# Finite union of compact sets is compact

• Apr 8th 2011, 09:47 AM
Sheld
Finite union of compact sets is compact
I think i have the proof, however I am worried about some assertions.

Pf: Let \$\displaystyle K = \cup_{i=1}^n K_i\$ where \$\displaystyle K_i \$ is compact \$\displaystyle \forall i
\$
Let \$\displaystyle C\$ be an open cover of \$\displaystyle K\$

\$\displaystyle C\$ is also an open cover for \$\displaystyle K_1,K_2,...,K_n
\$
Since \$\displaystyle K_i \$ is compact \$\displaystyle \forall i\$, \$\displaystyle \exists\$ a finite subcover \$\displaystyle C_i \: \forall i\$

\$\displaystyle \cup_{i=1}^n C_i\$ is a finite subcover for \$\displaystyle K\$

I am not sure if I can just say that an open cover of K exists. It makes sense, to me anyway, that I could cover any set with a bunch of open sets. Still, I am not sure if an open cover of K exists.

And the union of the finite subcovers remains finite?

Thanks for helping.
• Apr 8th 2011, 09:57 AM
Plato
Quote:

Originally Posted by Sheld
Pf: Let \$\displaystyle K = \cup_{i=1}^n K_i\$ where \$\displaystyle K_i \$ is compact \$\displaystyle \forall i\$
Let \$\displaystyle C\$ be an open cover of \$\displaystyle K\$
\$\displaystyle C\$ is also an open cover for \$\displaystyle K_1,K_2,...,K_n\$
Since \$\displaystyle K_i \$ is compact \$\displaystyle \forall i\$, \$\displaystyle \exists\$ a finite subcover \$\displaystyle C_i \: \forall i\$
\$\displaystyle \cup_{i=1}^n C_i\$ is a finite subcover for \$\displaystyle K\$
I am not sure if I can just say that an open cover of K exists. It makes sense, to me anyway, that I could cover any set with a bunch of open sets. Still, I am not sure if an open cover of K exists. And the union of the finite subcovers remains finite?

I think that I would point out that any open covering of \$\displaystyle K\$ must also by definition be an open covering for each \$\displaystyle K_i\$. So in each case there is a finite subcover, as you said. Also point out that the finite union of finite collections of sets is also a finite collect of sets.
• Apr 8th 2011, 12:20 PM
Sheld
• Apr 8th 2011, 07:43 PM
hatsoff
Quote:

Originally Posted by Sheld
I think i have the proof, however I am worried about some assertions.

Pf: Let \$\displaystyle K = \cup_{i=1}^n K_i\$ where \$\displaystyle K_i \$ is compact \$\displaystyle \forall i
\$
Let \$\displaystyle C\$ be an open cover of \$\displaystyle K\$

\$\displaystyle C\$ is also an open cover for \$\displaystyle K_1,K_2,...,K_n
\$
Since \$\displaystyle K_i \$ is compact \$\displaystyle \forall i\$, \$\displaystyle \exists\$ a finite subcover \$\displaystyle C_i \: \forall i\$

\$\displaystyle \cup_{i=1}^n C_i\$ is a finite subcover for \$\displaystyle K\$

Sure, that's fine. Maybe add Plato's comments if you want to be even more explicit, but either way it works.

Quote:

I am not sure if I can just say that an open cover of K exists. It makes sense, to me anyway, that I could cover any set with a bunch of open sets. Still, I am not sure if an open cover of K exists.
Two things...

First of all, an open cover of any set ALWAYS exists because the whole space is open by definition. For instance, suppose \$\displaystyle X\$ is a space and \$\displaystyle A\subset X\$. Then \$\displaystyle \{X\}\$ is an open cover of \$\displaystyle A\$.

Second, even if an open cover didn't exist, we don't need one to show that a set is compact. In fact, that would GUARANTEE that it was compact!

Recall that a set A is compact if and only if it satisfies the following:

For all Y, if Y is an open cover of A, then there is a finite subcover.

In other words, a set A is non-compact if and only if it satisfies the following:

Not for all Y, if Y is an open cover of A, then there is a finite subcover.

This is true if and only if

There exists Y such that Y is an open cover of A, and it has no finite subcover.

If there is no open cover of A, then it cannot satisfy this property and hence is compact.

Quote:

And the union of the finite subcovers remains finite?
Yup. Any finite union of finite sets is again finite.