When is it enough to show a property of a function holds on a basis of a topology? Say, we have a topological space $X$ and a function $f: X \to X$. Sometimes it is enough to show a property of a function holds on a basis of a topology. For example, it can be shown that if we want to show that $f$ is continuous it's enough to show that the preimages of the sets in a basis of $X$ are open.
Now take the following property $f$ might have:

A map $f: X \to X$ is called topologically mixing if for all non-empty open sets $U,V$ there exists $m \in \mathbb N$ such that  for all $n \geq m$ holds $f^n(U) \cap V \neq \emptyset.$

Several texts show this property on a basis without showing why it's enough.
Should one show for each property separately that it's enough to show it on a basis? Or is there some higher order result that legitimizes it for certain types on properties?
Added:
To clarify: I gave this particular property (mixingness) as an example. My question is: with a property defined in terms of topology, when is it enough to show that 

$f$ has property P

by showing it on the basis only?


*

*always

*sometimes: i.e. when the property satisfies such-and-such condition (which?)

*one can't say in advance, it needs to be shown for each particular property separately

 A: Using your definition of mixing let $U$ and $V$ be given. Assume $f$ is mixing whenever the two open sets are basic. By definition we can select basic $U' \subset U$ and $V' \subset V$. By our assumption we know $f^n(U') \cap V' \ne \varnothing$ for large $n \in \mathbb N$. Now replacing $U'$ with $U$ and $V'$ with $V$ only makes the LHS larger so we have $f^n(U) \cap V \ne \varnothing$ for large $n \in \mathbb N$ as well.
The above is an example of a more general phenomenon. Suppose $Q(U_1, U_2, \ldots, U_r)$ is some proposition involving the open sets $U_1, U_2, \ldots, U_r$. For topological mixing we have $r=2$ and . . . 

$Q(U_1,U_2) =$ there exists $m \in \mathbb N$ such that  for all $n \geq m$ holds $f^n(U_1) \cap U_2 \neq \emptyset.$

Suppose moreover that $Q$ has the property that whenever $Q(U_1, U_2, \ldots, U_r)$ is true and every $U_r \subset U_r'$ then $Q(U_1', U_2', \ldots, U_r')$ is true as well. 
Then $P=\forall U_1 \ldots \forall U_r Q(U_1, \ldots, U_r)$ holding over all basis elements implies that $P$ holds over all members of the topology.
