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?
Best Answer
Take a look at the proof of Tychonoff's theorem through filters;
It relies on the fact that if $X$ is a topological space on which every ultra filter converges, then $X$ is compact.
The notion of filter (and specifically filter convergence) makes no sense if we don't require closure under unions and finite intersections.
A more basic example is the fact that every compact Hausdorff space is normal. The proof I know relies on using both axioms to construct separating open sets.