[Math] (Dis)continuity of the floor function

Question

I was able to prove that the floor function is continuous on $\mathbb{R}$ \ $\mathbb{Z}$. ( with epsilon-delta criterion ).

But I need help to show that the floor function is discontinuous on $\mathbb{Z}$.
It is easy with right-hand limit and left-hand limit, but I'm searching for an alternative solution. Maybe epsilon-delta criterion?

Answer 1

Surely, for $a \in \mathbb Z$ , you would have said the following :

The left hand limit of the floor function at $a$, is $a-1$, while the right hand limit is $a$. Since these don't match, continuity doesn't happen.

How does one make this into an argument? Well, recall the $\epsilon - \delta$ definition(let $f$ denote the floor function):

For all $\epsilon > 0$, there exists a $\delta > 0$ such that if $|x-a| < \delta$ then $|f(x) - f(a)| < \epsilon$.

So, in $\epsilon-\delta$ terms, what does it mean, for the function to be non -continuous at the point $a$?

There is some $\epsilon > 0$ such that for all $\delta>0$, there is some $x$ such that although $|x-a| < \delta$, yet $|f(x) - f(a)| \mathbf{>} \epsilon$.

Note the change of sign. So how would we prove this? We need an $\epsilon > 0$, first.

The difference between right and left hand limit above is $a-(a-1) = 1$. I claim that any $\epsilon < 1$ does the job.

For example, let's take $\epsilon = \frac 12$. Now, for all $\delta$, we need an $x$, such that although $|x-a| < \delta$, yet $|f(x)-f(a)| < \frac 12$.

What better thing, than to take something just smaller than $a$? We know that the left hand limit of the floor function at $a$ is $a-1$, so from the left, as $x$ gets closer to $a$, $f(x)$ gets closer to(infact, is equal to) $a-1$, right?

Let's take $x = a - \frac \delta 2$. This is smaller than $a$. Furthermore, $|x-a| = \frac \delta 2 < \delta$. However, $f(x) = a-1$, since $a-1 < x < a$. Therefore, $|f(x) - f(a)| = 1 > 0.5$.

Hence, since we found an $\epsilon > 0$ working for all $\delta$, the discontinuity of $f$ at any integer point $a$ is proved.

This proves, along with continuity on $\mathbb R\backslash Z$, that the floor function is continuous precisely on $\mathbb R \backslash Z$.