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Let {s(a)} -such that (a) belongs to order set (A)- is a net from the point of (X) , the net {s(a)} converges to (x) if and only if the filter that generated by it converges to (x)

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So you have a net $s: A \rightarrow X$ from a directed set (presumably) $A$ to a space $X$. The filter generated from $s$, call it $\mathcal{F}_s$ is defined as $\{F \subset X: \exists a_0 \in A: \{s(a): a \ge a_0 \} \subseteq F \}$. Then $s \rightarrow x$ iff $\mathcal{F}_s \rightarrow x$.

This is quite clear from the definitions of convergence. Take any open neighbourhood $O$ of $x$. When the net converges to $x$, some tail of the net must be a subset of $O$, and so this $O$ is by definition (!) in $\mathcal{F}_s$. This proves one direction and the other is almost the same.

When you write out the definitions it's basically true by construction.

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