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## #1 2008-07-16 04:01:41

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### 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 by

is 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.

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## #2 2008-07-16 15:01:07

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Pasting lemma

I made a mistake in my proof.

What I have is actually

, not the other way round.

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 (2008-07-16 20:01:23)

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## #3 2008-07-16 20:01:05

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### 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
.

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## #4 2008-07-16 20:39:57

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

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