In Rudin, theorem 3.7 is that:(adsbygoogle = window.adsbygoogle || []).push({});

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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Alternative proof of a theorem in Rudin

**Physics Forums | Science Articles, Homework Help, Discussion**