Let be $X$ and we assume that $(x_\lambda)_{\lambda\in\Lambda}$ is a net such that there exists a cofinal and increasing map $\varphi$ form $\Bbb N$ to $\Lambda$ such that $\big(x_{\varphi(n)}\big)_{n\in\Bbb N}$ converges to any $x_0\in X$: so if $X$ is first countable then the convergence of $\big(x_{\varphi(n)}\big)_{n\in\Bbb N}$ to $x_0$ implies the convergence of $(x_\lambda)_{\lambda\in\Lambda}$ to $x_0$? If the answer to the last question is negative then there is a relevant reason different form this or this or rather this that show why into first countable space is sufficent to consider only sequence?
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Antonio Maria Di Mauro
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3A usual convergent subsequence of a non convergent sequence provides a counter example to your question. – Ruy May 17 '22 at 21:00
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@Ruy Okay, so there are no other reasons other than the ones I listed for considering sequences instead of nets? – Antonio Maria Di Mauro May 17 '22 at 21:26
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1Yes, these are the main reasons. In a 1st countable compact space every sequence has a converging subsequence but unfortunately the converse is false. – Ruy May 17 '22 at 23:03
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As Ruy suggested above into the comments the conjecture is generally false: e.g. the sequence $(x_n)_{n\in\Bbb N}$ defined as $$ x_n:=\begin{cases}0,\,\text{if }n\,\text{is even}\\n,\,\text{if }n\,\text{is even}\end{cases} $$ is a sequence into a first countable space that does not converege but it has trivially a converging subsequence.
Antonio Maria Di Mauro
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