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Is the following true:

Suppose $(X_1, \tau_1)$ is Hausdorff while $(X_2, \tau_2)\space ...$ are not.

$x, y \in X_1 \space\space U,V\in\tau_1\space x \in U , y \in V \space\space V \cap U = \emptyset $ since $X_1$ is Hausdorff.

Lets try to test whether is $\prod(X_i, \tau_i) = (X, \tau)$ Hausdorff.

let $x, y \in X$ then $ x = (x_1, x_2, ....), y = (y_1, y_2, ....)$ let $ A=U \times X_2 \times ... $ and $ B=V \times X_2 \times ... $ where $U$ and $V$ are two neiborhoods of $x_1$ and $y_1$ shuch that $V \cap U = \emptyset $. Then $A \cap B = \emptyset $ since $ A \cap B = U\cap V \times X_2 \times ...$ So $(X, \tau)$ is Hausdorff since $x, y$ are arbitrary points and $A$ and $B$ are open. Where am I wrong ?

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    $\begingroup$ There are only disjoint neighborhoods of $x_1$ and $y_1$ if $x_1\neq y_1$, and $x\neq y$ does not imply $x_1\neq y_1$. $\endgroup$
    – MaoWao
    Jan 26, 2016 at 9:12

2 Answers 2

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The issue is, that $(x_1,x_2,...)$ and $(y_1,y_2,...)$ for $x_1\neq y_1$ is not general enough to represent every possible point.

Let $\tau_2 = \{X_2,\emptyset\}$ the trivial topology. Consider the points $x=(x_1,a,x_3,x_4,...)$, $y=(x_1,b,x_3,x_4,...)$. Pick any two open sets $A, B\in\tau$ such that $x\in A$, $y \in b$. Projected on their second component they have to be $X_2$. But then $x$ and $y$ lie at their intersection.

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    $\begingroup$ I mean that $x,y \in A \cap B$ So the intersection of $A$ and $B$ is not empty. My example can be improved for clarity. Let $X_i = \{p\}$ for $i\neq 2$ and $X_2= \{a,b\}$ and take the trivial topology on all of them. Now you can easily see that all spaces but $X_2 , \tau_2$ are Hausdorff-spaces (trivially since they contain only one element). But the only open set in the product space is $X_1 \times X_2 ...$ You can then not find disjoint environments for $(p, a, p, p ...)$ and $(p, b, p, p, ...)$. $\endgroup$
    – joedoe8585
    Jan 26, 2016 at 9:21
  • $\begingroup$ ouch! sorry for deleting the comment. which was "what do you mean by lie at their intersection ?". Thanks for clarifying ! $\endgroup$ Jan 26, 2016 at 9:24
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What you proved is essentially that the product of Hausdorff spaces is Hausdorff.

If $x\ne y$ in $X=\prod_i(X_i,\tau_i)$, then, for some $i$, $x_i\ne y_i$; take disjoint open neighborhoods $U_i$ and $V_i$ of $x_i$ and $y_i$ in $X_i$; then $\pi_i^{-1}(U_i)$ is an open set in the product containing $x$, $\pi_i^{-1}(V_i)$ is an open set in the product containing $y$ and they are disjoint.

(With $\pi_i\colon X\to X_i$ I denote the canonical projection.)

Now, suppose $X$ is Hausdorff and that none of the spaces $X_i$ is empty. For each $i$, fix $z_i\in X_i$ (the axiom of choice is necessary in case the index set is infinite, of course). We are thus able to define, for a fixed index $i_0$, a map $f\colon X_{i_0}\to X$ by $$ \pi_i(f(a))= \begin{cases} z_i & \text{if $i\ne i_0$} \\[3px] a & \text{if $i=i_0$} \end{cases} $$ This map is readily seen to be a topological embedding, in the sense $f$ provides a homeomorphism of $X_{i_0}$ with the image of $f$, endowed with the relative topology. Since a subspace of a Hausdorff space is Hausdorff, we get that $X_{i_0}$ is Hausdorff.

Thus we have proved the following result.

Theorem. The product $X=\prod_i(X_i,\tau_i)$ of nonempty spaces $(X_i,\tau_i)$ is Hausdorff if and only if each space $(X_i,\tau_i)$ is Hausdorff.

So no, just one is not enough.

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