In a metric space $X$, the collection of all subsequential limits of a sequence $\{p_n\}$ is closed (a few proofs can be found here and here). I was wondering if such a result holds for all topological spaces. If so, how could one prove it? If not, what would be a counterexample?
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Gabriel's answer demonstrates the result for any first-countable space, which makes sense as sequences interact with topology nicely in such spaces, and not necessarily so nicely outside of such spaces – FShrike Jul 25 '22 at 19:13
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See this for an example to supplement Eric's example, with $A=\Bbb N\setminus{1}$ a countable set not containing its limit point $1$ – FShrike Jul 25 '22 at 19:40
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If you had trouble understanding that example as I initially did, note that a sequence converging to $1$ must be unbounded, so there must be $a_{n_k}\gt 2^k$ for every $k$ in this sequence; then take $U=\Bbb N\setminus\bigcup_{k=1}^\infty a_{n_k}$ a neighbourhood of $1$ which contradicts convergence – FShrike Jul 25 '22 at 19:42
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Take any space $X$ with a countable subset $A$ which has an accumulation point $x$ that is not a limit of any sequence in $A$. (For instance, $X$ could be the Arens-Fort space.) Now take a sequence $(x_n)$ which takes each value in $A$ infinitely many times. The set of subsequential limits of $(x_n)$ then includes every element of $A$, but does not include $x$, and so is not closed.
Eric Wofsey
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I was searching for counterexamples along these lines... very nice. The key step is in taking a sequence attaining every value in $A$ infinitely many times; how can this be achieved? I have never seen such a construction – FShrike Jul 25 '22 at 19:39
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