Given $X×Y$ hausdorff. Show that $X$ hausdorff.

Assume $x_1≠x_2$ in $X$.
Then $(x_1,y_0)≠(x_2,y_0)$ for some $y_0∈Y$.
Then there exists disjoint open neighborhoods in $X×Y$.
As those neigborhoods are open there exist (disjoint) basis elements of $X×Y$ such that:
$(x_1,y_0)⊂U_1×V_1$ and $(x_2,y_0)⊂U_2×V_2$
where $U_1,U_2$ are open. If I can show that $U_1,U_2$ are disjoint, then I'm done, but I don't see how. Any hints ?

  • 2
    $\begingroup$ Suppose that $z \in U_1 \cap U_2$. What can you say about $(z,y_0)$? By the way, you need to require $Y \neq \varnothing$. $\endgroup$ Commented Mar 22, 2014 at 19:25

1 Answer 1


You need to require that $Y$ be nonempty to draw any conclusions about $X$ from properties of $X\times Y$.

If $Y\neq \varnothing$, it is frequently useful to note that one can embed $X$ as a subspace in the product, for any $y_0 \in Y$, the map

$$\iota \colon X \to X\times Y;\quad \iota(x) = (x,y_0)$$

is continuous, since both of its components are, and its inverse is the restriction of a coordinate projection, hence also continuous. Thus $\iota$ is an embedding.

Hence $X$ has all properties that $X\times Y$ has and that are inherited by subspaces, such as being $T_0$, $T_1$, Hausdorff, $T_3$, $T_{3\frac{1}{2}}$ and more. If $Y$ is a $T_1$ space (more generally, if $Y$ contains a point $y_1$ such that $\{y_1\}$ is closed), then $X$ embeds as a closed subspace into $X\times Y$, and hence also has all properties of $X\times Y$ that are inherited by closed subspaces, such as being $T_4$ or complete (in case of uniform spaces).

Of course we can also see that $X$ must be Hausdorff if $X\times Y$ is and $Y\neq \varnothing$ in the way you went:

$$\varnothing = (U_1 \times V_1) \cap (U_2 \times V_2) = (U_1\cap U_2) \times (V_1 \cap V_2),$$

and since $y_0 \in V_1 \cap V_2$, it follows that $U_1 \cap U_2 = \varnothing$.


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