Every bounded sequence in $\mathbb{R}^n$ possesses a convergent subsequence

Question

Can someone help me understand the proof for this theorem:

All bounded sequences in $R^n$ have a sub-sequence that is convergent

Here's the proof:

$Suppose \: n=2 \:and\: u_{j}=\binom{x_{j}}{y_{j}} \\ We\:suppose\:that\:u_{j}\:is\:bounded\:for\:||\cdot ||_{\infty} \:Which\:means: \\ \exists M\geq 0, \forall j,\:max(|x_{j}|,|y_{j}|)\geq M(so\: (x_{j})\:and\:(y_{j})\:are\:bounded)\\ 1)(x_{j})\:is\:bounded:\\ \exists\:subsequence \:\tau_{j}(that\:is\:strictly\:increasing)\\ 2)The\:sequence\:(y_{\sigma_{j} })\:is\:bounded,\:so\:it\:has\:a\:subsequence.\:The\:associated\:subsequence\:to\:(x_{\tau_{j}})\:converges$

I am having a difficult time to understand the second part. The $\delta$ and $\sigma$ sequences just seem to have no connection to me.

Answer 1

It is true that a bounded sequence has a monotonic subsequence, but (i) it need not be increasing, (ii) it need not be strictly monotonic, and (iii) in the first place it is impossible to get hold of such a subsequence before knowing the full sequence all the way to the end.

Instead use Bolzano's theorem that guarantees you a convergent subsequence for any bounded sequence in ${\mathbb R}$. You then can argue as follows: If the sequence ${\bf z}_n=(x_n,y_n)\in{\mathbb R}^2$ $(n\geq1)$ is bounded then so is the sequence $(x_n)_{n\geq1}$ in ${\mathbb R}$. It follows that there is a subsequence $x_k':=x_{n_k}$ in ${\mathbb R}$ with $\lim_{k\to\infty} x_k'=\xi\in{\mathbb R}$. The sequence $y_k':=y_{n_k}$ $(k\geq1)$ is a bounded sequence of real numbers as well, hence there is a subsequence $y''_l:=y'_{k_l}$ with $\lim_{l\to\infty} y''_l=\eta\in{\mathbb R}$. Put $x''_l:=x'_{k_l}$ and ${\bf z}''_l:=(x''_l,y''_l)$. Then $({\bf z''}_l)_{l\geq1}$ is a subsequence of the given sequence $\bigl({\bf z}_n\bigr)_{n\geq1}$ with $$\lim_{l\to\infty}{\bf z}_l=(\xi,\eta)\in{\mathbb R}^2\ .$$

Answer 2

It might be illuminating to study an example. Consider the sequence $$ \left((1, 1), (2, 2), (1, 3), (2, 1), (1, 2), (2, 3), (1, 1), (2, 2), (1, 3), (2, 1), (1, 2), \dots\right) $$ where the first coordinate alternates between $1$ and $2$, and the second coordinate iterates through the cycle $(1,2,3)$.

This sequence is bounded.

To construct a convergent subsequence, firstly consider only the first coordinate. We can construct a subsequence that is convergent in the first coordinate by selecting every other element of the original sequence to obtain $$ \left((1,1), (1, 3), (1, 2), (1,1), (1, 3), (1, 2), (1,1), (1, 3), (1, 2), \dots\right). \tag{*}\label{star} $$

The first coordinate of this subsequence trivially converges to $1$.

Now turn your attention to the second coordinate, and construct a subsequence of \eqref{star} that is convergent in the second coordinate. We can do this by selecting, say, every third element to obtain $$ \left((1,2), (1, 2), (1, 2), \dots\right). $$

We have now constructed a convergent subsequence of the original sequence.

Answer 3

Allow me to add an answer using concepts from topology.

Let $(p_k)$ be an arbitrary bounded sequence in $\mathbb{R}^n$. Then, $\exists\ M > 0$ s.t. $p_k \in [-M,M]^n$ for all $k$.

• $[-M,M]^n$ is a closed and bounded subset of $\mathbb{R}^n$ $\implies $ it is compact.

• Compactness $\implies $ Limit point compactness.

• For first countable Hausdorff spaces, limit point compactness $\implies$ Sequential compactness.

Note that $\mathbb{R}^n$, being a metric space (equipped with some metric (*)), is first countable and Hausdorff, and so is its subset $[-M,M]^n$ in the subspace topology.

By above arguments, $[-M,M]^n$ is sequentially compact. Hence, by the definition of sequential compactness, the sequence $(p_k)$ has a subsequence which converges to a point in $[-M,M]^n$.

(*) Same metric which is being used to evaluate convergence.

P.S. A direct argument based on the following results is also possible:

• For metric spaces, compactness, limit point compactness and sequential compactness are all equivalent properties.
• Subset of a metric space is a metric subspace with metric inherited from the original space.