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Question. Is there a Hausdorff space $ X $ such that every Hausdorff space $ Y $ can be embedded into $ X^\Lambda $ (given the product topology) for some sets $ \Lambda $?

Concerning Tychonoff spaces, $ X = [0, 1] $ satisfies the condition; i.e., every Tychonoff space can be embedded into some powers of $ [0, 1] $. I wonder if there is a counterpart for Hausdorff spaces.

I tried to disprove the existence by proving that for every Hausdorff space $ X $ there exists a non-trivial Hausdorff space $ Y $ such that there are only a few continuous maps from $ Y $ into $ X $, but I got stuck.

o-ccah
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  • Does every space $Y$ have a different power $\Lambda(Y)$? – Keen-ameteur Jun 04 '20 at 14:21
  • For any space $X$ there is a universal hausdorff space $X_\text{Haus}$ in the sense that every continuous map $X\rightarrow Y$ into a hausdorff space $Y$ factors uniquely via $X_\text{Haus}$. But this doesn’t seem to be the kind of universal object you are asking for... – Jonas Linssen Jun 04 '20 at 14:26
  • @Keen-ameteur Yes, $ Λ $ can depend on $ Y $. – o-ccah Jun 04 '20 at 14:47
  • Every subspace of a completely regular space is also completely regular, so non-completely regular spaces cannot be embedded into a product of intervals. – Hanul Jeon Jun 04 '20 at 16:56
  • I think the answer should be no based on the fact that there is no Hausdorff space $Y$ such that for every Hausdorff space $X$ the points of $X$ can be separated by maps into $Y$, so for any fixed Hausdorff $Y$ and $\lambda$ there is an Hausdorff space $X$ such that there no continuous injections $X\to Y^\lambda$, but I'm not seeing how to reach a clear contradiction – Alessandro Codenotti Jun 04 '20 at 17:22
  • @AlessandroCodenotti How can I prove your former statement that for each Hausdorff $Y$ there is a Hausdorff space $X$, whose points are not separated by maps into $Y$? – Hanul Jeon Jun 04 '20 at 17:47
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    @HanulJeon I learned that fact from this question a while ago, there is a proof in the answer – Alessandro Codenotti Jun 04 '20 at 17:57
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    See also my answer at https://math.stackexchange.com/a/1950171/86856 which is answering a rather different question but implies a negative answer to your question (at most one of the two spaces $X$ and $X'$ can embed in a power of $Y$, since they have the same "universal" map to a power of $Y$). – Eric Wofsey Jun 04 '20 at 22:46

1 Answers1

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In this nice answer the following fact is shown:

Given any Hausdorff space $X$, there exists a Hausdorff $Y$ and distinct $a,b\in Y$ such that each continuous $f\colon Y\to X$ satisfies $f(a)=f(b)$.

So if such a universal space $X$ existed, find $Y$ as promised by this fact.

Being universal, there must be an embedding $e: Y \to X^I$ for some index set $I$. But with $a,b$ from the fcat we see right away that for each $i \in I$, $\pi_i(e(a)) = \pi_i(e(b))$ (as $\pi_i \circ e$ is continuous from $Y$ to $X$) and thus $e(a)= e(b)$ and this contradicts the injectivity of $e$.

So there can be no universal Hausdorff space.

Henno Brandsma
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    For $T_1$ (aka Fréchet) spaces no such space can exist as well, I sketch a construction from paper by Herrlich here. $T_0$ spaces have Sierpiński space as a universal power space, and like the OP said $[0,1]$ works for completely regular $T_1$ spaces. The classes inbetween have not. – Henno Brandsma Jun 04 '20 at 22:38
  • Ah of course, I should have seen this argument after remembering the fact, very neat – Alessandro Codenotti Jun 05 '20 at 07:42