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First we show the forward implication. Assume \(x\in \overline{A}\text{.}\) Let \(I\) be the set of all pairs \(i=(U,a)\text{,}\) where \(U\) is an open neighborhoods of \(x\text{,}\) and \(a\in U\cap A\text{.}\) The set \(I\) is nonempty since \(x\in \overline{A}\text{.}\) For elements \(i=(U,a)\) and \(i'=(U',a')\) of \(I\) define \(i\leq i'\) if and only if \(U\supseteq U'\text{.}\) It is easy to see that \(I\) is a quasi-ordered and directed set with respect to this relation. Lastly, define \(f\colon I\rightarrow A\) as \(f((U,a))=a\text{.}\) The net \(f\) converges to \(x\text{.}\) Indeed, for any open set \(V\) containing \(x\text{,}\) there is an element \(a\in V\cap A\text{.}\) Setting \(i_0=(V,a)\text{,}\) if \(j=(U,a')\) satisfies \(j\geq i_0\text{,}\) then \(U\subseteq V\) and hence \(f(j)=f((U,a'))=a'\in V\text{,}\) as desired.
For the reverse implication, assume \(x_i\rightarrow x\) for some net \((x_i)_{i\in I}\) in \(A\text{.}\) Given any open set \(U\) containing \(x\text{,}\) there is an \(i_0\in I\) such that \(x_j\in U\) for all \(j\geq i_0\text{.}\) In particular, we have \(x_{i_0}\in U\text{.}\) Since \((x_i)_{i\in I}\) is a net in \(A\text{,}\) we conclude that \(x_{i_0}\in\cap A\cap U\text{.}\)
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Assume \(f\) is continuous and \((x_i)_{i\in I}\) is a net satisfying \(x_i\rightarrow x\text{.}\) Given any open set \(U\) containing \(y\text{,}\) the net \((x_i)_{i\in I}\) is eventually in \(f^{-1}(U)\text{,}\) since this is an open set containing \(x\text{.}\) But then clearly \((f(x_i))_{i\in I}\) is eventually in \(f(f^{-1}(U))\subseteq U\text{.}\)
Inversely, if \(f\) is not continuous, then there is an \(x\in X\) and open set \(U\) containing \(y=f(x)\) such that for all open sets \(V\) containing \(x\text{,}\) there is an element \(x'\in V\) such that \(f(x')\in f(V)-U\text{.}\) Let \(I\) be the set of all pairs \((V, z)\text{,}\) where \(V\) is an open neighborhood of \(x\) and \(f(z)\in f(V)-U\text{.}\) The set \(I\) is nonempty and is directed under the quasi-ordering \((V,z)\leq (V',z')\) if and only if \(V\supseteq V'\text{.}\) Define the net \(h\colon I\rightarrow X\) as \(h((V,z))=z\text{.}\)
The net \(h\) converges to \(x\text{:}\) given any open neighborhood \(V\) of \(x\text{,}\) there is a \(z\in V\) such that \(i_0=(V,z)\) is an element of \(I\text{;}\) for all \(j=(V',z')\geq i_0=(V,z)\text{,}\) we have \(V\subseteq V'\) and hence \(h(j)=h((V',z'))=z'\in V'\subseteq V\text{.}\)
On the other hand, the net \(f\circ h\text{,}\) defined as \(f\circ h((V,z))=f(z)\) does not converge to \(f(x)\text{.}\) Indeed by construction, for all \(j=(V,z)\in J\) we have \(f(z)\notin U\text{.}\)
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Assume \(X\) is Hausdorff and that the net \((x_i)_{i\in I}\) converges to \(x\in X\text{.}\) Given any \(y\ne x\text{,}\) pick disjoint open sets \(U, V\) containing \(x\) and \(y\text{,}\) respectively. I claim that \((x_i)_{i\in I}\) is not eventually in \(V\text{.}\) Indeed, since \(x_i\rightarrow x\text{,}\) there is an \(i_0\in I\) such that \(x_i\in U\) for all \(i\geq i_0\text{.}\) If, similarly, there were a \(j_0\in I\) such that \(x_j\in V\) for all \(j\geq j_0\text{,}\) then we could find an index \(k\) satisfying \(k\geq i_0\) and \(k\geq j_0\text{,}\) in which case \(x_k\in U\cap V=\emptyset\text{.}\) Contradiction! We conclude that \((x_i)_{i\in I}\) is not eventually in \(V\text{,}\) and hence that \((x_i)_{i\in I}\) does not converge to \(y\text{.}\)
Now assume that \(X\) is not Hausdorff, as witnessed by elements \(x\ne y\in X\text{:}\) this means for all open sets \(U, V\) containing \(x\) and \(y\text{,}\) respectively, we have have \(U\cap V\ne \emptyset\text{.}\) Let \(I\) be the set of all triples \((U,V, z)\text{,}\) where \(U, V\) are open neighborhoods of \(x\) and \(y\text{,}\) respectively, and \(z\in U\cap V\text{.}\) Declare \((U,V,z)\leq (U',V', z')\) if and only if \(U\supseteq U'\) and \(V\supseteq V'\text{.}\) The set \(I\) is directed: given \(i=(U, V, z_i), j=(U',V',z_j)\in I\text{,}\) we can take \(k=(U\cap U, V\cap V', z_k)\text{,}\) for some \(z_k\in U\cap U'\cap V\cap V'\text{.}\) Now, let \(f=(z_i)_{i\in I}\) be the net in \(X\) defined as follows: given \(i=(U, V,z)\text{,}\) define \(f(i)=z\text{.}\) I claim that \(z_i\rightarrow x\) and \(z_i\rightarrow y\text{.}\) Indeed given any open set \(U\) containing \(x\) and open set \(V\) containing \(y\text{,}\) there is an element \(z\in U\cap V\text{.}\) Let \(i_0=(U,V,z)\in I\text{.}\) For any \(i'=(U',V', z')\geq i_0\) we have \(f(i')=z'\in U'\cap V'\subseteq U\cap V\text{.}\) Thus \(f=(z_i)_{i\in I}\) is eventually in \(U\) and eventually in \(V\text{,}\) as desired.