Re: Topology with locally connected and expansion.


"William Elliot" <marsh@xxxxxxxxxxxxxxxxxx> wrote in message
news:Pine.BSI.4.58.0711192352400.24780@xxxxxxxxxxxxxxxxxxxx
On Mon, 19 Nov 2007, mina_world wrote:

A, B is open in X.

A \/ B, A /\ B are connected.
(\/ : union, /\ : intersection)

Show that A and B are connected.

This is also true when A,B are both closed.

A, B is closed in X.
A \/ B, A /\ B are locally connected.

Show that A and B are locally connected.

This was discussed awhile ago, most likely here. If not, then at
Ask-a-Topologists. Here's from the terse notes I took. They rely upon
the above theorem and that open subsets of a locally connected space are
locally connected and that a connected component of a locally connected
space is open. (Exercises left for you.) Locally connected A /\ B is
necessary, cf counter example below.

closed A,B, locally connected A /\ B, S = A \/ B
==> A,B locally connected. If x in A.open U /\ A:
case x in A\B: A\B open; some open connected V
with x in V subset A\B /\ U subset A /\ U
case x in A /\ B: some open V with
x in connected V /\ A/\B subset U /\ A/\B
let W = component of V /\ U with x in W; W open
V /\ A/\B subset W subset V /\ U; W = W/\A \/ W/\B
A /\ W, B /\ W closed in connected W
W /\ A/\B = V /\ A/\B connected
x in connected W /\ A subset V/\U /\ A subset U /\ A

Oh, my god...huge proof.
I have a question.
I can't understand that W /\ A is connected at last sentence.

Since A, B are closed, A, B are open. so, W /\ A /\ B is open in A.
so, x in connected W /\ A/\B subset W /\ A subset V/\U /\ A subset U /\ A.

No ?


.

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