If $X$ is not compact, then it has an infinite countable, closed, discrete subspace?

How can we prove that, basically how can we approach this and also that do we mean by discrete subspace?

A discrete subspace let's say $E\subset X$ is a space that every $A\subset E$ is open in $\mathcal{T}_{E}$ topology.

  • $\begingroup$ You are wrong about discrete subspace. A discrete subspace is a subspace such that the induced topology is the discrete topology. $\endgroup$ Feb 25, 2018 at 14:45
  • $\begingroup$ Oh, you mean $A\in \mathcal{T_{E}}$ if $A=E\cap U$ with $U\in \mathcal{P}(X)$ $\endgroup$ Feb 25, 2018 at 14:48
  • 2
    $\begingroup$ Discrete means every subset of $E$ is open in $E$. $\endgroup$
    – lush
    Feb 25, 2018 at 14:48
  • $\begingroup$ @Jonathan1234 A discrete subspace of $X$ is a subspace $E$ of $X$ such that every subset of $E$ can be expressed as $E\cap O$ for some open subset of $X$. $\endgroup$ Feb 25, 2018 at 14:50

1 Answer 1


If $X$ is not limit point compact it has by definition an infinite set $A$ without a limit point. Taking a countably infinite subset $B \subseteq A$ we have that $B$ is closed (as $B' \subseteq A' = \emptyset$) and the same holds for all its subsets, so $B$ has the discrete topology (in its subspace topology; all its subsets are closed in $X$ hence closed in $B$, so all its subsets are open in $B$ as well).

So the question your asking is implied by not being limit point compact, and is in fact equivalent to it. So, if $X$ is a $T_1$ space, this is equivalent to not being "countably compact" (every countable open cover has a finite subcover).

$X = \omega_1$ in the order topology is an example of a non-compact space that is limit point compact, so it has no countable closed discrete subspace.

For metric spaces the equivalence does hold, but not for general spaces.


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