Thread: Open Set Proving

The question is as follows:

Let f:R->R be continuous. Show that {x:f(x)>0} is an open subset of R.

I have figured out how to prove {x:f(x)=0} is a closed subset of R, but I can not solve the question {x:f(x)>0} is an open one.

Any help for that?
Thanks.

If is continuous at in the definition of continuity use .

It follows that for all near .

Thanks Plato, I am considering what you said.

Originally Posted by Plato

If is continuous at in the definition of continuity use .

It follows that for all near .

Thanks, and I get what you said about f(x) > f(a)/2 for all x near a, but why does it support {x:f(x)>0} is an open set?

In order to show that a set is open, you must know the definition.

One characterization of an open is that if a is in then every point near a is also in the set.
Another is that open sets are union of balls.
Another is that its complement is closed.

I get it! Thanks very much. The proving part is tricky.