analysis (differential/integral calculus, functional analysis, topology)
metric space, normed vector space
open ball, open subset, neighbourhood
convergence, limit of a sequence
compactness, sequential compactness
continuous metric space valued function on compact metric space is uniformly continuous
…
…
topology (point-set topology, point-free topology)
see also differential topology, algebraic topology, functional analysis and topological homotopy theory
Basic concepts
fiber space, space attachment
Extra stuff, structure, properties
Kolmogorov space, Hausdorff space, regular space, normal space
sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact
Examples
Basic statements
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
open subspaces of compact Hausdorff spaces are locally compact
compact spaces equivalently have converging subnet of every net
continuous metric space valued function on compact metric space is uniformly continuous
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
injective proper maps to locally compact spaces are equivalently the closed embeddings
locally compact and second-countable spaces are sigma-compact
Theorems
Analysis Theorems
For general topological spaces the condition of being compact neither implies nor is implied by being sequentially compact. However for metric spaces the two conditions happen to be equivalent
Using excluded middle and countable choice, then:
If $(X,d)$ is a metric space, regarded as a topological space via its metric topology, then the following are equivalent:
$(X,d)$ is a compact topological space.
$(X,d)$ is a sequentially compact topological space.
Assume first that $(X,d)$ is a compact topological space. Let $(x_k)_{k \in \mathbb{N}}$ be a sequence in $X$. We need to show that it has a sub-sequence which converges.
Consider the topological closures of the sub-sequences that omit the first $n$ elements of the sequence
and write
for their open complements.
Assume now that the intersection of all the $F_n$ were empty
or equivalently that the union of all the $U_n$ were all of $X$
hence that $\{U_n \to X\}_{n \in \mathbb{N}}$ were an open cover. By the assumption that $X$ is compact, this would imply that there is a finite subset $\{i_1 \lt i_2 \lt \cdots \lt i_k\} \subset \mathbb{N}$ with
This in turn would mean that $F_{i_k} = \empty$, which contradicts the construction of $F_{i_k}$. Hence we have a proof by contradiction that assumption $(\ast)$ is wrong, and hence that there must exist an element
By definition of topological closure this means that for all $n$ the open ball $B^\circ_x(1/(n+1))$ around $x$ of radius $1/(n+1)$ must intersect the $n$th of the above subsequence:
Picking one point $(x'_n)$ in the $n$th such intersection for all $n$ hence defines a sub-sequence, which converges to $x$.
This proves that compact implies sequentially compact for metric spaces.
For the converse, assume now that $(X,d)$ is sequentially compact. Let $\{U_i \to X\}_{i \in I}$ be an open cover of $X$. We need to show that there exists a finite sub-cover.
Now by the Lebesgue number lemma, there exists a positive real number $\delta \gt 0$ such that for each $x \in X$ there is $i_x \in I$ such that $B^\circ_x(\delta) \subset U_{i_x}$. Moreover, since sequentially compact metric spaces are totally bounded, there exists then a finite set $S \subset X$ such that
Therefore $\{U_{i_s} \to X\}_{s \in S}$ is a finite sub-cover as required.
In the proof of prop. 1 the implication that a compact topological space is sequentially compact requires less of $(X,d)$ than being a metric space. Actually, the proof works for any first-countable space that is a countably compact space, i. e. any countable open cover admits a finite sub-cover. Hence countably compact metric spaces are equivalently compact metric spaces.