locally constant

**Sogan** · Oct 28th 2010, 11:06 PM

Hello,

I want to show the following :
Let U be a open subset of IR^2.
If f is constant on each connected component, then f:U->\IR is locally constant.

My Idea was this:
Let x be a arbitrary point of U. We have to show that there is a (open) nbh. V of x, s.t.
f|V is constant.
We know x is an elm. of a conn. component V' of U and f is constant on V'.

Now i couldn't show that there exist an (open) nbh. V of x, s.t. V is a subset of V'.

Is this the right way? how can i show this property of conn. components.

Thank you for your help

**Plato** · Oct 29th 2010, 02:55 AM

You do have the correct idea. Just recall that a component is a maximally connected subset. So there is no problem getting a neighborhood as a subset.

**Tinyboss** · Oct 29th 2010, 05:49 PM

Are connected components in this space open or closed? Or...?

**HallsofIvy** · Oct 30th 2010, 05:20 AM

Connectedness components in any topological space are both open and closed.

**Drexel28** · Oct 31st 2010, 07:35 PM

Originally Posted by HallsofIvy

Connectedness components in any topological space are both open and closed.

This is not always true. It's true that if $X$ is a topological space and $C\subseteq X$ is a component that $C$ must be closed since $\overline{C}$ is a connected set containing $C$ and so by maximality we may conclude that $C=\overline{C}$ . That said, the following theorem is true

Theorem: Let $X$ be a topological space. Then, every connected component of $X$ is open if and only if $X$ is locally connected.
[

Thread: locally constant

LinkBack

Thread Tools

Display

locally constant

Similar Math Help Forum Discussions

[SOLVED] locally compact spaces

locally nilpotent operator.

locally connected (topology)

Constant plus natural log of constant. Find constant?

locally measurable functions

Search Tags