Question

Suppose {p} is a cauchy sequence in a metric space X and some subsequence {p} converges to a point a belongs to X . Prove that the full sequence {p} converges to a .

Answer 1

Suppose $\{{p_n} \}$ is a cauchy sequence in a metric space X and some subsequence $\{p_{n_k}\}$ converges to a point $a$ belongs to $X$.

Let $\epsilon > 0.$ There exists $N$ such that, for $n, m \geq N , d(p_n, p_m) <\epsilon$ ( $\{{p_n} \}$ is a cauchy sequence ). Then, consider any $n \geq N$ :

for $k$ sufficiently large (so that $n_k \geq N,$ ), $d(p_n, p_{n_k} ) \leq \epsilon$ .

Taking the limit as $k \rightarrow \infty$, it follows that $d(p_n, a)\leq \epsilon$ .

Or: if $k$ is sufficiently large then $d(p_{n_k}, a) <\epsilon$ (by the assumption $p_{n_k }\rightarrow a$ ), so $d(p_n, a) \leq d(p_n, p_{n_k}) + d(p_{n_k}, a) < 2 \epsilon$.

In any case, we conclude that $d(p_n, a)$ becomes arbitrarily small for $n$ large, i.e. $p_n \rightarrow a$ .