Let $X$ and $Y$ two sets. Given a set $\mathcal{D}\in\mathcal{P}(\mathcal{P}(X))$ one can define a relation $R$ on $Y^X$ saying $fRg$ iff $\{x\in X|f(x)=g(x)\}\in\mathcal{D}$. It is not too hard to show that if $card(Y)\geq3$, $R$ is an equivalence relation iff $\mathcal{D}$ is a filter. Two functions are in the same equivalence classes if they are equal on some important subset of $X$. It makes then sense that we use filters to see the spot where a limit is taken, because "the value of the function does not matter outside of some important subsets".

Hence, some axioms for a topology make sense. In terms of neighbourhoods, we call a topology on $X$ a family $(\mathcal{V}_x)_{x\in X}$ of sets in $\mathcal{P}(\mathcal{P}(X))$ which verifies for all $x\in X$ :

+  $\mathcal{V}_x$ is a filter on $X$
+  $\forall V\in\mathcal{V}_x,x\in V$
+  $\forall V\in\mathcal{V}_x,\exists W\in \mathcal{V}_x, W\subset V\wedge \forall y\in W,W\in \mathcal{V}_y$

The first axiom lets one see that we can use the neighbourhood of $x$ as the spot where the limit is taken. The second one that "it is where $x$ is that matters" (because if the second axiom was false, there would be an important set which does not contain $x$).

What meaning do you give to the third axiom ? I see that it guarantees the equivalence between the usual axioms of a topology using open sets and the ones presented above. Maybe the question can be reformulated this way : why do we need open sets at all ?