# $\{x_n\}_{n=1}^\infty$, $\{y_n\}_{n=1}^\infty$ $\subset M$ (compact) then $\exists$ $\{x_{n_k}\}, \{y_{n_k}\}$ s.t. they both converge to same point.

Full question:

$$\left< M,\rho \right>$$ is a compact metric space with sequences $$\{x_n\}_{n=1}^\infty$$, $$\{y_n\}_{n=1}^\infty$$ and $$\displaystyle{\lim_{n \to \infty} \rho(x_n, y_n)=0}$$.

Prove $$\exists$$ subsequences $$\{x_{n_k}\}_{k=1}^\infty$$, $$\{y_{n_k}\}_{k=1}^\infty$$ and a point $$x \in M$$ s.t.

$$\displaystyle{\lim_{k \to \infty} \rho(x_{n_k}, x)=0}$$ and $$\displaystyle{\lim_{k \to \infty} \rho(y_{n_k}, x)=0}$$

Below is my attempt and I would like to know if there is any mistake or room for improvement.

Proof:

$$M$$ is compact so $$\{x_n\}_{n=1}^\infty$$ has a subsequence $$\{x_{n_k}\}_{k=1}^\infty$$ that converges to a point in $$M$$. Call this point $$x \in M$$.

So $$\displaystyle{\lim_{k \to \infty} x_{n_k}=x}$$ $$\implies$$ $$\displaystyle{\lim_{k \to \infty} \rho(x_{n_k}, x)=0}$$

So given an $$\epsilon \gt 0$$, $$\exists N_1 \in \mathbb{N}$$ s.t. if $$k \geq N_1$$ then $$\rho(x_{n_k}, x) \lt \frac{\epsilon}{2}$$

Also given $$\displaystyle{\lim_{n \to \infty} \rho(x_n,y_n)=0}$$, so $$\exists N_2 \in \mathbb{N}$$ s.t. if $$n \geq N_2$$ then $$\rho(x_n,y_n) \lt \frac{\epsilon}{2}$$

But $$n_k \geq n \ \ \forall n,k \in \mathbb{N}$$

So $$\rho(x_{n_k},y_{n_k}) \lt \frac{\epsilon}{2}$$ if $$k \geq max\{N_1,N_2\}$$

Now, let $$k \geq max\{N_1,N_2\}$$ then

$$\rho(y_{n_k}, x) \leq \rho(y_{n_k},x_{n_k}) + \rho(x_{n_k}, x)$$ $$\lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

So for any given $$\epsilon \gt 0$$ we have shown there is an index $$N$$ s.t. if $$n \geq N$$ then $$\rho(y_{n_k}, x) \lt \epsilon$$ and $$\rho(x_{n_k}, x) \lt \epsilon$$ which means

$$\displaystyle{\lim_{n \to \infty} \rho(x_n,x)=0}$$ and $$\displaystyle{\lim_{n \to \infty} \rho(y_n,x)=0}$$ $$\ \ \ \ \ \ \Box$$

It is almost correct. The only problem that I see in it lies in the assertion “But $$n_k\geqslant n\ \forall n,k\in\mathbb N$$”, which is false. I suppose that you meant to write that $$n_k\geqslant k\ \forall k\in\mathbb N$$.

• That makes sense. I meant to write that the terms in a subsequence appear later in the sequence compared to the main sequence. Which is the same as writing what you wrote. Thanks for pointing that out.
– Sun
Oct 22, 2019 at 4:54

Rephrasing:

0) $$M$$ compact, there exist a convergent subsequence $$(x_{n_k}) \in M$$.

1) Let $$x \in M$$ be $$\lim_{k \rightarrow \infty} x_{n_k}=x$$,

i.e. for $$\epsilon >0$$ there is a $$K_0$$ s.t $$k \ge K_0$$,

implies $$d(x_{n_k},x) <\epsilon$$.

2)Given:

$$\lim_{n \rightarrow \infty} d(x_n,y_n)=0$$, i.e.

there is a $$N$$ s.t. $$n \ge N$$ implies $$d(x_n,y_n) < \epsilon$$.

3) Let $$K_1$$ be s.t for $$k \ge K_1$$,

$$n_k \ge n_{K_1} \ge N$$, implies

$$d(x_{n_k},y_{n _k}) <\epsilon$$.

4) Choose $$K_2=\max(K_0,K_1)$$.

5) Then for $$k \ge K_2$$

$$d(x,y_{n_k}) \le d(x,x_{n_k})+d(x_{n_k},y_{n_k})<2\epsilon$$.