# X is an infinite set, then exists some subspace homeomorphic to $(\Bbb N,\tau)$ where $\tau$ is indiscrete or $T_0$

My attempt to prove the statement on the title: the basis for some topology on $X$ can be finite or infinite.

• If the basis is finite then exist some countable infinite subset that belongs only to an unique open set, so the induced topology is homeomorphic to $(\Bbb N,\tau)$ with $\tau$ indiscrete.

• If the basis is infinite then exist some countable infinite subset where each element belongs to a different open set of the basis so the induced subspace is homeomorphic to $(\Bbb N,\tau)$ with $\tau$ being $T_0$.

My questions:

1. It is my proof correct? It lacks something?

2. There is a different way to prove this? Can you outline to me some different strategy?

3. If my proof is fine, there is a way to express better the obvious statement that, for example, if the basis is finite and X is infinite then some infinite subspace is indiscrete?

1) Yes.

2) Let $X$ hasn't countable infinite subset such induced topology is indiscrete.

Let $A$ be set of maximal indiscrete subspaces.

If $|A|$ is finite then induced topology for some countable infinite subset of $X/ \cup_{A^* \in A} A^*$ is homeomorphic to $(\Bbb N,\tau)$ with $\tau$ being $T_0$.

If $|A|$ is infinite then induced topology for some countable infinite subset of $\cup_{A^* \in A} \{$some $a \in A^*\}$ is homeomorphic to $(\Bbb N,\tau)$ with $\tau$ being $T_0$.

3) I think not.

• For 1) your "yes" is referred to what, I lack something or it is ok? – Masacroso Dec 19 '15 at 11:54
• "Yes" is refered to "ok". – Alfred XXX Dec 19 '15 at 12:34