hole
&& M_2 %
\ar@{->}[rr]
\ar@{->}[dl]^{f_2}
\ar@{->}[dd]^{\phi_2}|!{[d];[d]}\hole
&& M_3 \ar@{->}[dl]^{f_3}
\ar@{->... | {{proof wanted}}
[[Category:Homological Algebra]]
88br8tds8012egp9su74479ll2ycs7b | Connecting Homomorphism is Functorial | https://proofwiki.org/wiki/Connecting_Homomorphism_is_Functorial | https://proofwiki.org/wiki/Connecting_Homomorphism_is_Functorial | [
"Homological Algebra"
] | [
"Definition:Ring with Unity",
"Definition:Commutative Diagram",
"Definition:Module over Ring",
"Definition:Exact Sequence of Modules",
"Snake Lemma",
"Definition:Commutative Diagram",
"Definition:Commutative Square"
] | [
"Category:Homological Algebra"
] |
proofwiki-7750 | Hall's Marriage Theorem/General Set | Let $\SS = \family {S_k}_{k \mathop \in I}$ be an indexed family of finite sets.
For each $F \subseteq I$, let $\ds Y_F = \bigcup_{k \mathop \in F} S_k$.
Let $Y = Y_I$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\SS$ satisfies the marriage condition: {{:Axiom:Marriage Condition}} }}
{{item|(2):|There exists an injection $... | === $(2)$ implies $(1)$ ===
Suppose that for some finite $F \subseteq I$, $\card F > \card {Y_F}$.
Then $\card F \not \le \card {Y_F}$.
By contrapositive of Injection implies Cardinal Inequality, there can be no injection from $F$ to $Y_F$.
Thus there can be no injection from $I$ to $Y$ satisfying the requirements.
{{q... | Let $\SS = \family {S_k}_{k \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of finite sets.
For each $F \subseteq I$, let $\ds Y_F = \bigcup_{k \mathop \in F} S_k$.
Let $Y = Y_I$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\SS$ satisfies the [[Axiom:Marriage Condition|marriage condition]]: {{:Axiom:... | === $(2)$ implies $(1)$ ===
Suppose that for some [[Definition:Finite Subset|finite]] $F \subseteq I$, $\card F > \card {Y_F}$.
Then $\card F \not \le \card {Y_F}$.
By [[Definition:Contrapositive Statement|contrapositive]] of [[Injection implies Cardinal Inequality]], there can be no injection from $F$ to $Y_F$.
Th... | Hall's Marriage Theorem/General Set | https://proofwiki.org/wiki/Hall's_Marriage_Theorem/General_Set | https://proofwiki.org/wiki/Hall's_Marriage_Theorem/General_Set | [
"Hall's Marriage Theorem"
] | [
"Definition:Indexing Set/Family",
"Axiom:Marriage Condition",
"Definition:Injection"
] | [
"Definition:Finite Subset",
"Definition:Contrapositive Statement",
"Injection implies Cardinal Inequality",
"Definition:Injection",
"Definition:Finite Subset"
] |
proofwiki-7751 | Branch of Finite Propositional Tableau is Finite | Let $T$ be a finite propositional tableau.
Let $\Gamma$ be a branch of $T$.
Then $\Gamma$ is a finite branch. | By definition, a finite propositional tableau $T$ is formed by applying the tableau extension rules a finite number of times.
Each tableau extension rule extends $T$ finitely.
Therefore $T$ is a finite tree.
The result follows from Branch of Finite Tree is Finite.
{{qed}}
Category:Propositional Tableaux
kog39iuejrvkl50... | Let $T$ be a [[Definition:Finite Propositional Tableau|finite propositional tableau]].
Let $\Gamma$ be a [[Definition:Branch (Graph Theory)|branch]] of $T$.
Then $\Gamma$ is a [[Definition:Finite Branch|finite branch]]. | By definition, a [[Definition:Finite Propositional Tableau|finite propositional tableau]] $T$ is formed by applying the [[Definition:Tableau Extension Rules|tableau extension rules]] a finite number of times.
Each [[Definition:Tableau Extension Rules|tableau extension rule]] extends $T$ finitely.
Therefore $T$ is a [... | Branch of Finite Propositional Tableau is Finite | https://proofwiki.org/wiki/Branch_of_Finite_Propositional_Tableau_is_Finite | https://proofwiki.org/wiki/Branch_of_Finite_Propositional_Tableau_is_Finite | [
"Propositional Tableaux"
] | [
"Definition:Propositional Tableau/Construction/Finite",
"Definition:Rooted Tree/Branch",
"Definition:Rooted Tree/Branch/Finite"
] | [
"Definition:Propositional Tableau/Construction/Finite",
"Definition:Propositional Tableau/Construction/Finite",
"Definition:Propositional Tableau/Construction/Finite",
"Definition:Tree (Graph Theory)/Finite",
"Branch of Finite Tree is Finite",
"Category:Propositional Tableaux"
] |
proofwiki-7752 | Tableau Confutation is Finished | Let $T$ be a tableau confutation.
Then $T$ is a finished tableau. | By definition of tableau confutation, every branch of $T$ is contradictory.
The result follows by definition of finished propositional tableau.
{{qed}} | Let $T$ be a [[Definition:Tableau Confutation|tableau confutation]].
Then $T$ is a [[Definition:Finished Propositional Tableau|finished tableau]]. | By definition of [[Definition:Tableau Confutation|tableau confutation]], every [[Definition:Branch (Graph Theory)|branch]] of $T$ is [[Definition:Contradictory Branch|contradictory]].
The result follows by definition of [[Definition:Finished Propositional Tableau|finished propositional tableau]].
{{qed}} | Tableau Confutation is Finished | https://proofwiki.org/wiki/Tableau_Confutation_is_Finished | https://proofwiki.org/wiki/Tableau_Confutation_is_Finished | [
"Propositional Tableaux"
] | [
"Definition:Tableau Confutation",
"Definition:Finished Propositional Tableau"
] | [
"Definition:Tableau Confutation",
"Definition:Rooted Tree/Branch",
"Definition:Contradictory/Branch",
"Definition:Finished Propositional Tableau"
] |
proofwiki-7753 | Expression of Vector as Linear Combination from Basis is Unique/General Result | Let $V$ be a vector space over a division ring $R$.
Let $B$ be a basis for $V$.
Let $x \in V$.
Then there is a unique finite subset $C$ of $R \times B$ such that:
:$\ds x = \sum_{\tuple {r, v} \mathop \in C} r \cdot v$
:$\forall \tuple {r, v} \in C: r \ne 0_R$ | === Existence ===
The existence of $C$ follows immediately from the definition of a basis.
{{qed|lemma}} | Let $V$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Division Ring|division ring]] $R$.
Let $B$ be a [[Definition:Basis (Linear Algebra)|basis]] for $V$.
Let $x \in V$.
Then there is a unique [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $C$ of $R \times B$ such that:
:$\ds x =... | === Existence ===
The existence of $C$ follows immediately from the definition of a [[Definition:Basis (Linear Algebra)|basis]].
{{qed|lemma}} | Expression of Vector as Linear Combination from Basis is Unique/General Result | https://proofwiki.org/wiki/Expression_of_Vector_as_Linear_Combination_from_Basis_is_Unique/General_Result | https://proofwiki.org/wiki/Expression_of_Vector_as_Linear_Combination_from_Basis_is_Unique/General_Result | [
"Vector Spaces",
"Linear Algebra"
] | [
"Definition:Vector Space",
"Definition:Division Ring",
"Definition:Basis (Linear Algebra)",
"Definition:Finite Set",
"Definition:Subset"
] | [
"Definition:Basis (Linear Algebra)",
"Definition:Basis (Linear Algebra)"
] |
proofwiki-7754 | Dependent Choice for Finite Sets | Let $\RR$ be a binary relation on a non-empty set $S$.
For each $a \in S$, let $C_a = \set {b \in S: a \mathrel \RR b }$
Suppose that:
:For all $a \in S$, $C_a$ is a non-empty finite set.
Let $s \in S$.
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that:
:$x_0 = s$
:$\forall n \in \N: x_... | Define $\sequence {D_n}$ recursively:
Let $D_0 = \set s$.
For each $n \in \N$ let $D_{n + 1} = \map \RR {D_n}$.
Now, for each $n \in \N$ let $E_n$ be the set of all enumerations of $D_n$.
Then $E_n$ is non-empty and finite for each $n$.
By the Axiom of Countable Choice for Finite Sets, there is a sequence $\sequence {e... | Let $\RR$ be a [[Definition:Binary Relation|binary relation]] on a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] $S$.
For each $a \in S$, let $C_a = \set {b \in S: a \mathrel \RR b }$
Suppose that:
:For all $a \in S$, $C_a$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite se... | Define $\sequence {D_n}$ [[Principle of Recursive Definition|recursively]]:
Let $D_0 = \set s$.
For each $n \in \N$ let $D_{n + 1} = \map \RR {D_n}$.
Now, for each $n \in \N$ let $E_n$ be the [[Definition:Set|set]] of all [[Definition:Enumeration|enumerations]] of $D_n$.
Then $E_n$ is [[Definition:Non-Empty Set|non... | Dependent Choice for Finite Sets | https://proofwiki.org/wiki/Dependent_Choice_for_Finite_Sets | https://proofwiki.org/wiki/Dependent_Choice_for_Finite_Sets | [
"Set Theory"
] | [
"Definition:Relation",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Sequence"
] | [
"Principle of Recursive Definition",
"Definition:Set",
"Definition:Enumeration",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Axiom:Axiom of Countable Choice for Finite Sets",
"Definition:Sequence",
"Principle of Recursive Definition",
"Definition:Element",
"Definition:Enumeration",
"De... |
proofwiki-7755 | Dependent Choice (Fixed First Element) | Let $\RR$ be a binary relation on a non-empty set $S$.
Suppose:
:$\forall a \in S: \exists b \in S: a \mathrel \RR b$
that is, that $\RR$ is a left-total relation (that is, a serial relation).
Let $s \in S$.
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that:
:$x_0 = s$
:$\forall n \in \... | Let $S' = \set {y \in S: s \mathrel {\RR^+} y}$, where $\RR^+$ is the transitive closure of $\RR$.
Let $\RR'$ be the restriction of $\RR$ to $S'$.
For each $x \in S'$, there is a $y \in S$ such that $x \mathrel \RR y$. But then $s \mathrel {\RR^+} y$, so $y \in S'$, so $x \mathrel {\RR'} y$.
Thus $\RR'$ is a left-tota... | Let $\RR$ be a [[Definition:Binary Relation|binary relation]] on a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] $S$.
Suppose:
:$\forall a \in S: \exists b \in S: a \mathrel \RR b$
that is, that $\RR$ is a [[Definition:Left-Total Relation|left-total relation]] (that is, a [[Definition:Serial Relation|s... | Let $S' = \set {y \in S: s \mathrel {\RR^+} y}$, where $\RR^+$ is the [[Definition:Transitive Closure of Relation|transitive closure]] of $\RR$.
Let $\RR'$ be the [[Definition:Restriction|restriction]] of $\RR$ to $S'$.
