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If $S$ or $T$ is bounded, then $S \cap T$ is bounded. |
If $S$ is bounded, then $S - T$ is bounded. |
If $f$ and $g$ are bounded functions on a set $S$, then the function $x \mapsto \|f(x) - g(x)\|$ is bounded on $S$. |
If $s$ and $t$ are bounded sets, then $s \times t$ is bounded. |
If $U$ is a compact set, then $U$ is bounded. |
If $S$ is a subset of the closure of $T$, $x \in S$, and $r > 0$, then $T$ intersects the ball of radius $r$ centered at $x$. |
If $S$ is a nonempty compact set, then there exist $x, y \in S$ such that for all $u, v \in S$, we have $d(u, v) \leq d(x, y)$. |
A sequence $S$ is Cauchy if and only if for every $\epsilon > 0$, there exists an $N$ such that for all $m, n \geq N$, we have $|S_m - S_n| < \epsilon$. |
If a set $S$ is sequentially compact, then for every $\epsilon > 0$, there exists a finite set $k$ such that $k \subseteq S$ and $S \subseteq \bigcup_{x \in k} B(x, \epsilon)$. |
If a set $S$ is sequentially compact, then it is compact. |
A metric space is compact if and only if it is sequentially compact. |
A set $S$ is compact if and only if every sequence in $S$ has a convergent subsequence. |
A set $S$ is compact if and only if every infinite subset of $S$ has a limit point in $S$. |
If every infinite subset of $S$ has a limit point in $S$, then $S$ is bounded. |
If $f$ is a contraction mapping on a complete metric space, then $f$ has a unique fixed point. |
If $S$ is a compact set with more than one point, and $g$ is a function from $S$ to $S$ such that $g(x)$ is closer to $g(y)$ than $x$ is to $y$ for all $x$ and $y$ in $S$, then $g$ has a unique fixed point. |
The diameter of the empty set is $0$. |
The diameter of a singleton set is zero. |
If $S$ is a nonempty set of vectors in a normed vector space, and if the distance between any two vectors in $S$ is at most $d$, then the diameter of $S$ is at most $d$. |
If $S$ is a bounded set in a metric space, then the distance between any two points in $S$ is less than or equal to the diameter of $S$. |
If $S$ is a bounded set in a metric space and $d$ is a positive number less than the diameter of $S$, then there exist two points $x$ and $y$ in $S$ such that $d < \text{dist}(x, y)$. |
If $S$ is a bounded set, then for all $x, y \in S$, we have $d(x, y) \leq \text{diameter}(S)$. Moreover, for all $d > 0$, if $d < \text{diameter}(S)$, then there exist $x, y \in S$ such that $d(x, y) > d$. |
A set $S$ is bounded if and only if there exists a real number $e$ such that for all $x, y \in S$, we have $|x - y| \leq e$. |
If $S$ is a nonempty compact set, then there exist $x, y \in S$ such that $d(x, y) = \text{diameter}(S)$. |
If $S$ is a bounded set, then the diameter of $S$ is nonnegative. |
If $S$ is a subset of $T$ and $T$ is bounded, then the diameter of $S$ is less than or equal to the diameter of $T$. |
If $S$ is a bounded set, then the diameter of the closure of $S$ is equal to the diameter of $S$. |
If $S$ is a compact set and $\mathcal{C}$ is a collection of open sets such that $S \subseteq \bigcup \mathcal{C}$, then there exists a positive real number $\delta$ such that for every subset $T$ of $S$ with diameter less than $\delta$, there exists a set $B \in \mathcal{C}$ such that $T \subseteq B$. |
If $S$ is a bounded closed set, then $S$ is sequentially compact. |
A set $S$ is compact if and only if it is bounded and closed. |
If each set in a nonempty collection of sets is compact, then the intersection of the collection is compact. |
A set is compact if and only if its closure is bounded. |
Let $f$ be a sequence of vectors in $\mathbb{R}^n$. For every finite subset $D$ of $\{1, \ldots, n\}$, there exists a subsequence $f_{r(n)}$ of $f$ and a vector $l$ such that for every $\epsilon > 0$, there exists $N$ such that for all $n \geq N$ and all $i \in D$, we have $|f_{r(n)}(i) - l(i)| < \epsilon$. |
If $s$ is a bounded set, then the projection of $s$ onto the first coordinate is also bounded. |
If $s$ is a bounded set, then the set of second coordinates of $s$ is bounded. |
If every Cauchy sequence in $s$ converges to a point in $s$, then $s$ is complete. |
