Statement:
stringlengths
7
24.3k
A net has a trivial limit if and only if every property holds eventually.
A net $f$ converges to $l$ if and only if either $f$ is a constant net or for every open set $S$ containing $l$, $f$ eventually takes values in $S$.
The predicate $P$ holds eventually at $x$ within the union of two sets $s$ and $t$ if and only if it holds eventually at $x$ within $s$ and within $t$.
The limit of a function $f$ at a point $x$ within the union of two sets $s$ and $t$ is the same as the limit of $f$ at $x$ within $s$ and the limit of $f$ at $x$ within $t$.
If $f$ converges to $l$ at $x$ within $S$ and $f$ converges to $l$ at $x$ within $T$, then $f$ converges to $l$ at $x$ within $S \cup T$.
If $f$ converges to $l$ at $x$ within $S$ and $f$ converges to $l$ at $x$ within $T$, and $S \cup T = \mathbb{R}$, then $f$ converges to $l$ at $x$.
If $f$ converges to $l$ at $x$, then $f$ converges to $l$ at $x$ within $S$.
If $x$ is an interior point of $S$, then the statement "$P$ holds eventually at $x$ within $S$" is equivalent to the statement "$P$ holds eventually at $x$".
If $x$ is in the interior of $S$, then the filter at $x$ within $S$ is the same as the filter at $x$.
If $X$ is a function defined on a topological space $T$ and $a$ is a point in $T$, then $X$ converges to $L$ at $a$ within $T$ if and only if for every sequence $S$ in $T$ that converges to $a$, the sequence $X \circ S$ converges to $L$.
If $f$ is monotone and bounded on the right half-open interval $(x, \infty) \cap I$, then $f$ converges to the infimum of its image on $(x, \infty) \cap I$.
A point $x$ is a limit point of a set $S$ if and only if there exists a sequence of points in $S$ that converges to $x$.
The identity function converges to $a$ at $a$ within $s$.
The identity function converges to $a$ at $a$.
The limit of a net at a point is the point itself.
If $x$ is an interior point of $S$, then the limit of $f$ at $x$ within $S$ is the same as the limit of $f$ at $x$.
If $x$ is in the interior of $S$, then the limit of any net converging to $x$ within $S$ is $x$.
A point $l$ is in the closure of a set $S$ if and only if there exists a sequence of points in $S$ that converges to $l$.
A set $S$ is closed if and only if every sequence in $S$ that converges to a limit is in $S$.
If $f$ and $g$ converge to $l$ at $x$ within the closures of $s$ and $t$, respectively, then the function that is $f$ on $s$ and $g$ on $t$ converges to $l$ at $x$ within $s \cup t$.
If $f$ is a continuous function from a compact set $S$ to $\mathbb{R}^n$ such that $f(x) \neq 0$ for all $x \in S$, then there exists a positive real number $d$ such that $d \leq \|f(x)\|$ for all $x \in S$.
If $s$ is a compact set and $t$ is an infinite subset of $s$, then $t$ has a limit point in $s$.
If a sequence converges to a limit $l$ and no term of the sequence is equal to $l$, then the sequence is infinite.
If every infinite subset of $s$ has a limit point in $s$, then $s$ is closed.
The closure of a set $S$ is the union of $S$ and the closure of $S$.
The union of two compact sets is compact.
The union of a finite collection of compact sets is compact.
If $A$ is a finite set and each $B_x$ is compact, then $\bigcup_{x \in A} B_x$ is compact.
The intersection of a closed set and a compact set is compact.
The intersection of two compact sets is compact.
The singleton set $\{a\}$ is compact.
If $s$ is a compact set, then $s \cup \{x\}$ is compact.
Any finite set is compact.
If $s$ is an open set, then $s - \{x\}$ is open.
A point $x$ is in the closure of a set $X$ if and only if for every open set $S$ containing $x$, the intersection of $S$ and $X$ is nonempty.
