nLab countably compact topological space





topology (point-set topology, point-free topology)

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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 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 XX is countably compact and AA is a countable closed discrete subspace. For each aAa \in A, choose an open neighborhood U aU_a such that U aA={a}U_a \cap A = \{a\}, and let V aV_a be the open subset U a¬AU_a \cup \neg A. Clearly we still have V aA={a}V_a \cap A = \{a\}. Also, {V a:aA}\{V_a: a \in A\} is a countable cover of XX, hence admits a finite subcover V a 1,,V a nV_{a_1}, \ldots, V_{a_n}. But then

A=( i=1 nV a i)A= i=1 n(V a iA)={a 1,,a n}A = \left(\bigcup_{i=1}^n V_{a_i} \right) \cap A = \bigcup_{i=1}^n (V_{a_i} \cap A) = \{a_1, \ldots, a_n\}

as was to be shown.


A space that is T 1T_1 and limit point compact is countably compact.


Consider any countable open cover U 1,U 2,U_1, U_2, \ldots of XX. Put V n=U 1U nV_n = U_1 \cup \ldots \cup U_n so that V 1V 2V_1 \subseteq V_2 \subseteq \ldots, and discard repetitions, i.e., consider the maximal chain of strict inclusions

=V 0V n 1V n 2\emptyset = V_0 \subset V_{n_1} \subset V_{n_2} \subset \ldots

It suffices to show this maximal chain is finite. Rename it as =W 0W 1W 2\emptyset = W_0 \subset W_1 \subset W_2 \subset \ldots.

For each W nW_n occurring in this chain (n1n \geq 1), pick a point x nW nW n1x_n \in W_n \setminus W_{n-1}. Observe that if m<nm \lt n, then x nW mx_n \notin W_m.

Since XX is T 1T_1 (points are closed), the set W m¬{x 1,,x m1}W_m \cap \neg \{x_1, \ldots, x_{m-1}\} is an open neighborhood of x mx_m that does not contain x nx_n whenever n>mn \gt m, and does not contain x nx_n whenever n<mn \lt m. Thus every point x mx_m is open relative to A={x 1,x 2,}A = \{x_1, x_2, \ldots\}, i.e., AA is a discrete subspace.

Finally, any point xAx \notin A belongs to some W nW_n, and then W n¬{x 1,x n}W_n \cap \neg \{x_1, \ldots x_n\} is an open neighborhood of xx that doesn’t intersect AA. Thus AA is a closed discrete subspace, and is finite by limit point compactness. Therefore the maximal chain consisting of the sets W nW_n is finite, as was to be shown.


If XX is countably compact and f:XYf \colon X \to Y is a map in Top, then the direct image f(X)f(X) is a countably compact subset of YY.


The proof is just like the proof of the corresponding statement where the word “countably” is dropped. Suppose {U n:n}\{U_n: n \in \mathbb{N}\} is a countable cover of f(X)f(X), so that f(X) nU nf(X) \subseteq \bigcup_{n \in \mathbb{N}} U_n. This is equivalent to

Xf 1( nU n)= nf 1(U n).X \subseteq f^{-1}\left(\bigcup_{n \in \mathbb{N}} U_n\right) = \bigcup_{n \in \mathbb{N}} f^{-1}(U_n).

Thus {f 1(U n):n}\{f^{-1}(U_n): n \in \mathbb{N}\} is a countable open cover of XX, so by countable compactness there is a finite subcover f 1(U n 1),,f 1(U n k)f^{-1}(U_{n_1}), \ldots, f^{-1}(U_{n_k}) of XX:

X i=1 kf 1(U n i)=f 1( i=1 kU n i).X \subseteq \bigcup_{i=1}^k f^{-1}(U_{n_i}) = f^{-1}\left(\bigcup_{i=1}^k U_{n_i}\right).

This is equivalent to f(X) i=1 kU n if(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.