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.


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


2 Answers 2


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 $n_k\geqslant k\ \forall k\in\mathbb N$.

  • $\begingroup$ 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. $\endgroup$
    – Sun
    Oct 22, 2019 at 4:54


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


$\lim_{n \rightarrow \infty} d(x_n,y_n)=0$, 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 $k \ge K_2$

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


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