A set $U$ is compact if and only if for every filter $F$ that is not the empty filter, if $F$ converges to a point in $U$, then $F$ converges to a point in $U$.
If $s$ is a countably compact set, then there exists a finite subcover of any countable open cover of $s$.
If every countable open cover of $s$ has a finite subcover, then $s$ is countably compact.
Any compact space is countably compact.
If $U$ is countably compact and $B$ is a countable basis for the topology of $U$, then $U$ is compact.
Any countably compact set is compact.
A set $U$ is countably compact if and only if it is compact.
If every sequence in $S$ has a convergent subsequence, then $S$ is sequentially compact.
If $S$ is a sequentially compact set and $f$ is a sequence of elements of $S$, then there exists a subsequence $f_{r(n)}$ of $f$ that converges to an element $l \in S$.
If $s$ is a closed set and $f$ is a sequence of points in $s$ that converges to $l$, then $l$ is also in $s$.
If $s$ is a sequentially compact set and $t$ is a closed set, then $s \cap t$ is sequentially compact.
If $s$ is a closed subset of a sequentially compact set $t$, then $s$ is sequentially compact.
If a set $U$ is sequentially compact, then it is countably compact.
If $U$ is a compact set, then $U$ is sequentially compact.
If $s$ is countably compact, $t$ is a countable infinite subset of $s$, then $s$ has an accumulation point in $t$.
If every infinite countable subset of $S$ has an accumulation point in $S$, then $S$ is sequentially compact.
A set $U$ is sequentially compact if and only if it is countably compact.
A set $S$ is sequentially compact if and only if every infinite countable subset of $S$ has an accumulation point in $S$.
A set $U$ is sequentially compact if and only if it is compact.
If every infinite subset of $S$ has a limit point in $S$, then $S$ is sequentially compact.
If $s$ and $t$ are sequentially compact, then $s \times t$ is sequentially compact.
The product of two compact sets is compact.
If $K$ is a compact set and $W$ is an open set containing the set $\{x_0\} \times K$, then there exists an open set $X_0$ containing $x_0$ such that $X_0 \times K \subseteq W$.
Suppose $f$ is a continuous function from a product space $U \times C$ to a metric space $Y$, where $C$ is compact. Then for every $\epsilon > 0$ and every $x_0 \in U$, there exists an open set $X_0$ containing $x_0$ such that for all $x \in X_0 \cap U$ and all $t \in C$, we have $d(f(x, t), f(x_0, t)) \leq \epsilon$.
If $f$ is continuous at $x$, then $f$ is continuous at $x$ within $s$.
If a net converges to a constant, then it converges to any constant.
A function is continuous on a set if and only if it is continuous at every point of that set.
If $f$ is continuous at $x$ within $s$, then $f$ is continuous at $x$ within $t$ for any $t \subseteq s$.
If $f$ is continuous on a set $S$, and $x$ is an interior point of $S$, then $f$ is continuous at $x$.
If $f$ is a continuous function defined on a set $S$ and $f(x) = g(x)$ for all $x \in S$, then $g$ is continuous on $S$.
If $f$ is a function from a first-countable topological space $X$ to a topological space $Y$, and if for every sequence $(x_n)$ in $X$ that converges to $x \in X$ and is contained in $S \subseteq X$, the sequence $(f(x_n))$ converges to $f(x)$, then $f$ is continuous at $x$ within $S$.
If $f$ is continuous at $a$ on $s$, and $x_n \to a$ with $x_n \in s$ for all $n$, then $f(x_n) \to f(a)$.
If $f$ is continuous at $a$ within $s$, and $x_n \to a$ with $x_n \in s$ for all $n$, then $f(x_n) \to f(a)$.
A function $f$ is continuous at $a$ within $S$ if and only if for every sequence $x_n$ in $S$ that converges to $a$, the sequence $f(x_n)$ converges to $f(a)$.
