# infinite discrete subspace

In topological space $(X, ‎\tau )$ every compact subspace of $X$ is closed, so no infinite subspace of $X$ can have the cofinite topology.

Is it right to say:

Each infinite subspace of $X$ contains an infinite discrete subspace.

How can I prove it?

Thank you.

• Isn't $[0,1]\subseteq \Bbb{R}$ a counterexample? It's compact, but the only discrete subspaces are the finite sets. – mathmax Jun 6 '14 at 12:40
• @mathmax No. $\{1/n:n=1,2,\ldots\}$ is discrete. – David Mitra Jun 6 '14 at 12:49
• Leila means: suppose $X$ is a KC-space (every compact subset is closed). Does it follow that every infinite subspace of $X$ contains an infinite discrete subspace? This is well-known to be true for Hausdorff spaces, which are a special class of KC spaces. – Henno Brandsma Jun 6 '14 at 15:20

A nice "folklore" topology theorem that is often useful, see the paper: Minimal Infinite Topological Spaces, John Ginsburg and Bill Sands, The American Mathematical Monthly Vol. 86, No. 7 (Aug. - Sep., 1979), pp. 574-576.

Suppose $X$ is any infinite topological space. Then there exists a countably infinite subspace $A$ of $X$ such that $A$ is homeomorphic to one of the following five spaces:

1. $\mathbb{N}$ in the indiscrete topology.
2. $\mathbb{N}$ in the cofinite topology.
3. $\mathbb{N}$ in the upper topology (all non-trivial open sets are of the form $n^\uparrow = \{m \in \mathbb{N}: m \ge n\}$).
4. $\mathbb{N}$ in the lower topology (all non-trivial open sets are of the form $n^\downarrow = \{m \in \mathbb{N}: m \le n\}$).
5. $\mathbb{N}$ in the discrete topology.

Now, if $X$ is a KC-space (i.e. all compact subsets of $X$ are closed in $X$), then every subspace $Y$ of $X$ is also a KC-space. As the first 4 spaces are all non-KC (in the first 3 spaces all subsets are compact, and in the lower topology exactly all finite sets are compact, but not all of these are closed, e.g. $\{2\}$ is not closed as $3$ is in its closure) this means that every infinite (subset of a ) KC-space contains an infinite discrete subspace.

(Added) Note that this question and its answer give a (detailed explanation of a) direct proof for this. This uses recursion to construct the set, using at every stage that a countably infinite subspace $A$ of $X$ cannot have the cofinite topology. The latter is clear, as I remarked above: suppose $X$ is a KC-space, then $A \subset X$ is also a KC-space (a compact $K \subset A$ is also compact in $X$ so closed in $X$, so $K$ is closed in $A$ as well) and the cofinite countable space is not a KC-space, as all subsets are compact but only the finite subsets are closed (and the whole subspace).