# compact sets

• Jan 27th 2009, 11:16 AM
pandakrap
compact sets
Can anyone help me to understand why set S =(0,2) for each n in the Natural Numbers is not compact in the most basic way possible?

I know so far that S is compact iff every open cover of S contains a finite subcover.

and i also know that a Set (1/n,3) is a open cover of the set (0,2).

thank you.
• Jan 27th 2009, 11:33 AM
Plato
The collection $\left\{ {\left( {0,2 - \frac{1}{n}} \right):n \in \mathbb{Z}^ + } \right\}$ is an open covering of $(0,2)$.
Can you give us a finite subcover of that cover?
• Jan 27th 2009, 11:37 AM
pandakrap
i don't think you can.. because you have infinite many numbers in that set (0,2-(1/n))

but a cover of that would be (0,1) ?? but this would be infinite?? if i'm not wrong..
• Jan 27th 2009, 11:46 AM
pandakrap
wait i think what i said before is non sense..
couldn't the set (1/n,2), be a subcover of the set you gave??
• Jan 27th 2009, 12:17 PM
Plato
Quote:

Originally Posted by pandakrap
couldn't the set (1/n,2), be a subcover of the set you gave??

The set $\left(\frac{1}{n},2\right)$ is not even in the cover I gave you!
It must be a finite collection of subsets of the given collection.
• Jan 27th 2009, 12:28 PM
pandakrap
i'm sorry.. it's a bit difficult to picture this...

would (0,2-(2/n)) be a subcover of your cover (0,2-(1/n)) ??
• Jan 28th 2009, 10:28 PM
aliceinwonderland
Quote:

Originally Posted by pandakrap
i'm sorry.. it's a bit difficult to picture this...

would (0,2-(2/n)) be a subcover of your cover (0,2-(1/n)) ??

(0, 2-(2/n)) is not an open cover. $O = \{(0, 2-(2/n))\}_{n=2}^{\infty}$ is an open cover for (0,2). Same with $O = \{(0, 2-(1/n))\}_{n=2}^{\infty}$.

Even though the above $O$ is an open cover for (0,2), it has no finite subcover for (0,2). That means, n should go to infinity to cover (0,2).

The above one is a counterexample why (0,2) is not compact, because a subspace of a space X is compact if and only if every open cover of A by open sets in X has a finite subcover