## 9367 Q62

Forum for the GRE subject test in mathematics.
yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

### 9367 Q62

#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

Hi, can anyone give a quick crash course on what it means to be Hausdorff?

Also, isn't [0,1] close and bounded and therefore compact?

Also, isn't [0,1] connected?

Thanks for clearing my deep misunderstandings on topology! :S

Posts: 27
Joined: Sun Oct 17, 2010 4:57 am

### Re: 9367 Q62

You're thinking of the standard topology on R. The problem is giving you a specific topology to consider. Hausdorff means that for every pair of distinct points you can find a pair of non-intersecting open sets, one containing one of the points, one containing the other. The space is neither compact nor connected in this topology but it is Hausdorff. I think it's easy to see that it's not connected. To see that it's not compact, you can think of a typical geometric series (e.g. the open cover {[0, 1/2), [1/2, 3/4), ...} \cup {{1}}).

huayualice
Posts: 15
Joined: Thu Sep 11, 2014 7:19 am

### Re: 9367 Q62

I can't see why X is not connected. The definition says, if there exists disjoint non-empty open sets O1 and O2 such that O1 U O2 = X, then X is "disconnected". What would be the open sets be here? It seems that, if we want to cover the point x = 1, the open set has to end larger than 1, say, open set [a, 1.1). Then the union of such open sets would be larger than X, not equal to X.

Can anyone explain this?

Thanks!

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

### Re: 9367 Q62

In your case, you are viewing $X$ in the subspace topology, so that $X \cap [1,17) = \{1\}$