$f:M\to N\times N$ is continuous and $\Delta = \{(y,y):y\in N\}\subset N\times N$ then $f^{-1}(N\times N-\Delta)$ is an union of open balls in $M$

Question

I need to show the following:

But I have no idea of which things I must assume. I know that if I can prove that its open, then its automatically an union of open balls. However, I've read here that it suffices to show that $N\times N-\Delta$ is open. Is this it? And why does it follows that the inverse is open?

Related: $X$ is Hausdorff if and only if the diagonal of $X\times X$ is closed — Martin Sleziak, May 08 '16 at 02:06

score 1 · Accepted Answer · edited May 08 '16 at 02:07

1

An equivalent definition of continuity is that the inverse image of every open set is open. In fact, many people would take that as the true definition of continuity!

So if $N\times N-\Delta$ is open then the result follows.

Now it is open if and only if $\Delta$ is closed. The latter is easy to show. Take a sequence in it that converges and show the limit is in it too.

edited May 08 '16 at 02:07

Martin Sleziak

53,687

answered May 08 '16 at 00:34

Mark Joshi

5,604

$f:M\to N\times N$ is continuous and $\Delta = \{(y,y):y\in N\}\subset N\times N$ then $f^{-1}(N\times N-\Delta)$ is an union of open balls in $M$

1 Answers1