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
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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
A topological space is called countably compact if every open cover consisting of a countable set of open subsets (every countable cover) admits a finite subcover, hence if there is a finite subset of the open in the original cover which still cover the space.
A compact space is a fortiori countably compact.
A sequentially compact space is countably compact.
The long line is an example of a countably compact space that is not compact.
A countably compact space is a limit point compact space.
Recall that a space is limit point compact if every closed discrete subspace is finite; equivalently, if every countable closed discrete subspace is finite.
Suppose $X$ is countably compact and $A$ is a countable closed discrete subspace. For each $a \in A$, choose an open neighborhood $U_a$ such that $U_a \cap A = \{a\}$, and let $V_a$ be the open subset $U_a \cup \neg A$. Clearly we still have $V_a \cap A = \{a\}$. Also, $\{V_a: a \in A\}$ is a countable cover of $X$, hence admits a finite subcover $V_{a_1}, \ldots, V_{a_n}$. But then
as was to be shown.
A space that is $T_1$ and limit point compact is countably compact.
Consider any countable open cover $U_1, U_2, \ldots$ of $X$. Put $V_n = U_1 \cup \ldots \cup U_n$ so that $V_1 \subseteq V_2 \subseteq \ldots$, and discard repetitions, i.e., consider the maximal chain of strict inclusions
It suffices to show this maximal chain is finite. Rename it as $\emptyset = W_0 \subset W_1 \subset W_2 \subset \ldots$.
For each $W_n$ occurring in this chain ($n \geq 1$), pick a point $x_n \in W_n \setminus W_{n-1}$. Observe that if $m \lt n$, then $x_n \notin W_m$.
Since $X$ is $T_1$ (points are closed), the set $W_m \cap \neg \{x_1, \ldots, x_{m-1}\}$ is an open neighborhood of $x_m$ that does not contain $x_n$ whenever $n \gt m$, and does not contain $x_n$ whenever $n \lt m$. Thus every point $x_m$ is open relative to $A = \{x_1, x_2, \ldots\}$, i.e., $A$ is a discrete subspace.
Finally, any point $x \notin A$ belongs to some $W_n$, and then $W_n \cap \neg \{x_1, \ldots x_n\}$ is an open neighborhood of $x$ that doesn’t intersect $A$. Thus $A$ is a closed discrete subspace, and is finite by limit point compactness. Therefore the maximal chain consisting of the sets $W_n$ is finite, as was to be shown.
If $X$ is countably compact and $f \colon X \to Y$ is a map in Top, then the direct image $f(X)$ is a countably compact subset of $Y$.
The proof is just like the proof of the corresponding statement where the word “countably” is dropped. Suppose $\{U_n: n \in \mathbb{N}\}$ is a countable cover of $f(X)$, so that $f(X) \subseteq \bigcup_{n \in \mathbb{N}} U_n$. This is equivalent to
Thus $\{f^{-1}(U_n): n \in \mathbb{N}\}$ is a countable open cover of $X$, so by countable compactness there is a finite subcover $f^{-1}(U_{n_1}), \ldots, f^{-1}(U_{n_k})$ of $X$:
This is equivalent to $f(X) \subseteq \bigcup_{i=1}^k U_{n_i}$, as required.
Last revised on May 30, 2023 at 01:48:55. See the history of this page for a list of all contributions to it.