I want to prove the following statement
The set $A\subset R^n $ is compact if and only if A is closed and bounded.
This statement is the Heine Borel theorem.
I am not a mathematician. I am an international-relation student. I found many proof ways and explanations on the Google . But these are very complicated for me.
Please can you show its proof in the simplest way? Or can you suggest a resource which explains its proof in a simplest way?
I am preparing an exam. And so I really need to understand it. Thanks a lot.
Depending on the definition of compactness that you started with, there is an easy proof of this result. I'll include it here for others who might be interested. I'll use the following definition.
Definition 1.1 A set $A \subseteq \mathbb{R}^n$ is compact iff every sequence in $A$ has a convergent subsequence with limit in $A$.
This turns out to give a very easy proof of the Heine-Borel Theorem if you are aware of some of the other (nice) properties that sequences have in $\mathbb{R}^n$.
Theorem 1.1 Let $A \subseteq \mathbb{R}^n$ be a set. $A$ is compact iff $A$ is closed and bounded.
Proof:
$\implies:$ Assume that $A$ is compact. Let $(x_n)_{n \in \mathbb{N}}$ be any convergent sequence in $A$. We know that $A$ is closed iff for any convergent sequence in $A$, the limit lies in $A$. Since $A$ is compact, our sequence $(x_n)_{n \in \mathbb{N}}$ has a convergent subsequence whose limit lies in $A$. But the limit of this subsequence is also the limit of the sequence and this implies that the sequence converges in $A$. So, $A$ is closed.
Next, assume that $A$ is not bounded. Then, for every $n \in \mathbb{N}$, we can find a $x_n \in A$ such that $\left|\left|x_n\right|\right| \geq n$. But notice that this sequence $(x_n)_{n \in \mathbb{N}}$ cannot have a convergent subsequence and this contradicts the compactness of $A$. So, $A$ is closed and bounded.
$\impliedby:$ Now, assume that $A$ is closed and bounded. We want to show that $A$ is compact. So, let $(x_n)_{n \in \mathbb{N}}$ be a sequence in $A$. By the Bolzano-Weierstrass Theorem, this has a convergent subsequence since it is a bounded sequence (by virtue of $A$ being a bounded set). But since $A$ is closed, this convergent subsequence must converge in $A$. That proves compactness. $\Box$