# A result in topology

1. Sep 27, 2007

### quasar987

I just discovered the following. But since half the things I find in topology turn out to be wrong, I feel I better check with you guys.

What I convinced myself of this time is that if you have a function f:(X,T)-->(Y,S) btw topological spaces, and S' is a basis for S, then to show f is continuous, is suffices to show that $f^{-1}(S') \subset T$.

In words, that is because every open set of S can be written as a union of sets of S', and the operations of f^{-1} and union commute. (and that a union of open sets is open)

2. Sep 27, 2007

### Hurkyl

Staff Emeritus
That looks right. In fact, it's rather often used -- e.g. look at the definition of continuity you learned in calc I.

3. Sep 28, 2007

### SiddharthM

it's a fact!

4. Sep 28, 2007

Thanks.