* proof:- let & be a connected Subset of a space x that satisfies the Frechet properly. Assuming that E contains more thars One point show that E is an Infinite set. E be a connected set in a space let X Suppose to the contrary that the closure of £, =)E is not connected. then there exist- two sets A and B Such that. E=AUB and AB = = AB E being Connected we know that AUBE so there exist PE ELE we also know that E = EUE! (where E' is the set of the limit points of E must be a limit point of £ (but not in E) Taking the set of all such p. P we obtain the set- E" = { P(PEE", P&E } we find then that: E" = A Or E" = B. Say E" = A Then ECB. and we see that ANB‡Ø which is contradiction. So E must be connected E is an Infinite sef- - so
* proof:- let & be a connected Subset of a space x that satisfies the Frechet properly. Assuming that E contains more thars One point show that E is an Infinite set. E be a connected set in a space let X Suppose to the contrary that the closure of £, =)E is not connected. then there exist- two sets A and B Such that. E=AUB and AB = = AB E being Connected we know that AUBE so there exist PE ELE we also know that E = EUE! (where E' is the set of the limit points of E must be a limit point of £ (but not in E) Taking the set of all such p. P we obtain the set- E" = { P(PEE", P&E } we find then that: E" = A Or E" = B. Say E" = A Then ECB. and we see that ANB‡Ø which is contradiction. So E must be connected E is an Infinite sef- - so
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:proof:-
let
E be a connected Subset of a space x that satisfies
the Frechet properly. Assuming that & contains more than
One point show that E is an Infinite set.
let
E be a connected set in a space X
Suppose to the contrary that the closure of £,
=)E is not connected.
then there exist- two sets A and B Such that.
E=AUB and AOB = =AOB
E
E being connected we know that AUB + £ so there exist
PE ELE we also know that E = EUE!
E)
(where E' is the set of the limit points of E
must be a limit point of £ (but not in E)
=) so p
Taking the set of all such p.
11
we obtain the set £" = { P/PEE", P&E }
we find then that:
E" = A Or E" = B. Say E" = A Then ECB.
and we see that ANB&O which is contradiction.
So E must be connected
=>
E is an Infinite sef-
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