1. ## Compact sets

Find an infinite collection {Sn : n is an element of N} of compact sets R such that the union from n=1 to infinity Sn is compact.

2. Are you sure that you have copied this correctly? As written, it is almost trivial.
$\displaystyle S_n = \left[ {\frac{{ - 1}}{n},\frac{1}{n}} \right],\;\;N = \mathbb{Z}^ +$

3. I had the proper equation all written out but it didn't transfer over to the post, so I wrote it out in words. I'm sure that's how it reads in words.

4. Originally Posted by noles2188
I had the proper equation all written out but it didn't transfer over to the post, so I wrote it out in words. I'm sure that's how it reads in words.
Do you know that my example give you this?
[MATh]\bigcup\limits_{n = 1}^\infty {\left[ {\frac{{ - 1}}{n},\frac{1}{n}} \right]} = \left[ { - 1,1} \right][/tex]

5. In fact, I just read it again and the end should read "not compact" instead of "compact". Sorry for the confusion.

6. Originally Posted by noles2188
In fact, I just read it again and the end should read "not compact" instead of "compact". Sorry for the confusion.
Each singleton set is compact in the standard topology on R.
Take an infinite union of singleton sets, let's say, $\displaystyle \bigcup_{k \in N}\{k\}$ is an infinite union of compact sets, which is a set of natural numbers N.

Take an open interval, let's say, $\displaystyle I_x=(x-1/2, x+1/2), x \in N$. Then $\displaystyle C=\{I_x:x \in N\}$ is an open cover for N, but it has no finite subcover.

You can find other examples in other topological spaces, such as a discrete topology on N.