If $s$ is a complete metric space and $(f_n)$ is a Cauchy sequence in $s$, then there exists $l \in s$ such that $f_n \to l$. |
Any compact metric space is complete. |
A set $S$ is compact if and only if it is complete and totally bounded. |
If a sequence is Cauchy, then its range is bounded. |
The set of all elements of a complete space is complete. |
If a metric space is complete, then it is closed. |
If $S$ is a complete metric space and $t$ is a closed subset of $S$, then $S \cap t$ is a complete metric space. |
If $S$ is a closed subset of a complete metric space $t$, then $S$ is complete. |
A set $S$ is complete if and only if it is closed. |
A sequence $S$ converges if and only if it is Cauchy. |
If a sequence converges, then it is bounded. |
If $S$ is a compact set, then the frontier of $S$ is a subset of $S$. |
If $f$ is a continuous function defined on a closed set $S$, and $\sigma$ is a Cauchy sequence in $S$, then $f \circ \sigma$ is a Cauchy sequence. |
If $f$ is a function from a complete metric space to itself such that $f$ is Lipschitz with constant $c < 1$, then $f$ has a unique fixed point. |
If $S$ is a closed set and $\mathcal{F}$ is a finite collection of bounded closed sets, then $S$ intersects the intersection of the sets in $\mathcal{F}$. |
If $S$ is a closed set and each $T \in \mathcal{F}$ is compact, then $S \cap \bigcap \mathcal{F} \neq \emptyset$. |
If $S$ is a closed set, $T$ is a bounded set in a family $\mathcal{F}$ of closed sets, and $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has a nonempty intersection. |
If every finite intersection of compact sets is nonempty, then the infinite intersection of all compact sets is nonempty. |
If a sequence converges to a limit, then the set of all terms of the sequence and the limit is compact. |
The closed ball of radius $e$ around $x$ is compact. |
If $S$ is a bounded set, then the frontier of $S$ is compact. |
If $S$ is a compact set, then its frontier is compact. |
If $s$ is a nonempty compact set, then there exists $x \in s$ such that for all $y \in s$, we have $d(a, y) \leq d(a, x)$. |
If $s$ is a nonempty closed set, then there exists $x \in s$ such that $x$ is the closest point in $s$ to $a$. |
The image of a set under the distance function is bounded below. |
If $A$ is nonempty, then $\infdist(x, A)$ is equal to $\inf_{a \in A} \{ \text{dist}(x, a) \}$. |
The infimum distance from a point $x$ to a set $A$ is nonnegative. |
If $a \in A$, then the infimum of the distances from $x$ to $A$ is less than or equal to the distance from $x$ to $a$. |
If $x$ is within distance $d$ of some point in $A$, then the infimum distance of $x$ to $A$ is at most $d$. |
If $a$ is in $A$, then the infimum of the distance from $a$ to $A$ is $0$. |
If $A$ and $B$ are nonempty sets, then the infimum of the distance from $x$ to $A \cup B$ is the minimum of the infimum of the distance from $x$ to $A$ and the infimum of the distance from $x$ to $B$. |
The infimum distance from a point $x$ to a set $A$ is less than or equal to the sum of the infimum distance from $y$ to $A$ and the distance from $x$ to $y$. |
The difference between the infdist of $x$ and $y$ to a set $A$ is bounded by the distance between $x$ and $y$. |
A point $x$ is in the closure of a set $A$ if and only if the infimum of the distance from $x$ to $A$ is zero. |
If $A$ is a nonempty closed set, then $x \in A$ if and only if $\inf_{y \in A} |x - y| = 0$. |
If $S$ is a nonempty closed set and $x \notin S$, then the infimum of the distance from $x$ to $S$ is positive. |
If $X$ is a nonempty closed set, then there exists $x \in X$ such that $\inf_{y \in X} \|y - x\| = \inf_{y \in X} \|y - z\|$ for all $z \in X$. |
If $f$ converges to $l$, then $\inf_{a \in A} |f(x) - a|$ converges to $\inf_{a \in A} |l - a|$. |
If $f$ is a continuous function from a topological space $F$ to a metric space $X$, then the function $x \mapsto \inf_{y \in A} d(f(x), y)$ is continuous. |
If $f$ is a continuous function defined on a set $S$, then the function $x \mapsto \inf_{y \in A} |f(x) - y|$ is continuous on $S$. |
If $A$ is a nonempty compact set, then the set of points whose distance to $A$ is at most $e$ is compact. |
