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The question I am asking is actually I have asked here.

I have solved the problem in the following way:

Proof: We have every convergent net in $X$ to be eventually constant....$(1)$

We have to show that the Topology with respect to which the nets are convergent on $X$ is Discrete i.e every singleton is open.

Let us define $\mathcal T=\{G\subset X\,:\,\text{ No net in $X\setminus G$ converges in $G$}\}$.

Then obviously $\mathcal T$ is a topology on $X$. Now from the construction of $\mathcal T$, and the hypothesis $(1)$ we can say that no net in $X\setminus\{x\}$ can converge in $\{x\}$, (because if any net from $X\setminus\{x\}$ converges to {$x$} then $S$ is eventually equal to $x$, hence $S$ can't be in $X\setminus\{x\}$) i.e to say that $\{x\}$ is open. Hence $\mathcal T$ is Discrete.

Is my proof okay ? Am I missing something ? If yes then please let me know. Thank you

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  • $\begingroup$ Why is it true that $\emptyset\in\mathcal T$? $\endgroup$ – José Carlos Santos Aug 5 '17 at 14:09
  • $\begingroup$ No net in $X$ $-$ $\emptyset$ i.e in $X$ can converge in $\emptyset$. $\endgroup$ – hiren_garai Aug 5 '17 at 14:23
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    $\begingroup$ Right. I misunderstood the definition of $\mathcal T$. Please use $\setminus$ instead of $-$ when working with sets. $\endgroup$ – José Carlos Santos Aug 5 '17 at 14:27
  • $\begingroup$ Assuming that $\mathcal T$ is a topology, it looks like all that you proved was that it's the discrete topology. How do you deduce from that that the original topology is discrete? $\endgroup$ – José Carlos Santos Aug 5 '17 at 14:36
  • $\begingroup$ Ok sir..Is my proof ok ? $\endgroup$ – hiren_garai Aug 5 '17 at 14:36
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It's not true necessarily that $(x_i) \to x$ and the net is eventually constant than that constant must be $x$. So your proposed proof does not work as it stands.

E.g. in the space $X= \{0,1,2\}$ with open sets $\{\emptyset, \{1,2\}, X\}$ the constant net with only the value $2$ converges to $1$ and $0$ as well...

Note that this is a "local counterexample" (Lakatos' term), it shows we need a stronger assumption for the proof you propose to work:

(1) For every net $(x_i)_i$ that converges to some $x \in X$, there is some $i_0 \in I$ such that for all $i \ge i_0$: $x_i = x$ (so not just constant but constant with constant value the limit, which is stricter, but indeed holds in discrete spaces).

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  • $\begingroup$ But I have said that the net is eventually constant,not a constant...still I am wrong ? $\endgroup$ – hiren_garai Aug 6 '17 at 0:39
  • $\begingroup$ @HirenGarai you have to strengthen it to "eventually constant and with value = limit", otherwise you cannot show that singletons are open in your way. $\endgroup$ – Henno Brandsma Aug 6 '17 at 5:35
  • $\begingroup$ But if a convergent net is eventually constant and equal to $x$ say then will it not be case that it's limit is $x $ ? $\endgroup$ – hiren_garai Aug 6 '17 at 6:21
  • $\begingroup$ @HirenGarai no, see my example.limits are not unique. $\endgroup$ – Henno Brandsma Aug 6 '17 at 6:22
  • $\begingroup$ I got you sir !! I haven't thought that deep.Thanks for pointing me..So if every convergent net is eventually constant and equal to it's limit,then the space is Discrete ? $\endgroup$ – hiren_garai Aug 6 '17 at 6:30

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