## GR9367 Q62

Forum for the GRE subject test in mathematics.
huayualice
Posts: 15
Joined: Thu Sep 11, 2014 7:19 am

### GR9367 Q62

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

I. X is compact.
II. X is Hausdorff.
III. X is connected.

I don't get I . Isn't [0,1] closed and bounded so compact? For closedness: Since its complement (-infinity, 0) U (1,+infinity) is open in standard topology R, so it is also open in lower limit topology R_L, therefore itself, [0,1], is closed; and [0,1] is certainly bounded. What did I miss?

Thanks!

Austin
Posts: 26
Joined: Fri Jun 01, 2012 9:32 am

### Re: GR9367 Q62

You're using the Heine-Borel theorem, which assumes the standard topology on $\mathbb{R}$; that is, the topology generated by $\left\lbrace (a,b) : a.

To see that $[0,1]$ is not compact in our new topology (which is often called the lower-limit topology), we must find an open cover of $[0,1]$ which has no finite refinement. I believe the cover $\left\lbrace \left[\frac{n}{n+1},\frac{n+1}{n+2}\right) : n=0,1,2,\ldots\right\rbrace \cup \lbrace [1,e^2+\pi) \rbrace$ will do the trick.

(For what it's worth, the right endpoint of the interval $[1,e^2+\pi)$ really doesn't matter, so long as it's greater than 1. We just need an open set whose intersection with $[0,1]$ is $\lbrace 1 \rbrace$.)

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

Thanks a lot!