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The set of subsequential limits of a sequence [itex]\{p_n\}[/itex] in a metric space X is a closed subset of X.

I tried proving this myself, and my proof came out to be similar to the one in Rudin, yet to me, simpler. I wanted to post it and ask if it seems correct.

Like in Rudin, call the set of subsequential limits of [itex]\{p_n\}[/itex] [itex]E^*[/itex]. Then if q is a limit point of [itex]E^*[/itex], we need to show it is in [itex]E^*[/itex], meaning that there is some subsequence of [itex]\{p_n\}[/itex] which converges to q. Now, for every [itex]n\in\mathbb{N}[/itex], choose some [itex]q_n\in E^*[/itex] such that [itex] d(q_n, q)<\frac{1}{n}[/itex]. This is possible since we assumed that q was a limit point of [itex]E^*[/itex].

Now, since each [itex]q_n\in E^*[/itex], this means that there is some [itex] p_{n_k}[/itex] such that [itex]d(p_{n_k}, q_n)<\frac{1}{n}[/itex] (this is from the definition of convergence). Thus, [itex]d(q, p_{n_k})\le d(q, q_n)+d(q_n, p_{n_k})<\frac{2}{n}[/itex], showing that this subsequence [itex]\{p_{n_k}\}[/itex] converges to q, proving the theorem.

I'm just curious to see if this proof is valid, since it seems to me a bit simpler than the one given in the textbook. Thanks!