# Let $(x_n)$ and $(y_n)$ be sequences such that $x_n$ in $[0, 1]$ and $y_n$ in $[1, 2]$

Use the Bolzano-Weierstrass Theorem to show that there are numbers $$m_1, m_2, ...$$ such that $$x_{m_j}$$ and $$y_{m_j}$$ converge for j=1,2,...

So I have trouble approaching this problem. I can apply the Bolzano Weierstrass to the sequences separately to get $$x_{m_k}$$ converging to some $$x$$ and $$y_{m_j}$$ converging to some $$y$$, but this doesn't help.

I have also thought of adding $$x_n$$ and $$y_n$$ and subtracting $$x_n$$ and $$y_n$$ to get two bounded sequences each of which converges by Bolzano Weierstrass, but not sure this leads anywhere.

How do you approach this problem?

## 1 Answer

The Bolzano–Weierstrass theorem holds not only for bounded sequences in $$\Bbb R$$, but also for bounded sequences in the finite-dimensional space $$\Bbb R^d$$. This can be seen by repeatedly picking convergent subsequences in each coordinate.

In your case ($$d=2$$) that would work as follows:

First apply Bolzano-Weierstrass to the (bounded) sequence $$(x_n)_n$$, that gives a convergent subsequence $$(x_{n_k})_k$$.

Then apply Bolzano-Weierstrass to the (bounded) subsequence $$(y_{n_k})_k$$ of $$(y_n)_n$$, that gives a convergent subsequence $$(y_{n_{k_j}})_j$$. Note that $$(x_{n_{k_j}})_j$$ is also convergent as a subsequence of $$(x_{n_k})_k$$.

Therefore, with $$m_j = n_{k_j}$$, $$(x_{m_j})_j$$ and $$(y_{m_j})_j$$ are “common” convergent subsequences of $$(x_n)_n$$ and $$(y_n)_n$$ respectively.