[Math] If a set is compact then it is closed

general-topology

Show that if a set is compact then it is closed.

definitions: Let $A\subset \mathbb{R}$. A point $p\in\mathbb{R}$ is an accumulation point or limit point of $A$ if and only if every open set $G$ containing $p$ contains a point of $A$ different from $p$. A subset $A$ of $\mathbb{R}$ is closed if and only if $A$ contains each of its points of accumulation.

proof: Suppose $A\subset \mathbb{R}$ is compact and $A\subset \bigcup O_\alpha$ where $O_\alpha$ is open. Since, $A\subset \bigcup O_\alpha$ there exists $\alpha_1,\dots,\alpha_n$ such that $$A\subset O_{\alpha_{1}}\cup O_{\alpha_{2}}\cup \ldots \cup O_{\alpha_{n}}$$
Now suppose we have a point $p\in\mathbb{R}$. Let $G = \bigcup_{i=1}^{n}O_{\alpha_{i}}$, and $p\subset G$. Then $p$ is an accumulation point of $A$ since every open set $G$ contains $p$ can also contain a point of $A$ different from $p$.

I am not sure if this is the right approach but it makes sense to me, I just don't know how to go on from here. Any suggestions would be greatly appreciated.

Best Answer

Several mistakes here:

  1. What is $\alpha$? How are $O_\alpha$? Are they just any collection of sets covering $A$?
  2. You take a point $p\in \mathbb R$ and prove that if $P\in G$, then $p$ is an accumulation point of $A$, which is not true:

Take $A=[0,1]$ and take $O_1=(-1,1), O_2=(0,2)$. Then, $G=(-1,2)$ and we can take $p=-\frac12$, but $p$ is then not an accumulation point of $A$.

  1. Even if it were true, you would then prove that every point in $G$ is an accumulation point of $A$, which does not prove that $A$ is closed. For example, if $G=(0,1)$ and $A=(0,1)$, then every point $p\in G$ is an accumulation point of $A$, but $A$ is not closed.

I believe your proof is wrong from the beginning.

First of all, you are not proving what you need to prove. You need to prove that $A$ is closed, and you may do this by proving that every accumulation point of $A$ is in $A$.

Second of all, you are using compactness in the wrong place.

Compactness tells you that:

For EVERY collection $\{O_\alpha\}$ which covers $A$, there exists a finite subcollection which also covers $A$.

Therefore, the idea here is not to take the collection at the beginning.

The idea is this:

  1. Take a poin $p$ which is an accumulation point of $A$.
  2. Prove that $p\in A$. You may for example do this by assuming that $a\notin A$, then using compactness of $A$ to reach a contradiction.
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