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#1 2008-07-17 02:01:41
- JaneFairfax
- Legendary Member
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Pasting lemma
Let
and be topological spaces and suppose and are subsets of X such that . Let and be continuous functions such that for all . Then defined byis continuous.
Proof:
Let be Y-open and consider .
Then is A-open and is B-open and so there exist X-open sets and such that and .
Hence
Therefore is X-open, showing that is continuous.
#2 2008-07-17 13:01:07
- JaneFairfax
- Legendary Member
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Re: Pasting lemma
I made a mistake in my proof. 
What I have is actually
I need to find an X-open set U such that .
And I may have to assume that and are closed in as well.
Last edited by JaneFairfax (2008-07-17 18:01:23)
#3 2008-07-17 18:01:05
- JaneFairfax
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Re: Pasting lemma
Okay, I think we can say that if
and are either both open or both closed in , then is continuous.If and are both open in , then and are both open in ; hence is open in .
If and are both closed in , we’ll use this result: is continuous if and only if given any closed subset , is closed in .
So if is closed in , is closed in ; where is closed in . Hence is closed in . Similarly (where is closed in ) is closed in . Thus is closed in .