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My textbook says "A Tichonov space is homeomorphic to a subset of a compact Hausdorff space."

Doesn't the subset also have to be compact Hausdorff?

Motivation:- For the subset of the compact Hausdorff space to be homeomorphic to a Tichonov space, it will have to be regular and normal. We know that any compact Hausdorff space is both normal and regular. Is this necessarily true for every subset also?

Thanks for your time!

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No, the subset need not be compact Hausdorff. For a very simple example, note that the Tikhonov space $(0,1)$ is a non-compact subspace of the compact Hausdorff space $[0,1]$.

A subspace of a compact Hausdorff space is compact iff it is closed, so in general you can find lots of non-compact subspaces of a compact Hausdorff space. It is also entirely possible to find non-normal subspaces of compact Hausdorff spaces. For example, let $X=\omega_1+1$ with the order topology, and let $Y=\omega+1$ with the order topology. Both spaces are compact and Hausdorff, so their product is as well. But $(X\times Y)\setminus\{\langle\omega_1,\omega\rangle\}$ is not normal: the closed sets $\omega_1\times\{\omega\}$ and $\{\omega_1\}\times\omega$ cannot be separated by disjoint open sets. For that matter $(X\times X)\setminus\{\langle\omega_1,\omega_1\rangle\}$ is another example: the closed sets $\omega_1\times\{\omega_1\}$ and $\{\langle\alpha,\alpha\rangle:\alpha\in\omega_1\}$ cannot be separated by disjoint open sets.

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Thanks. I think I got it. It is akin to saying every regular normal $T_{1}$ space need not be compact Hausdorff, although the converse is true. – Ayush Khaitan Jun 28 '13 at 7:28
@Ayush: Somewhat akin, but there’s a better way to think about it. Hausdorffness is a hereditary property: if $X$ has it, so does every subspace of $X$. Regularity is also a hereditary property. But compactness and normality are not hereditary. – Brian M. Scott Jun 28 '13 at 7:31
Can I say the subset of the compact hausdorff space has to be regular normal and $T_{1}$? – Ayush Khaitan Jun 28 '13 at 7:40
@Ayush: If $A$ is closed in $X^*$, then $X^*\setminus A$ is open in $X^*$, so there is an open $U$ in $X$ such that $U\cap X^*=X^*\setminus A$. But then $A'=X\setminus U$ is a closed set in $X$ such that $A'\cap X^*=A$. // The argument works for regularity because $x\notin A'$; it fails for normality because it’s possible that $A'\cap B'\ne\varnothing$. (And the proof at that link is excessively clumsy.) – Brian M. Scott Jun 28 '13 at 10:08
@Ayush: Every subspace of $[0,1]$ is a Tikhonov space. Every subspace of any Tikhonov space is Tikhonov. And every Tikhonov space is a subspace of some product of copies of $[0,1]$. – Brian M. Scott Jun 28 '13 at 12:11

No. A trivial counterexample: the open interval $(0,1)$ is Tychonoff (since it is a metric space), and it does embed into a compact Hausdorff spaces - e.g., $[0,1]$ - but regardless of the compact Hausdorff space chosen, and regardless of the embedding chosen, the image of the embedding will not be compact, because it is (by definition of "embedding") homeomorphic to $(0,1)$, and it isn't compact.

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Sorry I seem to have asked two different questions. Are you saying the subset does not have to be compact Hausdorff? – Ayush Khaitan Jun 28 '13 at 7:15
That's right, the image of the embedding need not be compact, since the original space itself need not be compact (of course it must be Hausdorff, as Hausdorff-ness is inherited by subsets). – Zev Chonoles Jun 28 '13 at 7:17

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