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
The finite intersection property on a set of subsets says that every finite number of them has non-empty intersection (def. below).
This property is typically used in order to re-formulate the condition on topological space to be compact in terms of closed subsets (prop. below).
The formulation of compactness in terms of the finite intersection property is useful for proving the Tychonoff theorem as well as for proving that compact spaces are equivalently those for which every net has a converging subnet.
(finite intersection property)
Let $X$ be a set. Then a set of subsets $\{S_i \subset X\}_{i \in I}$ is said to satify the finite intersection property if for every finite subset $J \subset I$ the corresponding intersection is inhabited: $\underset{i \in J \subset I}{\cap} S_i \neq \emptyset$.
Let $(X,\tau)$ be a topological space.
Assuming excluded middle, then:
The following are equivalent.
$(X, \tau)$ is a compact topological space in the sense that for every open cover $\{U_i \subset X\}_{i \in I}$ there is a finite subset $J \subset I$ such that $\{U_i \subset X\}_{i \in J \subset I}$ is still a cover.
Every set of closed subsets $\{C_i \subset X\}_{i \in I}$ which satisfies the finite intersection property (def. ) has even non-empty total intersection $\underset{i \in I}{\cap} C_i \neq \emptyset$.
In one direction, assume that $(X,\tau)$ is compact, and that $\{C_i \subset X\}_{i \in I}$ satisfies the finite intersection property. We need to show that then $\underset{i \in I}{\cap} C_i \neq \emptyset$.
Assume that this were not the case, hence assume that $\underset{i \in I}{\cap} C_i = \emptyset$. This would imply that the open complements $U_i \coloneqq X \backslash C_i$ were an open cover of $X$, because (using de Morgan's law)
But then by compactness of $(X,\tau)$ there were a finite subset $J \subset I$ such that $\{ U_i \subset X\}_{i \in J \subset I}$ were still an open cover, hence that $\underset{i \in J \subset I}{\cup} U_i = X$. Translating this back through the de Morgan's law again this would mean that
This would be in contradiction with the finite intersection property of $\{C_i \subset X\}_{i \in I}$, and hence we have proof by contradiction.
Conversely, assume that every set of closed subsets in $X$ with the finite intersection property has non-empty total intersection. We need to show that the every open cover $\{U_i \subset X\}_{i \in I}$ of $X$ has a finite subcover.
Write $C_i \coloneqq X \backslash U_i$ for the closed complements of these open subsets.
Assume that there were no finite subset $J \subset I$ such that $\underset{i \in J \subset I}{\cup} U_i = X$. By de Morgan's law this means equivalently that there were no finite subset $J$ such that $\underset{i \in J \subset I}{\cap} C_i = \emptyset$, hence it would mean that $\{C_i \subset X\}_{i \in I }$ satisfied the finite intersection property.
But by assumption this would then imply that $\underset{i \in I}{\cap} C_i \neq \emptyset$, which, again by de Morgan, would mean that $\underset{i \in I}{\cup} U_i \neq X$. But this contradicts the assumption that the $\{U_i \subset X\}_{i \in I}$ are a cover. Hence we have a proof by contradiction.
Last revised on May 10, 2017 at 09:54:08. See the history of this page for a list of all contributions to it.