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.

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    $\begingroup$ A set is closed iff its complement is open. This might be simpler than the accumulation-point caracterization. ALso you need your space to be Hausdorff. $\endgroup$ – Tlön Uqbar Orbis Tertius Apr 28 '15 at 12:19
  • $\begingroup$ The thing is I have not covered Hausdorff spaces, so I was wondering if my approach could be attainable. $\endgroup$ – Wolfy Apr 28 '15 at 12:26
  • $\begingroup$ @HerbertQuain: pretty sure $\mathbb{R}$ with the usual topology is indeed Hausdorff ;-) Looks as though the questioner is studying the reals, not topology in general, since the definitions have been given only for the reals. $\endgroup$ – Steve Jessop Apr 28 '15 at 17:46
  • $\begingroup$ @SteveJessop I agree with you, my comment was only here as an indication in case OP had further knowledge in topology. And after all, there are non-Hausdorff topologies on $\mathbb{R}$, even if they only exist for the purpose of counterexamples^^) $\endgroup$ – Tlön Uqbar Orbis Tertius Apr 28 '15 at 18:38

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.

Sorry, but your argument is completely wrong.

You can't work with a “generic” open cover, but you have to build one, starting from a point $p$ not belonging to $A$, in order to show it is not an accumulation point.

Since the statement is false if one doesn't assume the space is Hausdorff, a proof must exploit this property.

I'll prove it in any Hausdorff space, so also for metric spaces.

Let $p\in X$ and suppose $p\notin A$. For every $y\in A$ choose

  • an open neighborhood $U_y$ of $y$
  • an open neighborhood $V_y$ of $p$

such that $U_y\cap V_y=\emptyset$.

This is possible because of the Hausdorff property. In metric spaces, just take $U_y=B(y;d(y,x)/2)$ and $V_y=B(x;d(y,x)/2$, where $B(z;r)$ denotes the open ball with center $z$ and radius $r$.

The family $(U_y)_{y\in A}$ forms an open cover of $A$. Since $A$ is compact, we have $A\subseteq U_{y_1}\cup\dots\cup U_{y_n}$ for some $y_1,\dots,y_n\in A$.

Then $V=V_{y_1}\cap \dots \cap V_{y_n}$ is a neighborhood of $p$ and $V\cap A=\emptyset$. So $p$ does not belong to the closure of $A$. Therefore $A$ contains all the points that belong to its closure, so it is closed.



If $A \subseteq \mathbb{R}^p$ is compact then $A$ is closed


We will show that $\mathbb{R}^p \setminus A$ is open.

Let $x \in \mathbb{R}^p \setminus A$.

We must show that there exists a $r>0$ such that $B(x,r) \subset \mathbb{R}^p \setminus A$.

For each $m \in \mathbb{N}$, let $G_m = \{ y \in \mathbb{R}^p : ||y-x|| > \frac{1}{m} \}$. Each set $G_m$ is open in $\mathbb{R}^p$ (Try to show this)

Now we know $$\displaystyle \bigcup_{m \in \mathbb{N}} G_m =\mathbb{R}^p \setminus \{x\} $$

Since $x \not \in A$, we have $A \subseteq \mathbb{R}^p \setminus \{x\} = \displaystyle \bigcup_{m \in \mathbb{N}} G_m $.

In the view of the compactness of $A$, it follows that there exists a natural number $M$ such that

$$A \subseteq G_1 \cup G_2 \cup \dots \cup G_M$$

Since the sets $G_m$ increase with $m$ we have that $$G_1 \cup G_2 \cup \dots \cup G_M = G_M$$ and

$$A \subseteq G_M = \{ y \in \mathbb{R}^p : ||y-x|| > \frac{1}{M} \}$$

Hence the neighborhood $$B\bigg(x, \frac{1}{M}\bigg) = \{z \in \mathbb{R}^p : ||z - x|| < \frac{1}{M} \} $$ does not intersect with $A$. In other words

$$B(x, \frac{1}{M}) \subset \mathbb{R}^p \setminus A$$ Hence $\mathbb{R}^p \setminus A$ is open and consequently $A$ is closed


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