# Why exactly do we need union and finite intersection of open sets to be open?

If we define the continuity of a function $$f: X \rightarrow Y$$ at a point $$x$$ (with respective to the topology $$(X, T)$$) in the obvious way, except that “open” in the context means belonging to a given subfamily $$T\subset P(X)$$, we are still able to prove the following result:

$$\textbf{Proposition 1. }$$ A function $$f: (X, T_X) \rightarrow (Y, T_Y)$$ is continuous if and only if for any $$U$$ open in $$(Y, T_Y)$$, $$f^{-1}(U)$$ is a union of $$V_i(i\in I)$$ for some $$V_i\in T_X$$.

Note that composition of continuous functions is also continuous.

Moreover, to prove that connectedness and compactness are invariant under continuous functions we can utilize the above proposition. It turns out that both are true.

However, there are obviously some results that requires the topology axioms, for example, we need both union and finite intersection axioms when trying to prove the following propositions:

$$\textbf{Proposition 2. }$$ (Need all two) Every compact subset of a Hausdorff space is closed.

$$\textbf{Proposition 3. }$$ (Need finite intersection axiom) $$\overline{A}\cup \overline{B} = \overline{A\cup B}$$.

Are there any fundamental results which requires the full strength of topology axioms? Why exactly do we need all of them if most fundamental results still hold to be true?

• We might want to patch together functions $f_n$ such that they agree on the intersection of open sets, which is needed for example in the definition of topological manifolds. In fact, maybe more generally, coverings make more sense when you are allowed to take intersections and unions. – twnly Nov 10 '18 at 0:09
• How do you prove that the sum of two continuous functions is continuous? – bof Nov 10 '18 at 1:48
• Why do we need to generalize the notion of a topology, if the standard notion is already general enough for all the applications? Exactly what do we gain from your proposed generalization? What new applications do you have in mind? – bof Nov 10 '18 at 1:52

It relies on the fact that if $$X$$ is a topological space on which every ultra filter converges, then $$X$$ is compact.