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!

## GR9367 Q62

### Re: GR9367 Q62

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

To see that $$[0,1]$$ is

(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 $$\lbrace 1 \rbrace$$.)

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 $$\lbrace 1 \rbrace$$.)