# Thread: Finite union of compact sets is compact

1. ## 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.

2. 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.

4. 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.

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.

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