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

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This sounds like homework. What have you tried? Also, unless it is rewritten, this question might be considered too localised and be closed soon. Providing motivation which is of interest to research mathematicians may prevent closure. By the way, closure may be a useful hint. Gerhard "Obscure by Accident, Not Intention" Paseman, 2010.10.13 –  Gerhard Paseman Oct 14 '10 at 5:59

3 Answers 3

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.

Reference to the original paper: Minimal Infinite Topological Spaces, John Ginsburg and Bill Sands, The American Mathematical Monthly Vol. 86, No. 7 (Aug. - Sep., 1979), pp. 574-576. The Ramsey proof I saw somewhere else, though.

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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.

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