# Intuition on the Topological definition of continuity, considering the special case of the step function.

I'm trying to get an intuition for open sets and topological reasoning in general. One example I want to understand is the step function, and specifically why it would be considered discontinuous under the topological definition of continuity. For instance, suppose I define the step function as:

f(x) = the smallest integer larger than x.

The simple example here implies that f is a map from R->R (which is implicitly thought of as having the open ball topology? Could we think of the step function as continuous under some other topology?), then it seems trivially true that the step function is continuous because both spaces have the same open sets.

In contrast to the last statement, I do understand that it's not just about having the same open sets, but about the preimage of the open sets in the output space being open in the input space.

However, I get extremely confused here because there is no open set under the open ball topology which is covered by the mapping. That is to say that f(x) always outputs an integer, and a single integer in R does not constitute an open set given the open ball topology.

My initial thought was that perhaps this fact is the reason the function is called discontinuous, but then a constant function, for instance f(x) = 3, would also be considered discontinuous under identical reasoning as there is no open ball which contains only 3 in R.

I believe that my confusion is somehow related to nuances of the distinction between the image set and range of a mapping. As a follow up question, I could imagine adding further complications. Suppose I chose to think of my step function as a map from R->C where the image set is simply the integers with nonzero imaginary part. How could I show that the mapping is discontinuous in this case?

Take the left-continuous unit step function with the jump at zero. Consider the inverse image of $(-1/2,1/2)$. It is $(-\infty,0]$. Intuitively this means that you can't go any distance to the right of $0$ and still have values within $1/2$ of $f(0)=0$.
For the right continuous one, the inverse image of $(1/2,3/2)$ is $[0,\infty)$. Intuitively this means that you can't go any distance to the left of $0$ and still have values within $1/2$ of $f(0)=1$.
For ones which are neither left nor right-continuous, you can choose $\epsilon$ small enough that the inverse image of $(f(0)-\epsilon,f(0)+\epsilon)$ is $\{ 0 \}$. Intuitively this means that you can't go any distance in either direction from $0$ and have values within $\epsilon$ of $f(0)$.