For each $x \in S'$, there is a $y \in S$ such that $x \mathrel \RR y$. But then $s \mathrel {\RR... | Dependent Choice (Fixed First Element) | https://proofwiki.org/wiki/Dependent_Choice_(Fixed_First_Element) | https://proofwiki.org/wiki/Dependent_Choice_(Fixed_First_Element) | [
"Set Theory"
] | [
"Definition:Relation",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Left-Total Relation",
"Definition:Serial Relation",
"Definition:Sequence"
] | [
"Definition:Transitive Closure of Relation",
"Definition:Restriction",
"Definition:Left-Total Relation",
"Definition:Non-Empty Set",
"Definition:Left-Total Relation",
"Axiom:Axiom of Dependent Choice",
"Definition:Sequence/Infinite Sequence",
"Definition:Transitive Closure of Relation/Finite Chain",
... |
proofwiki-7756 | Locally Finite Connected Graph is Countable | Let $G = \struct {V, E}$ be a graph which is connected and locally finite.
Then $G$ has countably many vertices and countably many edges. | We first show that $V$ is countable.
If $V$ is finite, then it is surely countable.
Suppose instead that $V$ is infinite.
Choose an arbitrary vertex $q \in V$.
Recursively define a sequence $\sequence {S_n}$:
:Let $S_0 = \set q$.
:Let $S_{n + 1}$ be the set of all vertices that are adjacent to some element of $S_n$ but... | Let $G = \struct {V, E}$ be a [[Definition:Graph (Graph Theory)|graph]] which is [[Definition:Connected Graph|connected]] and [[Definition:Locally Finite Graph|locally finite]].
Then $G$ has [[Definition:Countable Set|countably many]] [[Definition:Vertex of Graph|vertices]] and [[Definition:Countable Set|countably]] ... | We first show that $V$ is [[Definition:Countable Set|countable]].
If $V$ is [[Definition:Finite Set|finite]], then it is surely [[Definition:Countable Set|countable]].
Suppose instead that $V$ is [[Definition:Infinite Set|infinite]].
Choose an arbitrary [[Definition:Vertex of Graph|vertex]] $q \in V$.
[[Principle o... | Locally Finite Connected Graph is Countable | https://proofwiki.org/wiki/Locally_Finite_Connected_Graph_is_Countable | https://proofwiki.org/wiki/Locally_Finite_Connected_Graph_is_Countable | [
"Graph Theory"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Locally Finite Graph",
"Definition:Countable Set",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Countable Set",
"Definition:Graph (Graph Theory)/Edge"
] | [
"Definition:Countable Set",
"Definition:Finite Set",
"Definition:Countable Set",
"Definition:Infinite Set",
"Definition:Graph (Graph Theory)/Vertex",
"Principle of Recursive Definition",
"Definition:Sequence/Infinite Sequence",
"Definition:Set",
"Definition:Adjacent (Graph Theory)",
"Definition:Pa... |
proofwiki-7757 | Law of Cosines/Right Triangle | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that:
: $a$ is opposite $A$
: $b$ is opposite $B$
: $c$ is opposite $C$.
Let $\triangle ABC$ be a right triangle such that $\angle A$ is right.
Then:
:$c^2 = a^2 + b^2 - 2 a b \cos C$ | Let $\triangle ABC$ be a right triangle such that $\angle A$ is right.
:300px
{{begin-eqn}}
{{eqn | l = a^2
| r = b^2 + c^2
| c = Pythagoras's Theorem
}}
{{eqn | l = c^2
| r = a^2 - b^2
| c = adding $-b^2$ to both sides and rearranging
}}
{{eqn | r = a^2 - 2 b^2 + b^2
| c = adding $0 = b^2... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that:
: $a$ is [[Definition:Opposite (in Triangle)|opposite]] $A$
: $b$ is [[Definition:Opposite (in Triangle)|opposite]] $B$
: $c$ is [[Definition:Opposite (in Triangle)|opposite]] $C$.
Let $\triangle ABC$ be a [[Defin... | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] such that $\angle A$ is [[Definition:Right Angle|right]].
:[[File:CosineRule-Proof3-right.png|300px]]
{{begin-eqn}}
{{eqn | l = a^2
| r = b^2 + c^2
| c = [[Pythagoras's Theorem]]
}}
{{eqn | l = c^2
| r = a^2 - b^2
| c = addi... | Law of Cosines/Right Triangle | https://proofwiki.org/wiki/Law_of_Cosines/Right_Triangle | https://proofwiki.org/wiki/Law_of_Cosines/Right_Triangle | [
"Law of Cosines"
] | [
"Definition:Triangle (Geometry)",
"Definition:Triangle (Geometry)/Opposite",
"Definition:Triangle (Geometry)/Opposite",
"Definition:Triangle (Geometry)/Opposite",
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Right Angle"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Right Angle",
"File:CosineRule-Proof3-right.png",
"Pythagoras's Theorem"
] |
proofwiki-7758 | Minimally Inductive Set is Minimal | The minimally inductive set $\omega$ is a subset of every inductive set. | Let $A$ be an inductive set.
Let $B$ be another arbitrary inductive set.
Then from Intersection is Subset, $A \cap B \subseteq A$.
From Intersection of Inductive Sets, $A \cap B$ is also an inductive set.
This set $A \cap B$ is one of the subsets
By the definition of $\omega$ it follows that $\omega \subseteq A \cap B$... | The [[Definition:Minimally Inductive Set|minimally inductive set]] $\omega$ is a [[Definition:Subset|subset]] of every [[Definition:Inductive Set|inductive set]]. | Let $A$ be an [[Definition:Inductive Set|inductive set]].
Let $B$ be another arbitrary [[Definition:Inductive Set|inductive set]].
Then from [[Intersection is Subset]], $A \cap B \subseteq A$.
From [[Intersection of Inductive Sets]], $A \cap B$ is also an [[Definition:Inductive Set|inductive set]].
This set $A \cap... | Minimally Inductive Set is Minimal | https://proofwiki.org/wiki/Minimally_Inductive_Set_is_Minimal | https://proofwiki.org/wiki/Minimally_Inductive_Set_is_Minimal | [
"Minimally Inductive Set"
] | [
"Definition:Minimally Inductive Set",
"Definition:Subset",
"Definition:Inductive Set"
] | [
"Definition:Inductive Set",
"Definition:Inductive Set",
"Intersection is Subset",
"Intersection of Inductive Sets",
"Definition:Inductive Set",
"Definition:Subset"
] |
proofwiki-7759 | Law of Cosines/Proof 3/Acute Triangle | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that:
:$a$ is opposite $A$
:$b$ is opposite $B$
:$c$ is opposite $C$.
Let $\triangle ABC$ be an acute triangle.
Then:
:$c^2 = a^2 + b^2 - 2a b \cos C$ | Let $\triangle ABC$ be an acute triangle.
:300px
Let $BD$ be dropped perpendicular to $AC$.
Let:
{{begin-eqn}}
{{eqn | l = h
| r = BD
}}
{{eqn | l = e
| r = CD
}}
{{eqn | l = f
| r = AD
}}
{{end-eqn}}
We have that $\triangle CDB$ and $\triangle ADB$ are right triangles.
Hence:
{{begin-eqn}}
{{eqn | n ... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that:
:$a$ is [[Definition:Opposite (in Triangle)|opposite]] $A$
:$b$ is [[Definition:Opposite (in Triangle)|opposite]] $B$
:$c$ is [[Definition:Opposite (in Triangle)|opposite]] $C$.
Let $\triangle ABC$ be an [[Definit... | Let $\triangle ABC$ be an [[Definition:Acute Triangle|acute triangle]].
:[[File:CosineRule-Proof3-acute.png|300px]]
Let $BD$ be dropped [[Definition:Perpendicular|perpendicular]] to $AC$.
Let:
{{begin-eqn}}
{{eqn | l = h
| r = BD
}}
{{eqn | l = e
| r = CD
}}
{{eqn | l = f
| r = AD
}}
{{end-eqn}}
... | Law of Cosines/Proof 3/Acute Triangle | https://proofwiki.org/wiki/Law_of_Cosines/Proof_3/Acute_Triangle | https://proofwiki.org/wiki/Law_of_Cosines/Proof_3/Acute_Triangle | [
"Law of Cosines"
] | [
"Definition:Triangle (Geometry)",
"Definition:Triangle (Geometry)/Opposite",
"Definition:Triangle (Geometry)/Opposite",
"Definition:Triangle (Geometry)/Opposite",
"Definition:Triangle (Geometry)/Acute"
] | [
"Definition:Triangle (Geometry)/Acute",
"File:CosineRule-Proof3-acute.png",
"Definition:Right Angle/Perpendicular",
"Definition:Triangle (Geometry)/Right-Angled",
"Pythagoras's Theorem",
"Pythagoras's Theorem"
] |
proofwiki-7760 | Law of Cosines/Proof 3/Obtuse Triangle | Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that:
: $a$ is opposite $A$
: $b$ is opposite $B$
: $c$ is opposite $C$.
Let $\triangle ABC$ be an obtuse triangle such that $A$ is obtuse
Then:
:$c^2 = a^2 + b^2 - 2a b \cos C$ | Let $\triangle ABC$ be an obtuse triangle.
:300px
Let $AC$ be extended and $BD$ be dropped perpendicular to $AC$.
Let:
{{begin-eqn}}
{{eqn | l = h
| r = BD
}}
{{eqn | l = e
| r = CD
}}
{{eqn | l = f
| r = AD
}}
{{end-eqn}}
We have that $\triangle CDB$ and $\triangle ADB$ are right triangles.
Hence:
{{... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that:
: $a$ is [[Definition:Opposite (in Triangle)|opposite]] $A$
: $b$ is [[Definition:Opposite (in Triangle)|opposite]] $B$
: $c$ is [[Definition:Opposite (in Triangle)|opposite]] $C$.
Let $\triangle ABC$ be an [[Defi... | Let $\triangle ABC$ be an [[Definition:Obtuse Triangle|obtuse triangle]].
:[[File:CosineRule-Proof3-obtuse.png|300px]]
Let $AC$ be extended and $BD$ be dropped [[Definition:Perpendicular|perpendicular]] to $AC$.
Let:
{{begin-eqn}}
{{eqn | l = h
| r = BD
}}
{{eqn | l = e
| r = CD
}}
{{eqn | l = f
| r... | Law of Cosines/Proof 3/Obtuse Triangle | https://proofwiki.org/wiki/Law_of_Cosines/Proof_3/Obtuse_Triangle | https://proofwiki.org/wiki/Law_of_Cosines/Proof_3/Obtuse_Triangle | [
"Law of Cosines"
] | [
"Definition:Triangle (Geometry)",
"Definition:Triangle (Geometry)/Opposite",
"Definition:Triangle (Geometry)/Opposite",
"Definition:Triangle (Geometry)/Opposite",
"Definition:Triangle (Geometry)/Obtuse",
"Definition:Obtuse Angle"
] | [
"Definition:Triangle (Geometry)/Obtuse",
"File:CosineRule-Proof3-obtuse.png",
"Definition:Right Angle/Perpendicular",
"Definition:Triangle (Geometry)/Right-Angled",
"Pythagoras's Theorem",
"Pythagoras's Theorem"
] |
proofwiki-7761 | König's Lemma/Countable | Let $G = \struct {V, E}$ be a graph with countably infinitely many vertices which is connected and is locally finite.
Then every vertex of $G$ lies on a path of infinite length. | Let $r$ be a vertex of $G$.
Recursively define a sequence $\sequence {S_n}$:
Let $S_0 = \set r$.
