# Prove that an extension is a continuous extension

• Aug 11th 2011, 04:44 PM
davismj
Prove that an extension is a continuous extension
Here is my proof so far. I am lost on how to show that g is continuous! :(

http://i53.tinypic.com/20qgw1d.png

thank you so much!
• Aug 12th 2011, 12:00 AM
Drexel28
Re: Prove that an extension is a continuous extension
Quote:

Originally Posted by davismj
Here is my proof so far. I am lost on how to show that g is continuous! :(

http://i53.tinypic.com/20qgw1d.png

thank you so much!

I think you are making this slightly more difficult than it need to be, it's a little hard to follow precisely what's being said. It looks like (and this is the correct approach) that you're defining $\widetilde{f}:X\to \mathbb{R}$ by the rule $\widetilde{f}_E=f$ and by defining $f(\lim x_n)=\lim f(x_n)$ for $\lim x_n\in X-E$ where $x_n$ is some sequence in $E$. Is this correct? If not, could you explain more.

I think perhaps most of your questions could be answered by this blog post of mine.
• Aug 12th 2011, 06:50 AM
davismj
Re: Prove that an extension is a continuous extension
I'm following the hint given in baby Rudin. I do think its way more difficult, but I followed it because I was having trouble showing that the extension g is actually continuous. But that's exactly the idea. In particular, it uses sequences of sets instead of sequences of numbers. Take a set near x with things in E. You can take those sets as small as you want, and the closure of the image of those sets under f is always non-empty, and in fact there is only on element in every such set.

Thus, for x in X we associate the unique element y in R such that y is that unique element. It's pretty simple and straightforward, but the problem is showing that g is continuous. Why is it true that any open set around y is the image of an open set in x? Or why is it that for any e there is a d such that d(v,w) < d implies d(g(v),g(w)) < e.

Your stuff here looks like it answers my question exactly. I'm about to go out but when I get back I'll read through it.