Does anyone understand this proof: If $A$ is closed and bounded $\implies A$ is sequentially compact.

Question

This is how it goes, I will highlight the parts in yellow which I don;t understand why it is , or the idea behind it.

$A$ is bounded so $(\forall x \in A)(\exists M > 0)(\|x\|<M)$ Let ${x_k}\subset A \subset R^n \implies x_k=(x_1^k,...,x_n^k) \\ |x_1^k|<M,...,|x_n^k|<M ,k\in \mathbb N.$

From here we get that the coordinates of the sequence from $A \subset \mathbb R^n$ are bounded meaning that these sequences (real) are bounded:

$$x_1=(x_1^1,x_2^1,...,x_n^1) \\ x_2=(x^2_1, x^2_2,...,x_n^2)\\ ......\\...... \\ x_n=(x_1^n,x_2^n,..., x_n^n)$$For every of these sequences (in the coloumns) there exist convergent subsequences.(Why? Because of boundedness? Is there a theory I cannot see?)

$ \exists (x^{n_k}_1)$ sub-sequence of sequence $(x_1^k)$ that converges to $x_1$. Lets evaluate now the subsequence of sequence $\{x_k\}, \{x_{n_k}\} \subset A \subset \mathbb R^n$, $$x_{k_1}=(x_1^{k_1},x_2^{k_1},...,x_n^{k_1}) \\ x_{k_2}=(x^{k_2}_1, x^{k_2}_2,...,x_n^{k_2})\\ ......\\...... \\ x_{k_n}=(x_1^{k_n},x_2^{k_n},..., x_n^{k_n})$$ For the real sequence $\{x_2^{k_n}\}$ exists a convergent subset $\{x_2^{k_l}\}\text{such that} x_2^{k_l} \to x_2$. Repeating this process of singling out subsequences $n$ times we get a final subsequence $${x_{ks}}\text{ of }\{x_k\}$$ whose coordinates are in order $\to x_1,x_2,...,x_n$ so $x_{ks}\to x=(x_1,x_2,..,x_n)$ Since $A=\overline{A} \implies A $ is sequentially compact.

If anyone understood this, I would immensely appreciate an explanation. Thankful in advance.

Answer 1

Each of the sequence in the columns is finite/bounded, hence has a convergent subsequence by the Bolzano Weierstrass theorem.

Then $(x_1^k)$ has a subsequence $(x_1^{n_k})$ that converges to, say, $x_1$, $(x_2^k)$ has a subsequence $(x_2^{n_k})$ that converges to $x_2$, and so on, $(x_n^k)$ has a subsequence $(x_n^{n_k})$ that converges to $x_n$.

Let $x = (x_1, ..., x_n)$. We have that $x_n^{n_k} \to x \in A$ (since A is closed). This shows that any arbitrary sequence in $A$ has a convergent subsequence, so that $A$ is sequentially compact.

Answer 2

Hints:A metric space is sequentially compact if every sequence has a convergent subsequence. This is equivalent to every infinite subset of that space has a cluster point.

A metric space being totally bounded is equivalent to the fact that every sequence has a Cauchy subsequence.

Is it more obvious now?