Let $S_{n + 1}$ be the set of all vertices that are adjacent to some element of $S_n$ but ''not'' adjacent to any element of $S_k$ for $k < n$.
That is, $S_n$ is the set of vertices whose shortest path(s) to $r$ have $n$ ed... | Let $G = \struct {V, E}$ be a [[Definition:Graph (Graph Theory)|graph]] with [[Definition:Countably Infinite Set|countably infinitely many]] [[Definition:Vertex of Graph|vertices]] which is [[Definition:Connected Graph|connected]] and is [[Definition:Locally Finite Graph|locally finite]].
Then every [[Definition:Vert... | Let $r$ be a [[Definition:Vertex of Graph|vertex]] of $G$.
[[Principle of Recursive Definition|Recursively define]] a [[Definition:Infinite Sequence|sequence]] $\sequence {S_n}$:
Let $S_0 = \set r$.
Let $S_{n + 1}$ be the [[Definition:Set|set]] of all [[Definition:Vertex of Graph|vertices]] that are [[Definition:Adj... | König's Lemma/Countable | https://proofwiki.org/wiki/König's_Lemma/Countable | https://proofwiki.org/wiki/König's_Lemma/Countable | [
"König's Lemma"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Countably Infinite/Set",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Locally Finite Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Path (Graph Theory)",
"Definition:Infinite Set",
"De... | [
"Definition:Graph (Graph Theory)/Vertex",
"Principle of Recursive Definition",
"Definition:Sequence/Infinite Sequence",
"Definition:Set",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Adjacent (Graph Theory)/Vertices/Undirected Graph",
"Definition:Element",
"Definition:Adjacent (Graph Theory)/... |
proofwiki-7762 | Paths of Minimal Length from Vertex form Tree | Let $G = \struct {V, E}$ be a simple graph.
Let $r \in V$ be a vertex in $G$.
Let $P$ be the set of minimal length paths beginning at $r$.
Let $p, q \in P$.
Let $E'$ be defined as follows:
$\set {p, q} \in E'$ {{iff}} either:
:$q$ is formed by extending $p$ with one edge of $E$ and one vertex of $V$
or:
:$p$ is formed ... | Let $\tuple r$ be the $0$-length $G$-path whose only vertex is $r$. | Let $G = \struct {V, E}$ be a [[Definition:Simple Graph|simple graph]].
Let $r \in V$ be a [[Definition:Vertex of Graph|vertex]] in $G$.
Let $P$ be the [[Definition:Set|set]] of [[Definition:Minimal Length Path|minimal length paths]] beginning at $r$.
Let $p, q \in P$.
Let $E'$ be defined as follows:
$\set {p, q}... | Let $\tuple r$ be the $0$-[[Definition:Length of Walk|length]] [[Definition:Path (Graph Theory)|$G$-path]] whose only [[Definition:Vertex of Graph|vertex]] is $r$. | Paths of Minimal Length from Vertex form Tree | https://proofwiki.org/wiki/Paths_of_Minimal_Length_from_Vertex_form_Tree | https://proofwiki.org/wiki/Paths_of_Minimal_Length_from_Vertex_form_Tree | [
"Paths (Graph Theory)",
"Tree Theory"
] | [
"Definition:Simple Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Set",
"Definition:Minimal Length Path",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Tre... | [
"Definition:Walk (Graph Theory)/Length",
"Definition:Path (Graph Theory)",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Walk (Graph Theory)/Length",
"Definition:Walk (Graph Theory)/Length",
"Definition:Walk (Graph Theory)/Length",
"Definition:Path (Graph Theory)",
"Definition:Path (Graph Theo... |
proofwiki-7763 | Finite Sequences in Set Form Acyclic Graph | Let $S$ be a set.
Let $V$ be the set of finite sequences in $S$.
Let $E$ be the set of unordered pairs $\set {p, q}$ of elements of $V$ such that either:
:$q$ is formed by extending $p$ by one element or
:$p$ is formed by extending $q$ by one element.
That is:
:$\card {\Dom p \symdif \Dom q} = 1$, where $\symdif$ is sy... | {{proof wanted}}
Category:Graph Theory
ttrxo7lcehixqwf34y9hf4oiyifowb6 | Let $S$ be a [[Definition:set|set]].
Let $V$ be the set of [[Definition:Finite Sequence|finite sequences]] in $S$.
Let $E$ be the set of unordered pairs $\set {p, q}$ of elements of $V$ such that either:
:$q$ is formed by extending $p$ by one element or
:$p$ is formed by extending $q$ by one element.
That is:
:$\ca... | {{proof wanted}}
[[Category:Graph Theory]]
ttrxo7lcehixqwf34y9hf4oiyifowb6 | Finite Sequences in Set Form Acyclic Graph | https://proofwiki.org/wiki/Finite_Sequences_in_Set_Form_Acyclic_Graph | https://proofwiki.org/wiki/Finite_Sequences_in_Set_Form_Acyclic_Graph | [
"Graph Theory"
] | [
"Definition:set",
"Definition:Finite Sequence",
"Definition:Symmetric Difference",
"Definition:Acyclic Graph"
] | [
"Category:Graph Theory"
] |
proofwiki-7764 | Four Color Theorem for Finite Maps implies Four Color Theorem for Infinite Maps | Suppose that any finite planar graph can be assigned a proper vertex $k$-coloring such that $k \le 4$.
Then the same is true of any infinite planar graph. | Let $G$ be an infinite planar graph.
Let $C$ be a set of vertices of $G$.
For each $c \in C$ let $p_c^1, p_c^2, p_c^3, p_c^4$ be propositional symbols, where
: $p_c^i$ is true {{iff}} the color of vertex $c$ is $i$.
Let $\PP_0$ be the vocabulary defined as:
:$\PP_0 = \set {p_c^1, p_c^2, p_c^3, p_c^4: c \in C}$
Let $\ma... | Suppose that any [[Definition:Finite Graph|finite]] [[Definition:Planar Graph|planar graph]] can be assigned a [[Definition:Proper Vertex Coloring|proper vertex $k$-coloring]] such that $k \le 4$.
Then the same is true of any [[Definition:Infinite Graph|infinite]] [[Definition:Planar Graph|planar graph]]. | Let $G$ be an [[Definition:Infinite Graph|infinite]] [[Definition:Planar Graph|planar graph]].
Let $C$ be a [[Definition:Set|set]] of [[Definition:Vertex of Graph|vertices]] of $G$.
For each $c \in C$ let $p_c^1, p_c^2, p_c^3, p_c^4$ be [[Definition:Propositional Symbol|propositional symbols]], where
: $p_c^i$ is [[D... | Four Color Theorem for Finite Maps implies Four Color Theorem for Infinite Maps | https://proofwiki.org/wiki/Four_Color_Theorem_for_Finite_Maps_implies_Four_Color_Theorem_for_Infinite_Maps | https://proofwiki.org/wiki/Four_Color_Theorem_for_Finite_Maps_implies_Four_Color_Theorem_for_Infinite_Maps | [
"Graph Theory"
] | [
"Definition:Finite Graph",
"Definition:Planar Graph",
"Definition:Proper Coloring/Vertex Coloring",
"Definition:Infinite Graph",
"Definition:Planar Graph"
] | [
"Definition:Infinite Graph",
"Definition:Planar Graph",
"Definition:Set",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Language of Propositional Logic/Alphabet/Letter",
"Definition:True",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Language of Propositional Logic/Alphabet/Letter",
... |
proofwiki-7765 | Connected Vertices are Connected by Path | Let $G = \struct {V, E}$ be a simple graph.
Let $x, y \in V$.
Let there exist a walk $w: \N_n \to V$ from $x$ to $y$.
Then there exists a subsequence $z_n$ of $w$ which is a path from $x$ to $y$. | We represent a walk as a sequence of vertices.
However, the same argument will work for the representation as an alternating sequence of vertices and edges.
The proof proceeds by induction on the length of $w$.
If the length of $w$ is $0$, then it is trivially a path.
Suppose that every walk of length less than $n$ has... | Let $G = \struct {V, E}$ be a [[Definition:Simple Graph|simple graph]].
Let $x, y \in V$.
Let there exist a [[Definition:Walk (Graph Theory)|walk]] $w: \N_n \to V$ from $x$ to $y$.
Then there exists a [[Definition:Subsequence|subsequence]] $z_n$ of $w$ which is a [[Definition:Path (Graph Theory)|path]] from $x$ to ... | We represent a [[Definition:Walk (Graph Theory)|walk]] as a [[Definition:Sequence|sequence]] of [[Definition:Vertex of Graph|vertices]].
However, the same argument will work for the representation as an alternating sequence of [[Definition:Vertex of Graph|vertices]] and [[Definition:Edge of Graph|edges]].
The proof ... | Connected Vertices are Connected by Path | https://proofwiki.org/wiki/Connected_Vertices_are_Connected_by_Path | https://proofwiki.org/wiki/Connected_Vertices_are_Connected_by_Path | [
"Graph Theory"
] | [
"Definition:Simple Graph",
"Definition:Walk (Graph Theory)",
"Definition:Subsequence",
"Definition:Path (Graph Theory)"
] | [
"Definition:Walk (Graph Theory)",
"Definition:Sequence",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Edge",
"Principle of Mathematical Induction",
"Definition:Walk (Graph Theory)/Length",
"Definition:Walk (Graph Theory)/Length",
... |
proofwiki-7766 | Rooted Tree Corresponds to Arborescence | Let $T = \struct {V, E}$ be a rooted tree with root $r$.
Then there is a unique orientation of $T$ which is an $r$-arborescence. | Recall that a tree is connected and has no cycles.
Thus there is exactly one path from each node of $T$ to each other node of $T$.
{{explain|This is in fact a result that already exists and can be quoted directly.}}
Let $A$ be the set of all ordered pairs $x, y \in V$ such that:
:$\tuple {x, y} \in E$ and
:The unique p... | Let $T = \struct {V, E}$ be a [[Definition:Rooted Tree|rooted tree]] with root $r$.
Then there is a unique [[Definition:Orientation (Graph Theory)|orientation]] of $T$ which is an $r$-[[Definition:Arborescence|arborescence]]. | Recall that a [[Definition:Tree (Graph Theory)|tree]] is [[Definition:Connected Graph|connected]] and has no [[Definition:Cycle (Graph Theory)|cycles]].
Thus there is exactly one [[Definition:Path (Graph Theory)|path]] from each [[Definition:Node of Tree|node]] of $T$ to each other [[Definition:Node of Tree|node]] of ... | Rooted Tree Corresponds to Arborescence | https://proofwiki.org/wiki/Rooted_Tree_Corresponds_to_Arborescence | https://proofwiki.org/wiki/Rooted_Tree_Corresponds_to_Arborescence | [
"Rooted Trees"
] | [
"Definition:Rooted Tree",
"Definition:Orientation (Graph Theory)",
"Definition:Arborescence"
] | [
"Definition:Tree (Graph Theory)",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Cycle (Graph Theory)",
"Definition:Path (Graph Theory)",
"Definition:Tree (Graph Theory)/Node",
"Definition:Tree (Graph Theory)/Node",
"Definition:Set",
"Definition:Ordered Pair",
"Definition:Path (Graph Theor... |
proofwiki-7767 | Equivalence of Definitions of Arborescence | Let $G = \struct {V, A}$ be a digraph.
