#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.
9367
Re: 9367
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.
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.
Re: 9367
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.
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.
Re: 9367
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
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

 Posts: 16
 Joined: Sun Jul 19, 2009 12:29 pm
Re: 9367
You are so smart!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.