- #1

PsychonautQQ

- 784

- 10

## Homework Statement

Show that X is a Hausdorff space IFF the 'diagnol of x' given by t = {(x,x) | X * X} is closed as a subspace of X*X

## Homework Equations

## The Attempt at a Solution

So since X Is Hausdorff so is X*X and t, because the product of two Hausdorff spaces if Hausdoff and the subspace of any Hausdorff space is Hausdorf.

So I've been doing a lot of thinking about this and have some ideas, so here we go.

Since X is Hausdorff, for any x_1 and x_2 in X, there exists open sets (or neighborhoods) U_1 and U_2 that contain x_1 and x_2 respectively and their intersection is the empty set.

To show that t is closed as a subset of X*X, we must show that t contains all it's limit points. It is perhaps worth noting that if A is a subset of a Hausdorff space Y, then b is a limit point of A iff every neighborhood of b intersects A at infinitely many points.

First I will try to show that X being a Hausdorff space implies that t is closed:

I will proceed by contradiction

Suppse X is a Hausdorff space and that t is not closed. Then there exists an element in the closure of t (hereby known as ct) that is not in t, we will call this element y=(x_1,x_2) where x_1 does not equal x_2, for if it did then y would be in ct. Since y is in the closure of t, every neighborhood of y contains some element of t, or said otherwise, if U is a neighborhood of y, then the intersection of U and t is nonempty.

Question: y must be of the form (x_1,x_2), correct? I am confused by that notion because that would mean that y is an element of X*X, which is not necesarily true I don't think, but rather it is only gaurenteed to be on the closure of X*X.

If y must be of the form (x_1,x_2) where each x_i is an element of X then I smell a contradiction lingering closely...

Anyone have any thoughts/concerns?