Let $r \in V$.
{{TFAE|def = Arborescence}} | === Definition 1 implies Definition 3 ===
Let $G$ be an $r$-arborescence by definition 1.
Let $v \in V$ such that $v \ne r$.
Then there is exactly one directed walk $w$ from $r$ to $v$.
Since $v \ne r$, either:
:$w = \tuple {r, v}$
or:
:$\exists m \in V: w = \tuple {r, \ldots, m, v}$
Thus $v$ is the final vertex of the... | Let $G = \struct {V, A}$ be a [[Definition:Digraph|digraph]].
Let $r \in V$.
{{TFAE|def = Arborescence}} | === Definition 1 implies Definition 3 ===
Let $G$ be an [[Definition:Arborescence/Definition 1|$r$-arborescence by definition 1]].
Let $v \in V$ such that $v \ne r$.
Then there is exactly one [[Definition:Directed Walk|directed walk]] $w$ from $r$ to $v$.
Since $v \ne r$, either:
:$w = \tuple {r, v}$
or:
:$\exists ... | Equivalence of Definitions of Arborescence | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Arborescence | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Arborescence | [
"Arborescences"
] | [
"Definition:Digraph"
] | [
"Definition:Arborescence/Definition 1",
"Definition:Directed Walk",
"Definition:Arc of Digraph/Endvertex/Final Vertex",
"Definition:Digraph/Arc",
"Definition:Digraph/Arc",
"Definition:Arc of Digraph/Endvertex/Final Vertex",
"Definition:Distinct",
"Definition:Digraph/Arc",
"Definition:Directed Walk",... |
proofwiki-7768 | Negation as Implication of Bottom | $p \implies \bot \dashv\vdash \neg p$ | {{BeginTableau|p \implies \bot \vdash \neg p}}
{{Premise|1|p \implies \bot}}
{{Assumption|2|p}}
{{ModusPonens|3|1,2|\bot|1|2}}
{{Contradiction|4|1|\neg p|2|3}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|\neg p \vdash p \implies \bot}}
{{Premise|1|\neg p}}
{{Assumption|2|p}}
{{NonContradiction|3|1,2|1|2}}
{{Implication... | $p \implies \bot \dashv\vdash \neg p$ | {{BeginTableau|p \implies \bot \vdash \neg p}}
{{Premise|1|p \implies \bot}}
{{Assumption|2|p}}
{{ModusPonens|3|1,2|\bot|1|2}}
{{Contradiction|4|1|\neg p|2|3}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|\neg p \vdash p \implies \bot}}
{{Premise|1|\neg p}}
{{Assumption|2|p}}
{{NonContradiction|3|1,2|1|2}}
{{Implicatio... | Negation as Implication of Bottom | https://proofwiki.org/wiki/Negation_as_Implication_of_Bottom | https://proofwiki.org/wiki/Negation_as_Implication_of_Bottom | [] | [] | [] |
proofwiki-7769 | Clavius's Law implies Law of Excluded Middle | From Clavius's Law:
:$\neg p \implies p \vdash p$
follows the Law of Excluded Middle:
:$\vdash p \lor \neg p$ | {{BeginTableau|\vdash p \lor \neg p}}
{{Assumption|1|\neg (p \lor \neg p)}}
{{SequentIntro|2|1|\bot|1|Negation of Excluded Middle is False}}
{{Explosion|3|1|p \lor \neg p|2}}
{{Implication|4||\neg(p \lor \neg p) \implies p \lor \neg p|1|3}}
{{SequentIntro|5||p \lor \neg p|4|Clavius's Law}}
{{EndTableau}}
{{qed}} | From [[Clavius's Law]]:
:$\neg p \implies p \vdash p$
follows the [[Law of Excluded Middle]]:
:$\vdash p \lor \neg p$ | {{BeginTableau|\vdash p \lor \neg p}}
{{Assumption|1|\neg (p \lor \neg p)}}
{{SequentIntro|2|1|\bot|1|[[Negation of Excluded Middle is False/Form 1|Negation of Excluded Middle is False]]}}
{{Explosion|3|1|p \lor \neg p|2}}
{{Implication|4||\neg(p \lor \neg p) \implies p \lor \neg p|1|3}}
{{SequentIntro|5||p \lor \neg p... | Clavius's Law implies Law of Excluded Middle | https://proofwiki.org/wiki/Clavius's_Law_implies_Law_of_Excluded_Middle | https://proofwiki.org/wiki/Clavius's_Law_implies_Law_of_Excluded_Middle | [
"Law of Excluded Middle",
"Clavius's Law"
] | [
"Clavius's Law",
"Law of Excluded Middle"
] | [
"Negation of Excluded Middle is False/Form 1",
"Clavius's Law/Formulation 1"
] |
proofwiki-7770 | Parity of Integer equals Parity of its Square/Even | Let $p \in \Z$ be an integer.
Let $p$ be even.
Then $p^2$ is also even. | Let $r = 0$, so:
:$p = 2 k$
Then:
:$p^2 = \paren {2 k}^2 = 4 k^2 = 2 \paren {2 k^2}$
and so $p^2$ is even. | Let $p \in \Z$ be an [[Definition:Integer|integer]].
Let $p$ be [[Definition:Even Integer|even]].
Then $p^2$ is also [[Definition:Even Integer|even]]. | Let $r = 0$, so:
:$p = 2 k$
Then:
:$p^2 = \paren {2 k}^2 = 4 k^2 = 2 \paren {2 k^2}$
and so $p^2$ is [[Definition:Even Integer|even]]. | Parity of Integer equals Parity of its Square/Even | https://proofwiki.org/wiki/Parity_of_Integer_equals_Parity_of_its_Square/Even | https://proofwiki.org/wiki/Parity_of_Integer_equals_Parity_of_its_Square/Even | [
"Parity of Integer equals Parity of its Square"
] | [
"Definition:Integer",
"Definition:Even Integer",
"Definition:Even Integer"
] | [
"Definition:Even Integer"
] |
proofwiki-7771 | Parity of Integer equals Parity of its Square/Odd | Let $p \in \Z$ be an integer.
Let $p$ be odd.
Then $p^2$ is also odd. | Let $r = 1$, so:
:$p = 2 k + 1$
Then:
:$p^2 = \paren {2 k + 1}^2 = 4 k^2 + 4 k + 1 = 2 \paren {2 k^2 + 2 k} + 1$
and so $p^2$ is odd. | Let $p \in \Z$ be an [[Definition:Integer|integer]].
Let $p$ be [[Definition:Odd Integer|odd]].
Then $p^2$ is also [[Definition:Odd Integer|odd]]. | Let $r = 1$, so:
:$p = 2 k + 1$
Then:
:$p^2 = \paren {2 k + 1}^2 = 4 k^2 + 4 k + 1 = 2 \paren {2 k^2 + 2 k} + 1$
and so $p^2$ is [[Definition:Odd Integer|odd]]. | Parity of Integer equals Parity of its Square/Odd | https://proofwiki.org/wiki/Parity_of_Integer_equals_Parity_of_its_Square/Odd | https://proofwiki.org/wiki/Parity_of_Integer_equals_Parity_of_its_Square/Odd | [
"Parity of Integer equals Parity of its Square"
] | [
"Definition:Integer",
"Definition:Odd Integer",
"Definition:Odd Integer"
] | [
"Definition:Odd Integer"
] |
proofwiki-7772 | Negation of Excluded Middle is False/Form 1 | $\neg (p \lor \neg p) \vdash \bot$ | {{BeginTableau|\neg (p \lor \neg p) \vdash \bot}}
{{Assumption|1|\neg (p \lor \neg p)}}
{{SequentIntro|2|1|\neg p \land \neg \neg p|1|De Morgan's Laws}}
{{Simplification|3|1|\neg p|2|1}}
{{Simplification|4|1|\neg\neg p|2|2}}
{{NonContradiction|5|1|3|4}}
{{EndTableau}}
{{qed}} | $\neg (p \lor \neg p) \vdash \bot$ | {{BeginTableau|\neg (p \lor \neg p) \vdash \bot}}
{{Assumption|1|\neg (p \lor \neg p)}}
{{SequentIntro|2|1|\neg p \land \neg \neg p|1|[[De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Reverse Implication|De Morgan's Laws]]}}
{{Simplification|3|1|\neg p|2|1}}
{{Simplification|4|1|\neg\neg p|2|2}}
{{NonCo... | Negation of Excluded Middle is False/Form 1 | https://proofwiki.org/wiki/Negation_of_Excluded_Middle_is_False/Form_1 | https://proofwiki.org/wiki/Negation_of_Excluded_Middle_is_False/Form_1 | [
"Propositional Logic"
] | [] | [
"De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Reverse Implication"
] |
proofwiki-7773 | Negation of Excluded Middle is False/Form 2 | $\vdash \neg \neg (p \lor \neg p)$ | {{BeginTableau|\neg\neg (p \lor \neg p)}}
{{Assumption|1|\neg (p \lor \neg p)}}
{{SequentIntro|2|1|\bot|1|Negation of Excluded Middle is False: Form 1}}
{{Implication|3||\neg (p \lor \neg p) \implies \bot|1|2}}
{{SequentIntro|4||\neg \neg (p \lor \neg p)|3|Negation as Implication of Bottom}}
{{EndTableau}}
{{qed}}
Cate... | $\vdash \neg \neg (p \lor \neg p)$ | {{BeginTableau|\neg\neg (p \lor \neg p)}}
{{Assumption|1|\neg (p \lor \neg p)}}
{{SequentIntro|2|1|\bot|1|[[Negation of Excluded Middle is False/Form 1|Negation of Excluded Middle is False: Form 1]]}}
{{Implication|3||\neg (p \lor \neg p) \implies \bot|1|2}}
{{SequentIntro|4||\neg \neg (p \lor \neg p)|3|[[Negation as I... | Negation of Excluded Middle is False/Form 2 | https://proofwiki.org/wiki/Negation_of_Excluded_Middle_is_False/Form_2 | https://proofwiki.org/wiki/Negation_of_Excluded_Middle_is_False/Form_2 | [
"Propositional Logic"
] | [] | [
"Negation of Excluded Middle is False/Form 1",
"Negation as Implication of Bottom",
"Category:Propositional Logic"
] |
proofwiki-7774 | Injection iff Left Cancellable/Necessary Condition | Let $f: Y \to Z$ be an injection.
Then $f$ is left cancellable. | Let $f: Y \to Z$ be an injection.
Let $X$ be a set
Let $g_1: X \to Y, g_2: X \to Y$ be mappings such that:
: $f \circ g_1 = f \circ g_2$
Then $\forall x \in X$:
{{begin-eqn}}
{{eqn | l = \map f {g_1 \paren x}
| r = \map {f \circ g_1} x
| c = {{Defof|Composite Mapping}}
}}
{{eqn | r = \map {f \circ g_2} x
... | Let $f: Y \to Z$ be an [[Definition:Injection|injection]].
Then $f$ is [[Definition:Left Cancellable Mapping|left cancellable]]. | Let $f: Y \to Z$ be an [[Definition:Injection|injection]].
