Thread: locally constant

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

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.

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

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

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 $\displaystyle X$ is a topological space and $\displaystyle C\subseteq X$ is a component that $\displaystyle C$ must be closed since $\displaystyle \overline{C}$ is a connected set containing $\displaystyle C$ and so by maximality we may conclude that $\displaystyle C=\overline{C}$. That said, the following theorem is true

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