TITLE: Is there a “universal” Hausdorff space?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.