9367

Forum for the GRE subject test in mathematics.
goodtrain
Posts: 11
Joined: Sat Sep 19, 2009 2:57 pm

9367

Postby goodtrain » Sat Oct 31, 2009 1:45 am

#62
Let R be the set of real numbers with the topology generated by the basis {[a,b): a <b, a,b in R} If X is the subset [0,1] of R, which of the following must be true

1.x is compact
2.x is hausdorff
3.x is connected

(A) I only
(B) II only
(C) III only
(D) I AND II
(E) II and III

answer is B, can someone explain this.

esperant
Posts: 4
Joined: Thu Oct 29, 2009 2:10 pm

Re: 9367

Postby esperant » Sat Oct 31, 2009 2:00 am

Given any two distinct points in X (say 0.5 and 0.7) there is are disjoint open neighbourhoods of the two points (e.g. [.49,.51) and [.69,.71). Therefore II.

Consider the open cover of X given by [0,0.9), [0.9,0.99), [0.99, 0.999), etc and [1,2). This is obviously doesn't have a finite subcover. Therefore not I.

The cover from the above paragraph was disjoint, so X is not connected. Therefore not III.

Therefore B.

goodtrain
Posts: 11
Joined: Sat Sep 19, 2009 2:57 pm

Re: 9367

Postby goodtrain » Sat Oct 31, 2009 2:52 pm

thanks a lot, I don't get the hausdorff part.

Looking at the way you did it the open sets don't have to be in X, just have to be inthe topology generated on X otherwise you can't enclose 0, can you give me an example of a non hausdorff topology.

esperant
Posts: 4
Joined: Thu Oct 29, 2009 2:10 pm

Re: 9367

Postby esperant » Sat Oct 31, 2009 6:54 pm

X is Hausdorff because, given any two distinct points p and q in X we can find an open set [a,b) that contains p and an open set [c,d) that contains q such that [a,b) and [c,d) are disjoint.

The sets in the open cover only have to be 'open in X'. So the set [1] is open in X because [1]=[1,2) intersect X. But [1] is NOT open in R. etc etc etc

breezeintopl
Posts: 16
Joined: Sun Jul 19, 2009 12:29 pm

Re: 9367

Postby breezeintopl » Sun Nov 01, 2009 12:08 pm

esperant wrote:Given any two distinct points in X (say 0.5 and 0.7) there is are disjoint open neighbourhoods of the two points (e.g. [.49,.51) and [.69,.71). Therefore II.

Consider the open cover of X given by [0,0.9), [0.9,0.99), [0.99, 0.999), etc and [1,2). This is obviously doesn't have a finite subcover. Therefore not I.

The cover from the above paragraph was disjoint, so X is not connected. Therefore not III.

Therefore B.


You are so smart!




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