# Compact sets

• Mar 2nd 2009, 04:13 PM
noles2188
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.
• Mar 2nd 2009, 04:41 PM
Plato
Are you sure that you have copied this correctly? As written, it is almost trivial.
$S_n = \left[ {\frac{{ - 1}}{n},\frac{1}{n}} \right],\;\;N = \mathbb{Z}^ +$
• Mar 2nd 2009, 05:02 PM
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.
• Mar 2nd 2009, 05:33 PM
Plato
Quote:

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?
$$\bigcup\limits_{n = 1}^\infty {\left[ {\frac{{ - 1}}{n},\frac{1}{n}} \right]} = \left[ { - 1,1} \right]$$
• Mar 2nd 2009, 05:47 PM
noles2188
In fact, I just read it again and the end should read "not compact" instead of "compact". Sorry for the confusion.
• Mar 2nd 2009, 10:02 PM
aliceinwonderland
Quote:

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, $\bigcup_{k \in N}\{k\}$ is an infinite union of compact sets, which is a set of natural numbers N.

Take an open interval, let's say, $I_x=(x-1/2, x+1/2), x \in N$. Then $C=\{I_x:x \in N\}$ 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.