A set $S \subset \mathbb{R}^n$ is compact if and only if $S$ is bounded and closed. (Heine Borel)

Question

I would like to be sure abuot correctness proof and example. Could you help me? thank you!

A set $S \subset \mathbb{R}^n$ is compact if and only if $S$ is bounded and closed.

Proof.

$\Rightarrow$

(Compact $\Rightarrow$ closed and bounded.)

Let $S$ be compact (if every open covering admits a finite subcover) To show that $S$ is bounded, suppose that $S$ is unbounded. Then for every $n \in \mathbb{N}$ there is $x_n \in S$ such that $|x_n| > n$. Since $S$ is compact, the sequence $(x_n)$ has a convergent, hence bounded, subsequence $(x_{n_j})$. But $|x_{n_j} | > n_j$ with $n_j \to \infty$ as $j \to \infty$, a contradiction. So $S$ is bounded. \newline To see that $S$ is closed, we take a limit point $x$ of $S$ and a sequence $(x_n)$ with $x_n \in S$ and $x_n \neq x$ for all $n \in \mathbb{N}$ such that $(x_n) \to x$, and show that $x \in S$. The compactness of $S$ implies that there is a subsequence $(x_{n_j} )$ that converges to a point that is in $S$. Since $(x_n) \to x$, then $(x_{n_j} ) \to x$ as well, and so $x \in S$, and $S$ is closed.

$\Leftarrow$

(Compact $\Leftarrow$ closed and bounded.)

Suppose that $S$ is closed and bounded. Let $(x_n)$ be a sequence in $S$. Then there is a convergent subsequence $(x_{n_k} )$. Then by the closeness of $S$, $lim_{k \to \infty} x_{n_k} \in S$. Thus S is compact. The proof is finished.

Example.

Find an example of a bounded and closed set in metric space that is not compact. (Infinite dimension)

The discrete metric $d (x, y)$

Solution.

I.

Notice that any subset of a metric space with the discrete metric is closed and bounded.

Bounded. Let the metric space $(X,d)$ where $d$ is the discrete metric. It is bounded because $d(x, y) \leq 1$ for all $x, y \in X$, but it cannot be covered by a finitely many balls of radius $\frac{1}{2}$.

Closed. Every element is less than or equal to 1, and it is closed as a whole set.

Let $r=\frac{1}{2}$. Then the ball contains only the point itself. Every subset is closed because the complement of an open set is closed.

Not compact.

Since the open cover by singletons admits no finite subcover.

Let $S $ is an infinite subset of $X$, it follows $S$ is an infinite discrete metric space. Consider $G= \{\{k\} | k\in S\}$, which is an open cover of $S$. Clearly, $G$ has no finite subcover. Thus, $S$ is not compact

More generally, any infinite discrete space admits a proper subspace that is closed and bounded, but not compact.

Answer 1

(Compactness $\Leftarrow$ Closed and Bounded) Your proof is very short. And it is sequential compactness point of view. I don't know about it much. The proof I remember is set-theoretic and involved. It is as follows:

1. $[0,1]$ is compact. This is the key step.
2. $[-R,R]$ is compact for any $R\in\Bbb R$. Because $[-R,R]$ is homeomorphic to $[0,1]$ and compactness is a topological invariant.
3. $[-R,R]^n$ is compact for any $R\in\Bbb R$ and $n\in\Bbb N$. Because, the product space of compact spaces is compact.
4. Proof: Let $A\subset\Bbb R^n$ be a closed and bounded subset. Since $A$ is bounded, $\exists R>0$ such that $A\subset [-R,R]^n$. Since $A$ is closed, it is compact. Because, a closed subset of a compact space is compact.

I will think about your second question and the example you gave later.

Answer 2

A couple of issues.

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In your proof that compact implies bounded, it is not really necessary to make a proof by contradiction. Direct proofs are often more elegant and understandable. In particular, you could do something like this:

Define $C_i = B(0, i)$ (i.e., the open ball centered at $0$ with radius $i$). Then, clearly, $\mathbb R^n = \bigcup_{i=1}^\infty C_i$, which means that $S\subseteq\mathbb R^n=\bigcup_{i=1}^\infty C_i$ which means $\{C_i\}_{i=1}^\infty$ is a open covering of $S$. Therefore, because $S$ is compact, there exists a finite subcovering, i.e. there exists some finite set $N\subset\mathbb N$ such that $S\subseteq \bigcup_{i\in N} C_i$. Since $\bigcup_{i\in N} C_i$ is bounded (it is actually equal to $B(0, \max(N))$), that means $S$ is bounded.

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For your proof that closed and bounded implies compact, you make a very bad mistake right at the start. You say:

Suppose that $S$ is closed and bounded. Let $(x_n)$ be a sequence in $S$. Then there is a convergent subsequence $(x_{n_k} )$.

This is completely unjustified. If $S$ is closed and bounded, what property of $S$ implies that every sequence in $S$ has a convergent subsequence?