Let $X$ be a [[Definition:Set|set]]
Let $g_1: X \to Y, g_2: X \to Y$ be [[Definition:Mapping|mappings]] such that:
: $f \circ g_1 = f \circ g_2$
Then $\forall x \in X$:
{{begin-eqn}}
{{eqn | l = \map f {g_1 \paren x}
| r = \map {f \circ g_1} x
... | Injection iff Left Cancellable/Necessary Condition | https://proofwiki.org/wiki/Injection_iff_Left_Cancellable/Necessary_Condition | https://proofwiki.org/wiki/Injection_iff_Left_Cancellable/Necessary_Condition | [
"Injection iff Left Cancellable"
] | [
"Definition:Injection",
"Definition:Left Cancellable Mapping"
] | [
"Definition:Injection",
"Definition:Set",
"Definition:Mapping",
"Definition:By Hypothesis",
"Definition:Injection",
"Definition:Left Cancellable Mapping"
] |
proofwiki-7775 | Injection iff Left Cancellable/Sufficient Condition | Let $f: Y \to Z$ be a mapping which is left cancellable.
Then $f$ is an injection. | We use a Proof by Contraposition.
That is, we show that if $f: Y \to Z$ is not injective, then $f$ is not left cancellable.
Hence, suppose $f: Y \to Z$ is not injective.
Then:
: $\exists y_1 \ne y_2 \in Y: f \left({y_1}\right) = f \left({y_2}\right)$
Let the two mappings $g_1: Y \to Y, g_2: Y \to Y$ be defined as follo... | Let $f: Y \to Z$ be a [[Definition:Mapping|mapping]] which is [[Definition:Left Cancellable Mapping|left cancellable]].
Then $f$ is an [[Definition:Injection|injection]]. | We use a [[Proof by Contraposition]].
That is, we show that if $f: Y \to Z$ is not [[Definition:Injection|injective]], then $f$ is not [[Definition:Left Cancellable Mapping|left cancellable]].
Hence, suppose $f: Y \to Z$ is not [[Definition:Injection|injective]].
Then:
: $\exists y_1 \ne y_2 \in Y: f \left({y_1}\rig... | Injection iff Left Cancellable/Sufficient Condition | https://proofwiki.org/wiki/Injection_iff_Left_Cancellable/Sufficient_Condition | https://proofwiki.org/wiki/Injection_iff_Left_Cancellable/Sufficient_Condition | [
"Injection iff Left Cancellable",
"Proofs by Contraposition"
] | [
"Definition:Mapping",
"Definition:Left Cancellable Mapping",
"Definition:Injection"
] | [
"Proof by Contraposition",
"Definition:Injection",
"Definition:Left Cancellable Mapping",
"Definition:Injection",
"Definition:Mapping",
"Definition:Left Cancellable Mapping",
"Rule of Transposition"
] |
proofwiki-7776 | Peirce's Law implies Law of Excluded Middle | From Peirce's Law:
:$\paren {p \implies q} \implies p \vdash p$
follows the Law of Excluded Middle:
:$\vdash p \lor \neg p$ | {{BeginTableau|\vdash p \lor \neg p}}
{{Assumption|1|\paren {p \lor \neg p} \implies \bot}}
{{SequentIntro|2|1|\neg \paren {p \lor \neg p}|1|Negation as Implication of Bottom}}
{{SequentIntro|3|1|\bot|2|Negation of Excluded Middle is False}}
{{Explosion|4|1|p \lor \neg p|3}}
{{Implication|5||\paren {\paren {p \lor \neg... | From [[Peirce's Law]]:
:$\paren {p \implies q} \implies p \vdash p$
follows the [[Law of Excluded Middle]]:
:$\vdash p \lor \neg p$ | {{BeginTableau|\vdash p \lor \neg p}}
{{Assumption|1|\paren {p \lor \neg p} \implies \bot}}
{{SequentIntro|2|1|\neg \paren {p \lor \neg p}|1|[[Negation as Implication of Bottom]]}}
{{SequentIntro|3|1|\bot|2|[[Negation of Excluded Middle is False/Form 1|Negation of Excluded Middle is False]]}}
{{Explosion|4|1|p \lor \ne... | Peirce's Law implies Law of Excluded Middle | https://proofwiki.org/wiki/Peirce's_Law_implies_Law_of_Excluded_Middle | https://proofwiki.org/wiki/Peirce's_Law_implies_Law_of_Excluded_Middle | [
"Law of Excluded Middle",
"Peirce's Law"
] | [
"Peirce's Law",
"Law of Excluded Middle"
] | [
"Negation as Implication of Bottom",
"Negation of Excluded Middle is False/Form 1",
"Peirce's Law"
] |
proofwiki-7777 | Pseudocomplemented Lattice is Bounded | Let $\struct {L, \wedge, \vee, \preceq}$ be a pseudocomplemented lattice.
Then $\struct {L, \wedge, \vee, \preceq}$ is a bounded lattice. | By the definition of pseudocomplemented lattice, $L$ has a smallest element $\bot$.
Let $x \in L$.
Then:
:$x \wedge \bot = \bot$
{{explain}}
By the definition of pseudocomplemented lattice, there is a greatest element $\bot^*$ such that:
:$\bot \wedge \bot^* = \bot$
But then by the definition of greatest element:
:$\fo... | Let $\struct {L, \wedge, \vee, \preceq}$ be a [[Definition:Pseudocomplemented Lattice|pseudocomplemented lattice]].
Then $\struct {L, \wedge, \vee, \preceq}$ is a [[Definition:Bounded Lattice|bounded lattice]]. | By the definition of [[Definition:Pseudocomplemented Lattice|pseudocomplemented lattice]], $L$ has a [[Definition:Smallest Element|smallest element]] $\bot$.
Let $x \in L$.
Then:
:$x \wedge \bot = \bot$
{{explain}}
By the definition of [[Definition:Pseudocomplemented Lattice|pseudocomplemented lattice]], there is a ... | Pseudocomplemented Lattice is Bounded | https://proofwiki.org/wiki/Pseudocomplemented_Lattice_is_Bounded | https://proofwiki.org/wiki/Pseudocomplemented_Lattice_is_Bounded | [
"Pseudocomplemented Lattices",
"Bounded Lattices"
] | [
"Definition:Pseudocomplemented Lattice",
"Definition:Bounded Lattice"
] | [
"Definition:Pseudocomplemented Lattice",
"Definition:Smallest Element",
"Definition:Pseudocomplemented Lattice",
"Definition:Greatest Element",
"Definition:Greatest Element",
"Definition:Greatest Element",
"Definition:Smallest Element",
"Definition:Greatest Element",
"Definition:Bounded Lattice",
... |
proofwiki-7778 | Topology forms Complete Lattice | Let $\struct {X, \tau}$ be a topological space.
Then $\struct {\tau, \subseteq}$ is a complete lattice. | To show that $\struct {\tau, \subseteq}$ is a complete lattice, we must show that every subset of $\tau$ has a supremum and an infimum.
Let $S \subseteq \tau$.
By the definition of a topology:
:$\ds \bigcup S \in \tau$
By Union is Smallest Superset, $\ds \bigcup S$ is the supremum of $S$.
Let $I$ be the interior of $\d... | Let $\struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Then $\struct {\tau, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]]. | To show that $\struct {\tau, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]], we must show that every [[Definition:Subset|subset]] of $\tau$ has a [[Definition:Supremum of Set|supremum]] and an [[Definition:Infimum of Set|infimum]].
Let $S \subseteq \tau$.
By the definition of a [[Definition:Topolog... | Topology forms Complete Lattice | https://proofwiki.org/wiki/Topology_forms_Complete_Lattice | https://proofwiki.org/wiki/Topology_forms_Complete_Lattice | [
"Topology",
"Complete Lattices"
] | [
"Definition:Topological Space",
"Definition:Complete Lattice"
] | [
"Definition:Complete Lattice",
"Definition:Subset",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Topology",
"Union is Smallest Superset",
"Definition:Supremum of Set",
"Definition:Interior (Topology)",
"Intersection of Empty Set",
"Definition:Interior (Topology)",
"Inte... |
proofwiki-7779 | Law of Excluded Middle implies Peirce's Law | From the Law of Excluded Middle follows Peirce's Law:
:$\paren {p \lor \neg p} \vdash \paren {\paren {p \implies q} \implies p} \implies p$ | {{BeginTableau|\paren {p \lor \neg p} \vdash \paren {\paren {p \implies q} \implies p} \implies p}}
{{Premise|1|p \lor \neg p}}
{{Assumption|2|p}}
{{SequentIntro|3|2|\paren {\paren {p \implies q} \implies p} \implies p|2|True Statement is implied by Every Statement}}
{{Assumption|4|\neg p}}
{{SequentIntro|5|4|p \implie... | From the [[Law of Excluded Middle]] follows [[Peirce's Law]]:
:$\paren {p \lor \neg p} \vdash \paren {\paren {p \implies q} \implies p} \implies p$ | {{BeginTableau|\paren {p \lor \neg p} \vdash \paren {\paren {p \implies q} \implies p} \implies p}}
{{Premise|1|p \lor \neg p}}
{{Assumption|2|p}}
{{SequentIntro|3|2|\paren {\paren {p \implies q} \implies p} \implies p|2|[[True Statement is implied by Every Statement/Formulation 1|True Statement is implied by Every Sta... | Law of Excluded Middle implies Peirce's Law | https://proofwiki.org/wiki/Law_of_Excluded_Middle_implies_Peirce's_Law | https://proofwiki.org/wiki/Law_of_Excluded_Middle_implies_Peirce's_Law | [
"Peirce's Law",
"Law of Excluded Middle"
] | [
"Law of Excluded Middle",
"Peirce's Law"
] | [
"True Statement is implied by Every Statement/Formulation 1",
"False Statement implies Every Statement/Formulation 1"
] |
proofwiki-7780 | Inverse of Composite Bijection/Proof 2 | Let $f$ and $g$ be bijections.
Then:
:$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$
and $f^{-1} \circ g^{-1}$ is itself a bijection. | Let $f: X \to Y$ and $g: Y \to Z$ be bijections.
Then:
{{begin-eqn}}
{{eqn | l = \paren {g \circ f} \circ \paren {f^{-1} \circ g^{-1} }
| r = g \circ \paren {\paren {f \circ f^{-1} } \circ g^{-1} }
| c = Composition of Mappings is Associative
}}
{{eqn | r = g \circ \paren {I_Y \circ g^{-1} }
| c = Com... | Let $f$ and $g$ be [[Definition:Bijection|bijections]].
Then:
:$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$
and $f^{-1} \circ g^{-1}$ is itself a [[Definition:Bijection|bijection]]. | Let $f: X \to Y$ and $g: Y \to Z$ be [[Definition:Bijection|bijections]].
