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A filter $\mathcal F $ converges to $x$ in a topological space ($X$ , $T$) if and only if the net associated with the filter converges to $x$.

I have done the first part i.e if $\mathcal F $ converges to $x$ ,then the associated net $\mathcal S_\mathcal F $ converges to $x$.

(Note : The associated net $\mathcal S_\mathcal F$ is defined from the directed set ($D$ ,$\ge$) to $X$ by $\mathcal S(x,F)$ =$x$, where $D$ $=$ {($x, F$) $\in$ $X \times \mathcal F $ : $x \in F$} also ($x,F$)$ \ge$ ($y,G$) iff $F \subseteq G$ . )

I am stuck at the converse part.

Any insight. Thank you.

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$\newcommand{\SF}{\mathcal{S}_\mathcal{F}}$First, suppose $\SF$ converges to $x$. What does the definition say ? For every neighborhood $V$ of $x$, what do we have ? Try to get from that an element of $\mathcal{F}$ contained in $V$.

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  • $\begingroup$ But I think we have to show that the neighborhood system of $x$ is in $\mathcal F$ . Isn't it ? $\endgroup$ – hiren_garai Jul 29 '17 at 4:57
  • $\begingroup$ Yes, but remember that $\mathcal{F}$ is a filter. What properties does it have ? $\endgroup$ – Idéophage Jul 29 '17 at 4:59
  • $\begingroup$ Oh ! I got it...in that case V will be in $\mathcal F$.. $\endgroup$ – hiren_garai Jul 29 '17 at 5:01
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Suppose $S_{\mathcal{F}} \to x$ for some filter $\mathcal{F}$.

Let $O$ be a neighbourhood of $x$. Then by the definition of net convergence there is some $d_0 \in D$ such that

$$\forall d \ge d_0: \mathcal{S}(d) \in O\text{.}$$

$d_0 = (p, A)$ for some $A \in \mathcal{F}$ and $p \in A$.

But then let $a \in A$ arbitrary: then (as $A\subseteq A$) $(a,A) \ge (p,A) = d_0$ so $a = \mathcal{S}((a,A)) \in O$.
As this works for all $a \in A$ we have $A \subseteq O$, and as $A \in \mathcal{F}$ we also have $O \in \mathcal{F}$.
But this works for all neighbourhoods $O$ of $x$ so $\mathcal{F} \to x$.

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  • $\begingroup$ Got it.Thanks for the answer.. $\endgroup$ – hiren_garai Jul 29 '17 at 14:24

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