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!

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- Aug 11th 2011, 04:44 PMdavismjProve 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 AMDrexel28Re: Prove that an extension is a continuous extension
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 by the rule and by defining for where is some sequence in . 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 AMdavismjRe: 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.