Then:
{{begin-eqn}}
{{eqn | l = \paren {g \circ f} \circ \paren {f^{-1} \circ g^{-1} }
| r = g \circ \paren {\paren {f \circ f^{-1} } \circ g^{-1} }
| c = [[Composition of Mappings is Associative]]
}}
{{eqn | r = g \circ \paren {I_Y... | Inverse of Composite Bijection/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Composite_Bijection/Proof_2 | https://proofwiki.org/wiki/Inverse_of_Composite_Bijection/Proof_2 | [
"Inverse of Composite Bijection"
] | [
"Definition:Bijection",
"Definition:Bijection"
] | [
"Definition:Bijection",
"Composition of Mappings is Associative",
"Composite of Bijection with Inverse is Identity Mapping",
"Identity Mapping is Left Identity",
"Composite of Bijection with Inverse is Identity Mapping",
"Composition of Mappings is Associative",
"Composite of Bijection with Inverse is I... |
proofwiki-7781 | Equivalence of Definitions of Set Partition | Let $S$ be a set
{{TFAE|def = Set Partition}} | === Definition 1 implies Definition 2 ===
Let $\Bbb S$ be a set of subsets $\Bbb S$ of $S$ such that:
:$(1): \quad$ $\Bbb S$ is pairwise disjoint: $\forall S_1, S_2 \in \Bbb S: S_1 \cap S_2 = \O$ when $S_1 \ne S_2$
:$(2): \quad$ The union of $\Bbb S$ forms the whole set $S$: $\ds \bigcup \Bbb S = S$
:$(3): \quad$ None ... | Let $S$ be a [[Definition:Set|set]]
{{TFAE|def = Set Partition}} | === [[Definition:Set Partition/Definition 1|Definition 1]] implies [[Definition:Set Partition/Definition 2|Definition 2]] ===
Let $\Bbb S$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] $\Bbb S$ of $S$ such that:
:$(1): \quad$ $\Bbb S$ is [[Definition:Pairwise Disjoint|pairwise disjoint]]: $\forall S_1,... | Equivalence of Definitions of Set Partition | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Set_Partition | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Set_Partition | [
"Set Partitions"
] | [
"Definition:Set"
] | [
"Definition:Set Partition/Definition 1",
"Definition:Set Partition/Definition 2",
"Definition:Set",
"Definition:Subset",
"Definition:Pairwise Disjoint",
"Definition:Set Union/Set of Sets",
"Definition:Element",
"Definition:Empty Set",
"Definition:Non-Empty Set",
"Definition:Set Intersection",
"P... |
proofwiki-7782 | Sierpiński's Theorem | Let $\struct {S, \tau}$ be a compact connected Hausdorff space.
Let $\set {F_n: n \in \N}$ be a pairwise disjoint closed cover of $S$.
Then $F_n = S$ for some $n \in \N$. | {{ProofWanted}}
{{Namedfor|Wacław Franciszek Sierpiński|cat = Sierpiński}}
Category:Sierpiński's Theorem
Category:Compact Topological Spaces
Category:Connected Topological Spaces
Category:Hausdorff Spaces
3kz794mirt45g3uftjah3hiam1yrdsn | Let $\struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact]] [[Definition:Connected Topological Space|connected]] [[Definition:Hausdorff Space|Hausdorff space]].
Let $\set {F_n: n \in \N}$ be a [[Definition:Pairwise Disjoint|pairwise disjoint]] closed [[Definition:Cover of Set|cover]] of $S$.
Then $... | {{ProofWanted}}
{{Namedfor|Wacław Franciszek Sierpiński|cat = Sierpiński}}
[[Category:Sierpiński's Theorem]]
[[Category:Compact Topological Spaces]]
[[Category:Connected Topological Spaces]]
[[Category:Hausdorff Spaces]]
3kz794mirt45g3uftjah3hiam1yrdsn | Sierpiński's Theorem | https://proofwiki.org/wiki/Sierpiński's_Theorem | https://proofwiki.org/wiki/Sierpiński's_Theorem | [
"Sierpiński's Theorem",
"Compact Topological Spaces",
"Connected Topological Spaces",
"Hausdorff Spaces"
] | [
"Definition:Compact Topological Space",
"Definition:Connected Topological Space",
"Definition:T2 Space",
"Definition:Pairwise Disjoint",
"Definition:Cover of Set"
] | [
"Category:Sierpiński's Theorem",
"Category:Compact Topological Spaces",
"Category:Connected Topological Spaces",
"Category:Hausdorff Spaces"
] |
proofwiki-7783 | Sierpiński's Theorem/Lemma 1 | Let $\struct {S, \tau}$ be a compact connected Hausdorff space.
Let $A$ be a closed, non-empty proper subset of $S$.
Let $C$ be a component of $A$.
Then:
:$C \cap \partial A \ne \O$
where $\partial A$ denotes the boundary of $A$. | Let $p \in C$.
Let $\VV$ be the set of all subsets of $A$ containing $p$ that are clopen relative to $A$..
By Quasicomponents and Components are Equal in Compact Hausdorff Space and Quasicomponent is Intersection of Clopen Sets:
:$C$ is the intersection of $\VV$.
{{AimForCont}}:
:$C \cap \partial A = \O$
By Boundary of... | Let $\struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact]] [[Definition:Connected Topological Space|connected]] [[Definition:Hausdorff Space|Hausdorff space]].
Let $A$ be a [[Definition:Closed Set (Topology)|closed]], [[Definition:Non-Empty Set|non-empty]] [[Definition:Proper Subset|proper subset]] ... | Let $p \in C$.
Let $\VV$ be the [[Definition:Set of Sets|set]] of all [[Definition:Subset|subsets]] of $A$ containing $p$ that are [[Definition:Clopen Set|clopen]] relative to $A$..
By [[Quasicomponents and Components are Equal in Compact Hausdorff Space]] and [[Quasicomponent is Intersection of Clopen Sets]]:
:$C$ i... | Sierpiński's Theorem/Lemma 1 | https://proofwiki.org/wiki/Sierpiński's_Theorem/Lemma_1 | https://proofwiki.org/wiki/Sierpiński's_Theorem/Lemma_1 | [
"Sierpiński's Theorem"
] | [
"Definition:Compact Topological Space",
"Definition:Connected Topological Space",
"Definition:T2 Space",
"Definition:Closed Set/Topology",
"Definition:Non-Empty Set",
"Definition:Proper Subset",
"Definition:Component (Topology)",
"Definition:Boundary (Topology)"
] | [
"Definition:Set of Sets",
"Definition:Subset",
"Definition:Clopen Set",
"Quasicomponents and Components are Equal in Compact Hausdorff Space",
"Quasicomponent is Intersection of Clopen Sets",
"Definition:Set Intersection/Set of Sets",
"Boundary of Set is Closed",
"Definition:Closed Set/Topology",
"C... |
proofwiki-7784 | Quasicomponent of Compact Hausdorff Space is Connected | Let $\struct {X, \tau}$ be a compact Hausdorff space.
Let $C$ be a quasicomponent of $\struct {X, \tau}$.
Then $C$ is connected. | Let $p \in C$.
{{AimForCont}} $C$ is not connected.
Therefore, by definition of connected, there exist disjoint closed sets $A, B$ of $\struct {X, \tau}$ such that $C = A \cup B$.
By Compact Hausdorff Space is $T_4$, there exist disjoint open sets $U, V$ of $\struct {X, \tau}$ such that $U \supseteq A$ and $V \supseteq... | Let $\struct {X, \tau}$ be a [[Definition:Compact Topological Space|compact]] [[Definition:Hausdorff Space|Hausdorff space]].
Let $C$ be a [[Definition:Quasicomponent|quasicomponent]] of $\struct {X, \tau}$.
Then $C$ is [[Definition:Connected Set (Topology)|connected]]. | Let $p \in C$.
{{AimForCont}} $C$ is not [[Definition:Connected Set (Topology)|connected]].
Therefore, by definition of [[Definition:Connected Set (Topology)|connected]], there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]] $A, B$ of $\struct {X, \tau}$ such that $C = A \... | Quasicomponent of Compact Hausdorff Space is Connected | https://proofwiki.org/wiki/Quasicomponent_of_Compact_Hausdorff_Space_is_Connected | https://proofwiki.org/wiki/Quasicomponent_of_Compact_Hausdorff_Space_is_Connected | [
"Compact Topological Spaces",
"Connectedness Between Two Points"
] | [
"Definition:Compact Topological Space",
"Definition:T2 Space",
"Definition:Quasicomponent",
"Definition:Connected Set (Topology)"
] | [
"Definition:Connected Set (Topology)",
"Definition:Connected Set (Topology)",
"Definition:Disjoint Sets",
"Definition:Closed Set/Topology",
"Compact Hausdorff Space is T4",
"Definition:Disjoint Sets",
"Definition:Open Set/Topology",
"Quasicomponent is Intersection of Clopen Sets",
"Definition:Set In... |
proofwiki-7785 | Quasicomponents and Components are Equal in Compact Hausdorff Space | Let $T = \struct {S, \tau}$ be a compact Hausdorff space.
Then for each $A \subseteq S$:
$A$ is a component of $S$ {{iff}} $A$ is a quasicomponent of $S$. | {{ProofWanted|Follows from Quasicomponent of Compact Hausdorff Space is Connected}}
Category:Components (Topology)
Category:Quasicomponents
Category:Hausdorff Spaces
Category:Compact Topological Spaces
cos06oigo8cvp9miduovwh82etykfqa | Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact]] [[Definition:Hausdorff Space|Hausdorff space]].
Then for each $A \subseteq S$:
$A$ is a [[Definition:Component (Topology)|component]] of $S$ {{iff}} $A$ is a [[Definition:Quasicomponent|quasicomponent]] of $S$. | {{ProofWanted|Follows from [[Quasicomponent of Compact Hausdorff Space is Connected]]}}
[[Category:Components (Topology)]]
[[Category:Quasicomponents]]
[[Category:Hausdorff Spaces]]
[[Category:Compact Topological Spaces]]
cos06oigo8cvp9miduovwh82etykfqa | Quasicomponents and Components are Equal in Compact Hausdorff Space | https://proofwiki.org/wiki/Quasicomponents_and_Components_are_Equal_in_Compact_Hausdorff_Space | https://proofwiki.org/wiki/Quasicomponents_and_Components_are_Equal_in_Compact_Hausdorff_Space | [
"Components (Topology)",
"Quasicomponents",
"Hausdorff Spaces",
"Compact Topological Spaces"
] | [
"Definition:Compact Topological Space",
"Definition:T2 Space",
"Definition:Component (Topology)",
"Definition:Quasicomponent"
] | [
"Quasicomponent of Compact Hausdorff Space is Connected",
"Category:Components (Topology)",
"Category:Quasicomponents",
"Category:Hausdorff Spaces",
"Category:Compact Topological Spaces"
] |
proofwiki-7786 | Derived Set in T1 Space is Closed | Let $\struct {X, \tau}$ be a $T_1$ space.
Let $S \subseteq X$.
Let $S'$ be the derived set of $S$.
Then $S'$ is closed. | Let $x \in S' '$.
Let $U$ be an open neighborhood of $x$.
Then by the definition of derived set, $U$ contains an element $y$ of $S'$ such that $x \ne y$.
Then $U \setminus \set x$ is an open neighborhood of $y$ by the definition of a $T_1$ space.
The definition of derived set is applied once more to see that $U \setmin... | Let $\struct {X, \tau}$ be a [[Definition:T1 Space|$T_1$ space]].
Let $S \subseteq X$.
Let $S'$ be the [[Definition:Derived Set|derived set]] of $S$.
Then $S'$ is [[Definition:Closed Set (Topology)|closed]]. | Let $x \in S' '$.
Let $U$ be an [[Definition:Open Neighborhood of Point|open neighborhood]] of $x$.
Then by the definition of [[Definition:Derived Set|derived set]], $U$ contains an element $y$ of $S'$ such that $x \ne y$.
