Open subsets above and below f(x), proving continuity of f(x)

[frito898]

Homework Statement

Let f:R-->R be a function. Define A={(x,y) \in R2: y<f(x)}, B={(x,y) \in R2: y>f(x)}, i.e A is the subset of R2 under the graph of f and B is the subset above the graph of f. Show that if A and B are open subsets of R2, then f is continuous

Homework Equations

N/A

The Attempt at a Solution

I tried to prove this by contradiction
Working under the assumption that f is discontinuous of the first kind
case 1: f(x-)=f(x+) \neq f(x)
Assume without loss of generality that f(x)>f(x-)=f(x+)
there is a f(x-)=f(x+)<y<f(x)
There is no neighborhood around y with a radius greater than zero that doesn't contain points above the function
Therefore, y is not an interior point of A
Therefore, A is not open.
Case 2: f(x-) \neq f(x+)
...

I have proven the cases of simple discontinuities, but I have no idea how to approach discontinuities of the second kind. I don't think contradiction even works, so I am back at square 1. Any ideas? Thanks in advance!

 

[lanedance]

how about trying assuming A or B are not open, then there exists a point in A whose neighborhood contains a point of f or B