Question

does every infinite hausdorff space contains a countable infinite discrete subspace?

Answer 1

In a more general light:

folklore theorem: Every infinite topological space contains a homeomorphic copy of one (or more) of the following 5 spaces:

1. $\mathbf{N}$ in the indiscrete topology (only $\mathbf{N}$ and $\emptyset$ are open).
2. $\mathbf{N}$ in the co-finite topology (only $\mathbf{N}$ and all finite sets are closed).
3. $\mathbf{N}$ in the upper topology (the empty set and all sets $U(k) = \{ n \in \mathbf{N} : n \ge k \}$, $k \in \mathbf{N}$, are open).
4. $\mathbf{N}$ in the lower topology ($\mathbf{N}$, $\emptyset$, and all sets $L(k) = \{ n \in \mathbf{N} : n \le k \}$, $k \in \mathbf{N}$, are open).
5. $\mathbf{N}$ in the discrete topology (all subsets are open).

As each of the spaces has the property that every infinite subspace of it is homeomorphic to the whole space, this list is minimal.

And spaces 1-4 are not Hausdorff, which implies what you need, as being Hausdorff is hereditary.

The nicest proof of this I know uses Ramsey's theorem (off hand I do not know a reference, who does?) using a partition of the triples or pairs of X, IIRC.

Answer 2

Not sure whether this is "research-level", but: first show that any infinite Hausdorff space has a proper infinite closed subset. Then proceed by induction.

Answer 3

Yes. Lemma 1 in http://www.emis.de/journals/HOA/IJMMS/6/197.pdf