If $s$ is a closed set and $a \notin s$, then there exists a positive real number $d$ such that for all $x \in s$, we have $d \leq |a - x|$. |
If $s$ is a compact set and $t$ is a closed set with $s \cap t = \emptyset$, then there exists a positive real number $d$ such that for all $x \in s$ and $y \in t$, we have $d \leq \|x - y\|$. |
If $s$ is a closed set and $t$ is a compact set, and $s$ and $t$ are disjoint, then there exists a positive real number $d$ such that for all $x \in s$ and $y \in t$, we have $d \<le> \|x - y\|$. |
If $A$ is a compact set and $B$ is an open set such that $A \subseteq B$, then there exists an $\epsilon > 0$ such that the $\epsilon$-neighborhood of $A$ is contained in $B$. |
If $f$ is uniformly continuous on $S$, then for every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x, x' \in S$, if $|x - x'| < \delta$, then $|f(x) - f(x')| < \epsilon$. |
A function $f$ is uniformly continuous on a set $S$ if and only if for any two sequences $(x_n)$ and $(y_n)$ in $S$ such that $(x_n)$ converges to $x$ and $(y_n)$ converges to $y$, the sequence $(f(x_n))$ converges to $f(x)$ and the sequence $(f(y_n))$ converges to $f(y)$. |
Suppose $S$ is a compact set and $\mathcal{G}$ is a collection of open sets such that $S \subseteq \bigcup \mathcal{G}$. Then there exists an $\epsilon > 0$ such that for every $x \in S$, there exists a set $G \in \mathcal{G}$ such that $B(x, \epsilon) \subseteq G$. |
Suppose $\F$ is a family of functions from a compact set $S$ to a metric space $X$. If $\F$ is pointwise equicontinuous at every point of $S$, then $\F$ is uniformly equicontinuous. |
If $f$ is a continuous function from a compact set $S$ to a metric space, then $f$ is uniformly continuous on $S$. |
A function $f$ is continuous on the closure of a set $S$ if and only if for every $x \in \overline{S}$ and every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $y \in S$ with $|y - x| < \delta$, we have $|f(y) - f(x)| < \epsilon$. |
A function $f$ is continuous on the closure of a set $S$ if and only if for every sequence $x_n$ in $S$ that converges to a point $a$ in the closure of $S$, the sequence $f(x_n)$ converges to $f(a)$. |
If $f$ is uniformly continuous on a set $S$ and continuous on the closure of $S$, then $f$ is uniformly continuous on the closure of $S$. |
If $f$ is uniformly continuous on $X$, then it has a limit at every point of the closure of $X$. |
If $f$ is uniformly continuous on a set $X$, then there exists a uniformly continuous extension of $f$ to the closure of $X$. |
If $f$ is uniformly continuous on a bounded set $S$, then $f(S)$ is bounded. |
A subset $S$ of a topological space $T$ is open in $T$ if and only if $S$ is a subset of $T$ and for every $x \in S$, there exists an open ball around $x$ that is contained in $S$. |
A subset $S$ of a topological space $T$ is open in $T$ if and only if $S$ is a subset of $T$ and for every $x \in S$, there exists an open ball around $x$ that is contained in $S$. |
Suppose $S_n$ is a sequence of nonempty closed sets such that $S_n \subseteq S_m$ whenever $n \leq m$. If $S_0$ is bounded, then there exists a point $a$ such that $a \in S_n$ for all $n$. |
Suppose $S_n$ is a sequence of nonempty closed sets such that $S_n \subseteq S_{n+1}$ for all $n$. Then there exists a point $a$ such that $a \in S_n$ for all $n$. |
Suppose $S_n$ is a sequence of nonempty closed sets such that $S_n \subseteq S_m$ whenever $n \leq m$. If for every $\epsilon > 0$, there exists $n$ such that for all $x, y \in S_n$, we have $|x - y| < \epsilon$, then there exists a point $a$ such that $\bigcap_{n=1}^{\infty} S_n = \{a\}$. |
If $f$ is continuous at $x$ within $s$ and $f(x) \neq a$, then there exists an $\epsilon > 0$ such that for all $y \in s$, if $|x - y| < \epsilon$, then $f(y) \neq a$. |
If $f$ is continuous at $x$ and $f(x) \neq a$, then there exists an $\epsilon > 0$ such that $f(y) \neq a$ for all $y$ with $|x - y| < \epsilon$. |
If $f$ is a continuous function from a metric space to a T1 space, and $f(x) \neq a$, then there exists an open neighborhood of $x$ on which $f$ does not take the value $a$. |
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