If $f$ is a function from a first-countable space to a topological space, then $f$ is continuous at $a$ if and only if for every sequence $u$ converging to $a$, the sequence $f(u)$ converges to $f(a)$.
A function $f$ is continuous at $a$ if and only if for every sequence $x_n$ converging to $a$, the sequence $f(x_n)$ converges to $f(a)$.
If $f$ is a function from a first-countable space $X$ to a topological space $Y$, and if $f$ is sequentially continuous, then $f$ is continuous.
A function $f$ is continuous on a set $S$ if and only if for every point $a \in S$ and every sequence $(x_n)$ in $S$ that converges to $a$, the sequence $(f(x_n))$ converges to $f(a)$.
A function $f$ is continuous at $x$ if and only if for every open set $t$ containing $f(x)$, there exists an open set $s$ containing $x$ such that $f(s) \subseteq t$.
If $f$ is continuous at $x_0$ and $x \to x_0$, then $f(x) \to f(x_0)$.
If $f$ and $g$ are continuous functions from $S$ to $T$ and from $T$ to $S$, respectively, and $f$ and $g$ are inverses of each other, then $f$ is a homeomorphism from $S$ to $T$.
The map $x \mapsto x + a$ is a homeomorphism from $S + a$ to $S$.
The identity map is a homeomorphism.
If $f$ is a homeomorphism from $S$ to $T$ and $h$ is a homeomorphism from $T$ to $U$, then $h \circ f$ is a homeomorphism from $S$ to $U$.
If $f$ and $g$ are homeomorphisms, then so are $f'$ and $g'$ if $f'$ and $g'$ are equal to $f$ and $g$ on the relevant sets.
The identity map is a homeomorphism from the empty set to itself.
If $f$ is a homeomorphism from $S$ to $t$, then $g$ is a homeomorphism from $t$ to $S$.
A homeomorphism from $S$ to $t$ is the same as a homeomorphism from $t$ to $S$.
The empty set is homeomorphic to itself.
A topological space is homeomorphic to itself.
If $s$ is homeomorphic to $t$, then $t$ is homeomorphic to $s$.
If $S$ is homeomorphic to $T$ and $T$ is homeomorphic to $U$, then $S$ is homeomorphic to $U$.
Two topological spaces $S$ and $T$ are homeomorphic if and only if there exist continuous functions $f: S \to T$ and $g: T \to S$ such that $g \circ f$ is the identity function on $S$ and $f \circ g$ is the identity function on $T$.
If $f$ and $g$ are continuous functions such that $f$ maps $S$ onto $T$ and $g$ maps $T$ onto $S$, and $g$ is the inverse of $f$ and $f$ is the inverse of $g$, then $S$ and $T$ are homeomorphic.
If $f$ is a homeomorphism from $S$ to $T$, and $S'$ and $T'$ are subsets of $S$ and $T$ respectively, such that $f(S') = T'$, then $f$ is a homeomorphism from $S'$ to $T'$.
If $f$ is a homeomorphism from $S$ to $T$ and $x \in S$, then $g(f(x)) = x$.
If $f$ and $g$ are inverse homeomorphisms, then $f(g(x)) = x$.
If $f$ is a homeomorphism from $S$ to $T$, then $f(S) = T$.
If $f$ is a homeomorphism from $S$ to $T$, then $g$ is a homeomorphism from $T$ to $S$.
If $f$ is a homeomorphism from $S$ to $T$, then $f$ is continuous on $S$.
If $f$ is a homeomorphism from $S$ to $T$, then $g$ is continuous on $T$.
If $f$ is a function defined on a set $S$ such that no point of $S$ is a limit point of $S$, then $f$ is continuous on $S$.
If $S$ is a finite set, then any function $f$ defined on $S$ is continuous.
Two finite sets are homeomorphic if and only if they have the same number of elements.
If $f$ is a continuous injective map from a compact space $S$ to a Hausdorff space $T$, then $f$ is a homeomorphism.