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Theorem 3.7: The subsequential limits of a sequence $\{p_n\}$ in a metric space $X$ form a closed subset of $X$

And the proof starts as: Let $E^*$ be the set of all subsequential limits of $\{p_n\}$ and let $q$ be a limit point of $E^*$. We have to show that $q \in E^*$

Isn't this a strong assumption? How do we eliminate the possibility that no such point exists? On the other hand if a limit point of $E^*$ exists then $E^*$ can not be a finite set and thus there are infinite subsequential limits. But if $\{p_n\}$ is convergent then there is only one limit since all subsequences converge to that. Is the answer that we simply don't care for these cases?

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    $\begingroup$ If no limit point exists, then the set is vacuously a closed set. The set of integers is closed, but it has no limit points. You only need to show that every limit point of $E^*$ is in $E^*$; it doesn't matter how many there are. $\endgroup$ – Trevor Norton Jan 15 '16 at 20:50
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    $\begingroup$ You confuse the notions of limit point and accumulation point. Note also that, if $\{q_n\}$ is an enumeration of the rationals, then $E^*=\mathbb R$. $\endgroup$ – Yiorgos S. Smyrlis Jan 15 '16 at 20:51
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There are two cases, Rudin chooses to not bother mentioning the trivial case where $E^*$ is finite.

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