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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.

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)

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

.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Pasting lemma