# 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?

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.