# Prove the product of two Hausdorff spaces is Hausdorff

1. Feb 19, 2017

### sa1988

1. The problem statement, all variables and given/known data

Prove the product of two Hausdorff spaces is Hausdorff

2. Relevant equations

3. The attempt at a solution

For $X$, $Y$ Hausdorff spaces, need to find $s, t \in X \times Y$ where neighbourhoods $S, T$ of $s$ and $t$ are disjoint.

Firstly, by the definition of the Hausdorff property, $\exists \ x_1, x_2 \in X$ whose neighbourhoods $U_1$ and $U_2$ are such that $U_1 \cap U_2 = \emptyset$,

and the same can be said for some $y_1, y_2 \in Y$.

The product $X \times Y$ must therefore contain points $(x_1, y_1), (x_2, y_2)$ made of these elements, whose neighbourhoods are the sets:

$T = \{ (x_1 - \delta , x_1 + \delta), (y_1 - \delta , y_1 + \delta) \}$ with $0 < \delta < min(x_1,y_1)$

and

$S = \{ (x_1 - \epsilon , x_2 + \epsilon), (y_2 - \epsilon, y_2 + \epsilon) \}$ with $0 < \epsilon < min(x_2, y_2)$

Now by original construction, $x_1 \not= x_2$ and $y_1 \not= y_2$, so it can be said that $x_1 < x_2$ and $y_1 < y_2$ without loss of generality.

and by the original condition for the Hausdorff spaces $X$ and $Y$, the points $x_1$, $x_2$, $y_1$, and $y_2$ all have open and separate neighbourhoods in their respective spaces, so there must exist $\delta$ and $\epsilon$ such that

$x_1 + \delta < x_2 - \epsilon$
and
$y_1 + \delta < y_2 - \epsilon$

which means there are two points $t = (x_1, y_1)$ and $s = (x_2, y_2)$ in $X \times Y$ such that

$(x_1 + \delta , y_1 + \delta) < (x_2 - \epsilon , y_2 - \epsilon)$

meaning $t < s \ \forall \ s, t$ under the above construction,

whose neighbourhoods $T$ and $S$ cannot intersect.

Hence the product of two Hausdorff spaces is Hausdorff.

Well... that's my effort. Not sure how solid it is though. Feels like I invoked a circular argument along the way...

Thanks.

2. Feb 19, 2017

### PeroK

A Hausdorff space means every pair of distinct points have disjoint neighbourhoods. And, they are general topological spaces, where you appear to have assumed they are well-ordered metric spaces.

3. Feb 19, 2017

### sa1988

Bummer, yeah I did. I thought Hausdorff meant there was just one or 'some' pairs of distinct points that could be found, rather than every pair being distinct by definition.

I'll have a rethink. Cheers.

4. Feb 20, 2017

### sa1988

Answers have come out for the exercise sheet I was working on.

The proof I was trying to get to is surprisingly simple. Just thought I'd add it here for closure to this thread, and for my own practice. Learning through repetition, heh.

Suppose $X$ and $Y$ Hausdorff spaces.

For $x_1, \ x_2$ distinct in $X$, there exists $U_1, \ U_2 \subset X$ where $U_1 \cap U_2 = \emptyset$

Now make the sets $V_1 = U_1 \times Y$
and $V_2 = U_2 \times Y$

where $V_1, \ V_2$ are any two sets in $X \times Y$ which are distinct by construction via the usage of $x_1$ and $x_2$.

Then $V_1 \cap V_2 = (U_1 \cap U_2) \times Y = \emptyset \times Y = \emptyset$

Hence $X \times Y$ is Hausdorff.