Ε-δ PROOF of ceiling function

[arpitm08]

Homework Statement

Let f: R -> Z be the ceiling function defined by f(x) = ceil(x). Give a ε-δ proof that if a is a real number that is not an integer, then f is continuous at a.

The Attempt at a Solution

I can prove that f(x) is not continuous at any integer. But i don't know how to prove this. I can do proofs for continuous functions, but I've never done one for a piece wise function. Any help would be awesome. Thanks.

 

[Mark44]

Think about what the graph of this function looks like -- essentially steps that are 1 unit wide. If a is not an integer, it shouldn't be too hard to find a number δ > 0, such that |f(x) - f(a)| < ε. In fact, if x is close enough to a, |f(x) - f(a)| will be 0.

 

[HallsofIvy]

If a is not an integer, then there exist $\delta> 0$ such that every number in the interval from $a- \delta$ to $a+ \delta$ is not an integer. On that interval f(x) is a constant.