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I want to prove that any closed and bounded set in $\mathbb{R}^n$ is sequentially compact.

Note that I want to prove this without using the Heine-Borel theorem and the equivalence of sequential compactness and compactness. Ah and I can use that any closed and bounded set in $\mathbb{R}$ is sequentially compact.

The way I did it is as follows: Let $S$ be a closed and bounded subset of $\mathbb{R}^n$. Then $\exists a,b \in \mathbb{R}$ such that $S \subseteq [a,b]\times [a,b] \times ...[a,b]$. Now I will consider a sequence in $S$.

Let $x_p = (x_p^1, x_p^2,...,x_p^n) \forall p \in \mathbb{N}$.

Then I will construct $n$ sequences of components of $x_p$. Let $x^{(i)}=x_1^i,x_2^i...$ Then $x^{(1)}\subset [a,b]$, and $[a,b]$ is sequentially compact. so $x^{(1)}$ has a convergent subsequence. Now if we denote the indices of this sequence by $t_1,t_2...$ we know that $x^1_{t_1},x^1_{t_2}, x^1_{t_3}...$ converges to some $x$ in $[a,b]$. Now if we consider $x^2_{t_1},x^2_{t_2}, x^2_{t_3}...$ it is a sequence in $[a,b]$ a sequentially compact space. So it has a convergent subsequence. Again we take the indices of this subsequence and we consider them to build a subsequence of $x_3$ which converges.

After we have repeated this procedure $n$ times we are left with $n$ we have found a set of indices $f_1,f_2...$ such that $x^n_{f_1}, x^n_{f_2}...$ converges. But then by construction $x^p_{f_1}, x^p_{f_2}...$ converges $\forall p \in1..n$ as it is a subsequence of a convergent sequence.

Then $x_{f_1},x_{f_2} ..$ converges (as it is a sequence in $\mathbb{R}^n$ for which all individual components converge) and as it is included in a closed set, the limit of the sequence is part of the set.

Since we started with an arbitrary sequence, this proves that any closed and bounded set in $\mathbb{R}^n$ is sequentially compact.

Can you please have a look at this and tell me if something is wrong? It seems fine to me, but I really want to get the concept behind sequential compactness. I know that historically it was the first definition of compactness, so I find it important to see how knowledge evolved and how people initially proved it, without using compactness.

Also, please feel free to show me another proof, I am very interested in that as well.

Thank you!

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2 Answers 2

up vote 2 down vote accepted

Let me give first the definitions:

D A metric space $M$ (or subset of a metric space) is said to be precompact if every sequence in $M$ contains a Cauchy (or fundamental) subsequence.

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D Let $S$ be any subset of $\Bbb R^n$. Given $\epsilon >0$, we say that $N$ is an $\epsilon$-net for $S$ if the set of open balls

$$B_\epsilon(N)=\{B(x,\epsilon):x\in N\}$$

covers $S$. That is, the set of open balls of radius $\epsilon$ centered at the points of $N$ cover $S$.

Now, a theorem by Hausdorff

T Let $P$ be a metric space, $M$ a subset of $P$. Then $P$ is precompact if and only if, given $\epsilon >0$, $P$ contains a finite $\epsilon$-net for $M$.

P Suppose $M$ is precompact. Then given $\epsilon >0$, choose $x_1\in M$. If the point $x_1\in M$ is such that $d(x,x_1)<\epsilon$ for each $x\in M$; $\{x_1\}$ is itself an $\epsilon$-net, and we're done. If there exists points with $d(x,x_1)>\epsilon$, choose as $x_2$ one of these points. If now for every $x\in M$ we have either $d(x,x_1)<\epsilon$ or $d(x,x_1)<\epsilon$ , we're done. Else, continue the process, noting that the distance from each new $x_n$ to the $x_i\; ;1\leq i\leq n-1$ exceeds $\epsilon$. Thus, if the construction fails to terminate after a finite number of steps, we would obtain a sequence $\{x_n\}_{n\in \Bbb N }$ which contains no Cauchy subsequence, contrary to the hypothesis that $M$ is precompact. Thus, the construction must terminate, providing us with a finite $\epsilon$-net, $\{x_1,\dots,x_{\ell}\}$.

Conversely, suppose that for any $\epsilon>0$, $P$ contains a finite $\epsilon$-net for $M$, and let $A$ be any infinite subset of $M$, in particular, a sequence containing infinitely many distinct points of $M$. We can select a Cauchy subsequence as follows. Let $x_0\in A$ be any point, then take $\epsilon =1$ in the condition and cover $A$ with a finite number of balls of radius $1$. One of these balls, say $B_1$, must contain an infinite $A_1$ of $A$. Choose $x_1\in A_1\setminus \{x_0\}$. Choosing $\epsilon = 1/2$; we cover $A_1$ with a finite number of balls of radius $1/2$. One of them, say $B_2$ must contain an infinite subset $A_2$ of $A_1$. Choose $x_2\in A_2\setminus \{x_0,x_1\}$. Continuing this process, we obtain a sequence $A_0\supseteq A_1\supseteq A_2\cdots$, where each $A_n$ is contained in a ball of radius $1/n$ and a sequence of distinct points $x_0,x_1,x_2,\dots$ with $x_n\in A_n$. This sequence is Cauchy, for if $m<n$, $$B_m\supseteq A_m\supseteq A_n$$ which means $$d(x_n,x_m)<\frac 2 m$$Since $2/m\to 0$, we're done. It follows $M$ is precompact.

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Look here for

T Let $S$ be bounded in $\Bbb R^n$. Then $S$ is precompact.

Now, since $\Bbb R^n$ is complete, any Cauchy (sub)sequence will converge. The assumption that the set is closed means the limit $\ell$ will be in such set. Thus, every bounded closed subset of $\Bbb R^n$ contains a convergent subsequence, that is, it is sequentially compact.

NOTE Observe the similarity of the $\implies$ direction of Hausdorff's criterion with the proof of Weiertrass-Bolzano that Ragib suggests.

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Your proof seems fine, and is quite similar to the standard proof that is mentioned in the link Qiaochu commented above. I find this following reasoning more intuitive, but some work needs to be done to make it a proof:

Start as you did, bounding your sequence in some large cube and cut the cube in half to make two rectangular prisms. At least one of these prisms has infinitely many points of the sequence inside of it - cut that one in half again. Now repeat this argument and you can see why you must have an accumulation point.

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