Hint: More generally, characteristic function of a set $A$ in a given topological space $(\Omega,\tau)$ is continuous exactly on $\Omega\setminus \partial A$. We first note that $\partial A$, $\mathrm{int}(A)$ (interior of $A$) and $\mathrm{int}(\Omega \setminus A)$ (interior of complement of $A$) form a partition of $\Omega$.
If $\omega \notin \partial A$, then $\omega \in \mathrm{int}(A)$ or $\mathrm{int}(\Omega \setminus A)$. So there is a neighborhood of $\omega$ fully included in $A$ (or $\Omega \setminus A$) on which $\chi_A$ is constant, $1$ (or $0$). Its image through $\chi_A$ is included in any given neighborhood of $\chi_A(\omega)$.
If $\omega \in \partial A$, then any neighborhood of $\omega$ has elements in $A$ that map to $1$ and elements in $\Omega\setminus A$ that map to $0$. So, there are neighborhoods of $\chi_A(\omega)$ (those that include $1$, but not $0$, or vice versa) that cannot include any image through $\chi_A$ of any neighborhood of $\omega$.
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 $\mathbb R \backslash Z$.
Best Answer
Any function becomes continuous if you remove its points of discontinuity from the domain.
In your case, the only discontinuity is at $0$, so by removing $0$ from the domain you make the function continuous.