1. ## 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.

2. If $f$ is continuous at $x=a~~f(a)>0$ in the definition of continuity use $\varepsilon = \frac{{f(a)}}{2}$.

It follows that $f(x)>\frac{{f(a)}}{2}$ for all $x$ near $a$.

3. Thanks Plato, I am considering what you said.

Originally Posted by Plato
If $f$ is continuous at $x=a~~f(a)>0$ in the definition of continuity use $\varepsilon = \frac{{f(a)}}{2}$.

It follows that $f(x)>\frac{{f(a)}}{2}$ for all $x$ near $a$.

4. 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?

5. 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.

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