Then $U \setminus \set x$ is an [[Definition:Open Neighborhood of Point|open neighborhood]] of... | Derived Set in T1 Space is Closed | https://proofwiki.org/wiki/Derived_Set_in_T1_Space_is_Closed | https://proofwiki.org/wiki/Derived_Set_in_T1_Space_is_Closed | [
"Topology"
] | [
"Definition:T1 Space",
"Definition:Derived Set",
"Definition:Closed Set/Topology"
] | [
"Definition:Open Neighborhood/Point",
"Definition:Derived Set",
"Definition:Open Neighborhood/Point",
"Definition:T1 Space",
"Definition:Derived Set",
"Definition:Open Neighborhood/Point",
"Definition:Subset",
"Definition:Closed Set/Topology",
"Category:Topology"
] |
proofwiki-7787 | Equivalence of Axiom Schemata for Groups/Warning | Suppose we build an algebraic structure with the following axioms:
{{begin-axiom}}
{{axiom | n = 0
| lc= Closure Axiom
| q = \forall a, b \in G
| ml= a \circ b
| mo= \in
| mr= G
}}
{{axiom | n = 1
| lc= Associativity Axiom
| q = \forall a, b, c \in G
| ml=... | Let $\struct {S, \circ}$ be the algebraic structure defined as:
:$\forall x, y \in S: x \circ y = x$
That is, $\circ$ is the left operation.
From Element under Left Operation is Right Identity, every element serves as a right identity.
Then given any $a \in S$, we have that:
:$x \circ a = x$
As $x$ is an identity, axio... | Suppose we build an [[Definition:Algebraic Structure with One Operation|algebraic structure]] with the following axioms:
{{begin-axiom}}
{{axiom | n = 0
| lc= [[Definition:Closed Algebraic Structure|Closure Axiom]]
| q = \forall a, b \in G
| ml= a \circ b
| mo= \in
| mr= G
}}
{{... | Let $\struct {S, \circ}$ be the [[Definition:Algebraic Structure with One Operation|algebraic structure]] defined as:
:$\forall x, y \in S: x \circ y = x$
That is, $\circ$ is the [[Definition:Left Operation|left operation]].
From [[Element under Left Operation is Right Identity]], every element serves as a [[Definitio... | Equivalence of Axiom Schemata for Groups/Warning | https://proofwiki.org/wiki/Equivalence_of_Axiom_Schemata_for_Groups/Warning | https://proofwiki.org/wiki/Equivalence_of_Axiom_Schemata_for_Groups/Warning | [
"Group Theory"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Associative Operation",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Group"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Left Operation",
"Element under Left Operation is Right Identity",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"More than one Right Identity then no Left Identity",
"Defi... |
proofwiki-7788 | Multiplicative Inverse in Monoid of Integers Modulo m | Let $\struct {\Z_m, \times_m}$ be the multiplicative monoid of integers modulo $m$.
Then:
:$\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$
{{iff}}:
:$k \perp m$ | First, suppose $k \perp m$.
That is:
:$\gcd \set {k, m} = 1$
By Bézout's Identity:
:$\exists u, v \in \Z: u k + v m = 1$
Thus:
:$\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$
Thus:
:$\eqclass u m$ is an inverse of $\eqclass k m$
Suppose that:
:$\exists u \in \Z: \eqclass u m \eqc... | Let $\struct {\Z_m, \times_m}$ be the [[Definition:Multiplicative Monoid of Integers Modulo m|multiplicative monoid of integers modulo $m$]].
Then:
:$\eqclass k m \in \Z_m$ has an [[Definition:Inverse Element|inverse]] in $\struct {\Z_m, \times_m}$
{{iff}}:
:$k \perp m$ | First, suppose $k \perp m$.
That is:
:$\gcd \set {k, m} = 1$
By [[Bézout's Identity]]:
:$\exists u, v \in \Z: u k + v m = 1$
Thus:
:$\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$
Thus:
:$\eqclass u m$ is an [[Definition:Inverse Element|inverse]] of $\eqclass k m$
Suppose th... | Multiplicative Inverse in Monoid of Integers Modulo m | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Monoid_of_Integers_Modulo_m | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Monoid_of_Integers_Modulo_m | [
"Modulo Arithmetic"
] | [
"Definition:Multiplicative Monoid of Integers Modulo m",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Bézout's Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Bézout's Identity"
] |
proofwiki-7789 | Element Commutes with Square in Group | Let $\left({G, \circ}\right)$ be a group.
Let $x \in G$.
Then $x$ commutes with $x \circ x$. | {{begin-eqn}}
{{eqn | l = x \circ \paren {x \circ x}
| r = \paren {x \circ x} \circ x
| c = {{Group-axiom|1}}
}}
{{end-eqn}}
{{qed}} | Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
Let $x \in G$.
Then $x$ [[Definition:Commute|commutes]] with $x \circ x$. | {{begin-eqn}}
{{eqn | l = x \circ \paren {x \circ x}
| r = \paren {x \circ x} \circ x
| c = {{Group-axiom|1}}
}}
{{end-eqn}}
{{qed}} | Element Commutes with Square in Group/Proof 1 | https://proofwiki.org/wiki/Element_Commutes_with_Square_in_Group | https://proofwiki.org/wiki/Element_Commutes_with_Square_in_Group/Proof_1 | [
"Group Theory",
"Commutativity",
"Element Commutes with Square in Group"
] | [
"Definition:Group",
"Definition:Commutative/Elements"
] | [] |
proofwiki-7790 | Element Commutes with Square in Group | Let $\left({G, \circ}\right)$ be a group.
Let $x \in G$.
Then $x$ commutes with $x \circ x$. | By definition, a group is also a semigroup.
Thus the result Element Commutes with Square in Semigroup can be applied.
{{qed}} | Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
Let $x \in G$.
Then $x$ [[Definition:Commute|commutes]] with $x \circ x$. | By definition, a [[Definition:Group|group]] is also a [[Definition:Semigroup|semigroup]].
Thus the result [[Element Commutes with Square in Semigroup]] can be applied.
{{qed}} | Element Commutes with Square in Group/Proof 2 | https://proofwiki.org/wiki/Element_Commutes_with_Square_in_Group | https://proofwiki.org/wiki/Element_Commutes_with_Square_in_Group/Proof_2 | [
"Group Theory",
"Commutativity",
"Element Commutes with Square in Group"
] | [
"Definition:Group",
"Definition:Commutative/Elements"
] | [
"Definition:Group",
"Definition:Semigroup",
"Element Commutes with Square in Semigroup"
] |
proofwiki-7791 | Element Commutes with Square in Semigroup | Let $\struct {S, \circ}$ be a semigroup.
Let $x \in S$.
Then $x$ commutes with $x \circ x$. | {{Semigroup-axiom|0}} is taken for granted.
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = x \circ \paren {x \circ x}
| r = \paren {x \circ x}\circ x
| c = {{Semigroup-axiom|1}}
}}
{{end-eqn}}
{{qed}}
Category:Semigroups
f4pvi4mb2pc2hp7dhf8j8zjwdrqo6wu | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $x \in S$.
Then $x$ [[Definition:Commute|commutes]] with $x \circ x$. | {{Semigroup-axiom|0}} is taken for granted.
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = x \circ \paren {x \circ x}
| r = \paren {x \circ x}\circ x
| c = {{Semigroup-axiom|1}}
}}
{{end-eqn}}
{{qed}}
[[Category:Semigroups]]
f4pvi4mb2pc2hp7dhf8j8zjwdrqo6wu | Element Commutes with Square in Semigroup | https://proofwiki.org/wiki/Element_Commutes_with_Square_in_Semigroup | https://proofwiki.org/wiki/Element_Commutes_with_Square_in_Semigroup | [
"Semigroups"
] | [
"Definition:Semigroup",
"Definition:Commutative/Elements"
] | [
"Category:Semigroups"
] |
proofwiki-7792 | Sufficient Condition for Vector Equals Inverse iff Zero | Let $\struct {\mathbf V, +, \circ}_\GF$ be a vector space over a field $\GF$, as defined by the vector space axioms.
Let $\GF$ be infinite.
Then:
:$\forall \mathbf v, -\mathbf v \in \mathbf V: \mathbf v = - \mathbf v \iff \mathbf v = \mathbf 0$ | === Necessary Condition ===
{{begin-eqn}}
{{eqn | l = \mathbf v
| r = \mathbf 0
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = -1_\GF \circ \mathbf v
| r = -1_\GF \circ \mathbf 0
| c = scaling both sides by the negative of the unity of $\GF$
}}
{{eqn | ll= \leadsto
| l = -\mathbf... | Let $\struct {\mathbf V, +, \circ}_\GF$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $\GF$, as defined by the [[Axiom:Vector Space Axioms|vector space axioms]].
Let $\GF$ be [[Definition:Infinite Set|infinite]].
Then:
:$\forall \mathbf v, -\mathbf v \in \mathbf V... | === Necessary Condition ===
{{begin-eqn}}
{{eqn | l = \mathbf v
| r = \mathbf 0
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = -1_\GF \circ \mathbf v
| r = -1_\GF \circ \mathbf 0
| c = scaling both sides by the [[Definition:Field Negative|negative]] of the [[Definition:Unity of Field|... | Sufficient Condition for Vector Equals Inverse iff Zero | https://proofwiki.org/wiki/Sufficient_Condition_for_Vector_Equals_Inverse_iff_Zero | https://proofwiki.org/wiki/Sufficient_Condition_for_Vector_Equals_Inverse_iff_Zero | [
"Vector Algebra"
] | [
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Axiom:Vector Space Axioms",
"Definition:Infinite Set"
] | [
"Definition:Field Negative",
"Definition:Multiplicative Identity",
"Vector Inverse is Negative Vector",
"Zero Vector Scaled is Zero Vector",
"Vector Inverse is Negative Vector"
] |
proofwiki-7793 | Subsets of Disjoint Sets are Disjoint | Let $S$ and $T$ be disjoint sets.
Let $S' \subseteq S$ and $T' \subseteq T$.
Then $S'$ and $T'$ are disjoint. | Let $S \cap T = \O$.
Let $S' \subseteq S$ and $T' \subseteq T$.
{{AimForCont}} $S' \cap T' \ne \O$.
Then:
{{begin-eqn}}
{{eqn | l = \exists x
| o = \in
| r = S' \cap T'
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S'
| c = {{Defof|Set Intersection}}
}}
{{eqn | lo= \l... | Let $S$ and $T$ be [[Definition:Disjoint Sets|disjoint sets]].
Let $S' \subseteq S$ and $T' \subseteq T$.
Then $S'$ and $T'$ are [[Definition:Disjoint Sets|disjoint]]. | Let $S \cap T = \O$.
Let $S' \subseteq S$ and $T' \subseteq T$.
{{AimForCont}} $S' \cap T' \ne \O$.
Then:
{{begin-eqn}}
{{eqn | l = \exists x
| o = \in
| r = S' \cap T'
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S'
| c = {{Defof|Set Intersection}}
}}
{{eqn | ... | Subsets of Disjoint Sets are Disjoint | https://proofwiki.org/wiki/Subsets_of_Disjoint_Sets_are_Disjoint | https://proofwiki.org/wiki/Subsets_of_Disjoint_Sets_are_Disjoint | [
"Subsets",
"Disjoint Sets"
] | [
"Definition:Disjoint Sets",
"Definition:Disjoint Sets"
] | [
"Proof by Contradiction",
"Definition:Disjoint Sets"
] |
proofwiki-7794 | Logarithmic Integral as Non-Convergent Series | The logarithmic integral can be defined in terms of a non-convergent series.
