Let $(X,\tau)$ be a $T_1$ topological space and $Y\subset X$ a dense subspace which is completely metrizable. Are there any sufficient conditions to ensure that $(X,\tau)$ is Hausdorff using the known facts?

EDIT: Here is an example of such a topological space which isn't Hausdorf, as Valerio asked.

Take $(Y,\|\cdot\|_1)$ and $(Y,\|\cdot\|_2)$ banach spaces such that there exists a sequence $(x_n)_{n\in\mathbb{N}}$ that converges to $x_1$ with respect with $\|\cdot\|_1$ and $x_2$ with respect to $\|\cdot\|_2$ and $x_1\not=x_2$ (This norms exist). Then construct the normed vector space $(Y,\max\{\|\cdot\|_1,\|\cdot\|_2\})$; in this new normed vector space the sequence doesn't converge but is Cauchy, so we complete it to the banach space $(X,\|\cdot\|_3)$. Now construct a new topology $\tau=\{A\ |\ A$ is open in $(X,\|\cdot\|_3) \land A\cap Y$ is open in $(Y,\|\cdot\|_1)\}$. Now $(X,\tau)$ is an example of a topological space like the original problem. It is no Hausdorf because the sequence I define has two different limits, $x_1\in Y$ and $x_3\in X-Y$

I know it is a bit long and I didn't prove all the things I claimed to be true, but this is the exact space I was working on. I'm trying to check what pairs of norms in $Y$ produce that if a sequence converges with respect to both of them, then it converges to the same element in $Y$.

  • $\begingroup$ do you have an example with a non Hausdorff space $X$? $\endgroup$ – Valerio Capraro Aug 8 '11 at 21:43
  • $\begingroup$ Isn't it easier to build an example by doubling a point of your favorite Polish space (e.g., the line with two origins)? $\endgroup$ – Clinton Conley Aug 8 '11 at 23:38
  • $\begingroup$ I used this example because is the space that appeared in my problem. I put it because it may be easier to answer the question for this particular space than the general problem. Like Paul Fabel pointed out, the question isn't easy to answer in a general context. $\endgroup$ – dan232 Aug 8 '11 at 23:49

This answer begins with an easily understood fact and example followed by a more complicated example serving to illustrate why convenient answers to dan232's question can be challenging to find.

FACT: If the 0-dimensional space X is T1, then X is T2.

Pf. Fix distinct points x and y. Since X is T1, X\y is open, and now obtain a clopen set U such that x is in U, and U is a subset of X\y. Note U and X\U are the desired open sets which show X is T2.

Example 0: The indiscrete topology on a two point space shows a 0-dimensional space need not be T2.

Here is a more complicated example which answers Valerio's question, and shows a variety of nice properties can be inadequate to ensure that a T1 space is T2.

Example 1: X is a 1-dimensional space which enjoys the following properties:

Property 1: X is compact and dim(X)=1

Property 2: Every compact subspace of X is a closed subspace of X (and in particular X is T1).

Property 3: There exists a point p in X such that Y=X\p is completely metrizable, (and in particular Y is open and dense in X).

Property 4: X is locally contractible.

Y is the union of countably many rays joined at a common point 0, and X is the one-point-compactification of Y (in the sense of Alexandroff).

To be precise, to obtain Y, consider the subspace of infinite rays emanating from 0 and passing through the standard `unit basis vectors' e1,e2,.... in the familiar Hilbert space l2 of square summable sequences of real numbers. Notice Y is a closed subspace of l2, and hence Y is completely metrizable. By definition Y will be an open dense subspace of its one point (Alexandroff) compactification.

To obtain X, we create the special point at infinity p, and if U is a subset of Y union {p} such that p is in U, then U is open iff Y\U is a compact subspace of Y.

In particular, since Y is not locally compact at 0, X, the one point compactifcation of Y is not T2. However X enjoys the aformentioned listed properties.


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