A net on X is a map w from a directed set to X. In the proofs such as: Let (X, T) be a topological space. If every net in X converges to at most one point. Then the space is Hausdorff. To prove the contra-positive we must find a net. The proof goes that for the directed set we choose the open sets containing x (Dx), the open sets containing y (Dy) and take their direct product. We say (,)≤(,) iff ⊇ and ⊇. Then it can be shown that for each ∈ Dx and ∈ Dy, fix a point x ∈U∩V so that w(U,V)=x. Clearly, the net converges to both x and y.
My question for our directed set, we just take the whole topology, and let x≤y if x⊇y, and w(E) = an arbitrary point E, doesn't this net in fact converge to every point, since for every net E, all the nets Q greater than E are subsets of it, so that w(Q)∈E?
The problem is that if $E$ is the empty set, you can't choose $w(E)$ to be an element of $E$! And if you omit the empty set from the index set of your net, the index set will (typically) no longer be directed, since if $U$ and $V$ are two disjoint open sets, then they would have no upper bound.
(If the topology has the unusual property that any two nonempty open sets have nonempty intersection, then it would work to omit the empty set from the index set, and this would give a net that converges to every point. But such a net only exists for topological space that have this unusual property.)