That is:
:$\ds \map \li z = \sum_{i \mathop = 0}^{+\infty} \frac {i! \, z} {\ln^{i + 1} z} = \frac z {\ln z} \paren {\sum_{i \mathop = 0}^{+\infty} \frac {i!} {\ln^i z} }$ | From the definition of the logarithmic integral:
:$\ds \map \li z = \int_0^z \frac {\d t} {\ln t}$
Using Integration by Parts:
{{begin-eqn}}
{{eqn | l = \map \li z
| r = \intlimits {\frac t {\ln t} } 0 z - \int_0^t t \map \rd {\ln^{-1} t}
}}
{{eqn | r = \frac z {\ln z} + \int_0^z \frac {\d t} {\ln^2 t}
| c ... | The [[Definition:Logarithmic Integral|logarithmic integral]] can be defined in terms of a non-convergent series.
That is:
:$\ds \map \li z = \sum_{i \mathop = 0}^{+\infty} \frac {i! \, z} {\ln^{i + 1} z} = \frac z {\ln z} \paren {\sum_{i \mathop = 0}^{+\infty} \frac {i!} {\ln^i z} }$ | From the definition of the [[Definition:Logarithmic Integral|logarithmic integral]]:
:$\ds \map \li z = \int_0^z \frac {\d t} {\ln t}$
Using [[Integration by Parts]]:
{{begin-eqn}}
{{eqn | l = \map \li z
| r = \intlimits {\frac t {\ln t} } 0 z - \int_0^t t \map \rd {\ln^{-1} t}
}}
{{eqn | r = \frac z {\ln z} + \... | Logarithmic Integral as Non-Convergent Series | https://proofwiki.org/wiki/Logarithmic_Integral_as_Non-Convergent_Series | https://proofwiki.org/wiki/Logarithmic_Integral_as_Non-Convergent_Series | [
"Logarithmic Integral"
] | [
"Definition:Logarithmic Integral"
] | [
"Definition:Logarithmic Integral",
"Integration by Parts",
"Derivative of Function to Power of Function",
"Integration by Parts",
"Integration by Parts",
"Derivative of Function to Power of Function",
"Derivative of Natural Logarithm Function",
"Category:Logarithmic Integral"
] |
proofwiki-7795 | Network with Positive Integer Mapping is Multigraph | Let $N = \struct {V, E, w}$ be a network whose weights are all strictly positive integers.
Then $N$ can be represented as a multigraph.
Conversely, any multigraph can be expressed as a network whose weights are all strictly positive integers. | === Network as Multigraph ===
Let $N = \struct {V, E, w}$ be a network, either directed or undirected.
{{WLOG}}, suppose $N$ is undirected, and in the following argument allow the term edge to include arcs.
Let $e \in E$ be an edge of the underlying graph $\struct {V, E}$ of $N$.
Let $e = u v$, where $u, v \in V$ are v... | Let $N = \struct {V, E, w}$ be a [[Definition:Network (Graph Theory)|network]] whose [[Definition:Weight (Network Theory)|weights]] are all [[Definition:Strictly Positive Integer|strictly positive integers]].
Then $N$ can be represented as a [[Definition:Multigraph|multigraph]].
Conversely, any [[Definition:Multigr... | === Network as Multigraph ===
Let $N = \struct {V, E, w}$ be a [[Definition:Network (Graph Theory)|network]], either [[Definition:Directed Network|directed]] or [[Definition:Undirected Network|undirected]].
{{WLOG}}, suppose $N$ is [[Definition:Undirected Network|undirected]], and in the following argument allow the ... | Network with Positive Integer Mapping is Multigraph | https://proofwiki.org/wiki/Network_with_Positive_Integer_Mapping_is_Multigraph | https://proofwiki.org/wiki/Network_with_Positive_Integer_Mapping_is_Multigraph | [
"Network with Positive Integer Mapping is Multigraph",
"Network Theory",
"Graph Theory"
] | [
"Definition:Network (Graph Theory)",
"Definition:Network (Graph Theory)/Weight",
"Definition:Strictly Positive/Integer",
"Definition:Multigraph",
"Definition:Multigraph",
"Definition:Network (Graph Theory)",
"Definition:Network (Graph Theory)/Weight",
"Definition:Strictly Positive/Integer"
] | [
"Definition:Network (Graph Theory)",
"Definition:Network (Graph Theory)/Directed",
"Definition:Network (Graph Theory)/Undirected",
"Definition:Network (Graph Theory)/Undirected",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Digraph/Arc",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Under... |
proofwiki-7796 | Complete Graph is Regular | Let $K_p$ be the complete graph of order $p$.
Then $K_p$ is $p-1$-regular. | By definition of complete graph, $K_p$ has $p$ vertices.
Also by definition of complete graph, each vertex of $K_p$ is adjacent to all the other $p - 1$ vertices of $K_p$.
As $K_p$ is a simple graph, there can be only one edge joining any pair of vertices of $K_p$.
So each vertex of $K_p$ has $p - 1$ edges to which it ... | Let $K_p$ be the [[Definition:Complete Graph|complete graph]] of [[Definition:Order of Graph|order]] $p$.
Then $K_p$ is [[Definition:Regular Graph|$p-1$-regular]]. | By definition of [[Definition:Complete Graph|complete graph]], $K_p$ has $p$ [[Definition:Vertex of Graph|vertices]].
Also by definition of [[Definition:Complete Graph|complete graph]], each [[Definition:Vertex of Graph|vertex]] of $K_p$ is [[Definition:Adjacent Vertices (Undirected Graph)|adjacent]] to all the other ... | Complete Graph is Regular | https://proofwiki.org/wiki/Complete_Graph_is_Regular | https://proofwiki.org/wiki/Complete_Graph_is_Regular | [
"Complete Graph is Regular",
"Complete Graphs",
"Regular Graphs"
] | [
"Definition:Complete Graph",
"Definition:Graph (Graph Theory)/Order",
"Definition:Regular Graph"
] | [
"Definition:Complete Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Complete Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Adjacent (Graph Theory)/Vertices/Undirected Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Simple Graph",
"Definition:Unique",
"Def... |
proofwiki-7797 | No Simple Graph is Perfect | Let $G$ be a simple graph whose order is $2$ or greater.
Then $G$ is not perfect. | Recall that a perfect graph is one where each vertex is of different degree.
We note in passing that the simple graph consisting of one vertex trivially fulfils the condition for perfection.
{{AimForCont}} $G$ is a simple graph of order $n$ where $n \ge 2$ such that $G$ is perfect.
First, suppose that $G$ has no isolat... | Let $G$ be a [[Definition:Simple Graph|simple graph]] whose [[Definition:Order of Graph|order]] is $2$ or greater.
Then $G$ is not [[Definition:Perfect Graph|perfect]]. | Recall that a [[Definition:Perfect Graph|perfect graph]] is one where each [[Definition:Vertex of Graph|vertex]] is of different [[Definition:Degree of Vertex|degree]].
We note in passing that the [[Definition:Simple Graph|simple graph]] consisting of one [[Definition:Vertex of Graph|vertex]] trivially fulfils the con... | No Simple Graph is Perfect | https://proofwiki.org/wiki/No_Simple_Graph_is_Perfect | https://proofwiki.org/wiki/No_Simple_Graph_is_Perfect | [
"Simple Graphs",
"Perfect Graphs"
] | [
"Definition:Simple Graph",
"Definition:Graph (Graph Theory)/Order",
"Definition:Perfect Graph"
] | [
"Definition:Perfect Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Degree of Vertex",
"Definition:Simple Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Perfect Graph",
"Definition:Simple Graph",
"Definition:Graph (Graph Theory)/Order",
"Definition:Perfect Graph",
"Def... |
proofwiki-7798 | Continuum Property implies Well-Ordering Principle | The Continuum Property of the positive real numbers $\R_{\ge 0}$ implies the Well-Ordering Principle of the natural numbers $\N$. | Suppose that the set of positive real numbers $\R_{\ge 0}$ has the Continuum Property.
{{AimForCont}} $T \subseteq \N$ is a subset of $\N$ which has no smallest element.
Then:
:$\forall t \in T: \exists u \in T: u < t$
Let $A \in \R_{\ge 0}$.
For every $t \in T$, let $a_t = \dfrac A {2^t}$.
We have that $t \in \N$, so ... | The [[Continuum Property]] of the [[Definition:Positive Real Number|positive real numbers]] $\R_{\ge 0}$ implies the [[Well-Ordering Principle]] of the [[Definition:Natural Numbers|natural numbers]] $\N$. | Suppose that the [[Definition:Set|set]] of [[Definition:Positive Real Number|positive real numbers]] $\R_{\ge 0}$ has the [[Continuum Property]].
{{AimForCont}} $T \subseteq \N$ is a [[Definition:Subset|subset]] of $\N$ which has no [[Definition:Smallest Element|smallest element]].
Then:
:$\forall t \in T: \exists u... | Continuum Property implies Well-Ordering Principle | https://proofwiki.org/wiki/Continuum_Property_implies_Well-Ordering_Principle | https://proofwiki.org/wiki/Continuum_Property_implies_Well-Ordering_Principle | [
"Number Theory",
"Natural Numbers",
"Real Numbers"
] | [
"Continuum Property",
"Definition:Positive/Real Number",
"Well-Ordering Principle",
"Definition:Natural Numbers"
] | [
"Definition:Set",
"Definition:Positive/Real Number",
"Continuum Property",
"Definition:Subset",
"Definition:Smallest Element",
"Definition:Upper Bound of Set",
"Continuum Property",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Contradiction",
"Definition:Supremum of... |
proofwiki-7799 | Circuit of Simple Graph has Three Edges or More | Let $G$ be a simple graph.
Let $C$ be a circuit in $G$.
Then $C$ has at least $3$ edges. | By definition, a circuit contains at least one edge.
That is, a single vertex does not comprise a trivial degenerate circuit with no edges.
Suppose $C$ contains only one edge $e$.
As $G$ is a simple graph it contains no loops.
Therefore $e$ starts and ends at two distinct vertices.
Therefore $C$, consisting entirely of... | Let $G$ be a [[Definition:Simple Graph|simple graph]].
Let $C$ be a [[Definition:Circuit (Graph Theory)|circuit]] in $G$.
Then $C$ has at least $3$ [[Definition:Edge of Graph|edges]]. | By definition, a [[Definition:Circuit (Graph Theory)|circuit]] contains at least one [[Definition:Edge of Graph|edge]].
That is, a single [[Definition:Vertex of Graph|vertex]] does not comprise a trivial degenerate [[Definition:Circuit (Graph Theory)|circuit]] with no [[Definition:Edge of Graph|edges]].
Suppose $C$ ... | Circuit of Simple Graph has Three Edges or More | https://proofwiki.org/wiki/Circuit_of_Simple_Graph_has_Three_Edges_or_More | https://proofwiki.org/wiki/Circuit_of_Simple_Graph_has_Three_Edges_or_More | [
"Circuits (Graph Theory)",
"Simple Graphs"
] | [
"Definition:Simple Graph",
"Definition:Circuit (Graph Theory)",
"Definition:Graph (Graph Theory)/Edge"
] | [
"Definition:Circuit (Graph Theory)",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Circuit (Graph Theory)",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Simple Graph",
"Definition:Loop (Graph Theory)",
"De... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.