# Countably Compact Sets

• Aug 27th 2013, 01:26 PM
Aryth
Countably Compact Sets
I'm working on this problem, and I'm not sure how to solve it.

Let $\displaystyle (X,\Omega)$ be a topological space and let $\displaystyle A \subseteq X$ be countably compact. Prove that every subset of $\displaystyle A$ that is closed relative to $\displaystyle \Omega$ is also countably compact.

I have a similar proof where this problem is stated without the 'countably' part. I'm unsure how to proceed with countably compact sets... Any help would be appreciated.
• Aug 27th 2013, 01:47 PM
Plato
Re: Countably Compact Sets
Quote:

Originally Posted by Aryth
Let $\displaystyle (X,\Omega)$ be a topological space and let $\displaystyle A \subseteq X$ be countably compact. Prove that every subset of $\displaystyle A$ that is closed relative to $\displaystyle \Omega$ is also countably compact.

As always this depends upon the definitions in use.
If $\displaystyle B\subset A$ that is closed relative to $\displaystyle \Omega$. Can you show that any countable open cover of $\displaystyle B$ has a finite sub cover all relative to $\displaystyle \Omega~?$
• Aug 27th 2013, 05:44 PM
Aryth
Re: Countably Compact Sets
Ah, so it is similar to the proof with compactness... I'll just change the proof a little, then. If you could verify that it is correct I would grateful.

Suppose that $\displaystyle A \subseteq X$ is countably compact and let $\displaystyle B \subset A$ be closed. We will let $\displaystyle \mathcal{F} = \{F_i : i \in \mathbb{N}\}$ be a countable cover of $\displaystyle A$. Since $\displaystyle X \backslash B$ is open, it follows that $\displaystyle \mathcal{F}$ along with $\displaystyle X \backslash B$ is a countable open cover of $\displaystyle A$. We know that $\displaystyle A$ is countably compact, so $\displaystyle A$ can be covered by a finite number of sets from $\displaystyle \mathcal{F}$, $\displaystyle F_1, F_2, \cdots , F_n$, and possibly $\displaystyle X \backslash B$. Since $\displaystyle B \subset A$, $\displaystyle F_1, F_2, \cdots, F_n$ covers $\displaystyle B$. Therefore, $\displaystyle B$ is countably compact.
• Aug 27th 2013, 07:00 PM
Plato
Re: Countably Compact Sets
Quote:

Originally Posted by Aryth
Ah, so it is similar to the proof with compactness... I'll just change the proof a little, then. If you could verify that it is correct I would grateful.

Suppose that $\displaystyle A \subseteq X$ is countably compact and let $\displaystyle B \subset A$ be closed. We will let $\displaystyle \mathcal{F} = \{F_i : i \in \mathbb{N}\}$ be a countable cover of $\displaystyle A$. Since $\displaystyle X \backslash B$ is open, it follows that $\displaystyle \mathcal{F}$ along with $\displaystyle X \backslash B$ is a countable open cover of $\displaystyle A$. We know that $\displaystyle A$ is countably compact, so $\displaystyle A$ can be covered by a finite number of sets from $\displaystyle \mathcal{F}$, $\displaystyle F_1, F_2, \cdots , F_n$, and possibly $\displaystyle X \backslash B$. Since $\displaystyle B \subset A$, $\displaystyle F_1, F_2, \cdots, F_n$ covers $\displaystyle B$. Therefore, $\displaystyle B$ is countably compact.

Good. That will work.