$g(x)=\lfloor x \rfloor$ is continuous in $a$ $\Leftrightarrow$ $a \in \mathbb{R} \backslash \mathbb{Z}$

Question

$g(x)=\lfloor x \rfloor$ is continuous in $a$ $\Leftrightarrow$ $a\in \mathbb{R} \backslash \mathbb{Z}$

So this is quite obvious to me, but I don't know how to start proving this statement. My definition of a Continuous function is:

For $D\subset \mathbb{C}$, $a\in D$ and $f: D\to \mathbb{C}$ the following statements have to be true:

1. $f$ is continuous in $a$ if for every sequence $(a_n)_n \subset D$ with $\lim \limits_{n\to \infty} a_n=a \Rightarrow \lim \limits_{n\to \infty} f(a_n)=f(a)$

2. $f$ is called continuous in $D$ if $f$ is continous in every $z\in D$.

So I would need to prove that if $a \in \mathbb{Z}$, the implication in 1. can't be satisfied.

Answer 1

If $a \in \mathbb{Z}$ then $a_n = a - \frac{1}{n+1}$ is a sequence such that $\lim_{n \to \infty} a_n = a$ but as for all $n$: $f(a_n) = a-1$ (being in the interval $[a-1,a)$, so it's rounded down), we have $$\lim_{n \to \infty} f(a_n)=a-1 \neq f(a)=a$$ so $f$ is not continuous at $a$ by your criterion 1.

If $a \notin \mathbb{Z}$ with $f(a)=z \in \Bbb Z$ so that $a \in (z,z+1)$ and $a_n$ is a sequence converging to $a$, for large enough $n$ all $a_n \in (z,z+1)$ too (as $(z,z+1)$ is open), so $f(a_n)=z$ for all large enough $n$ and so 1 is satisfied and $f$ is continuous at $a$.

Answer 2

For $a \in \mathbb{Z}$ take the sequence $a_n:= a-\frac{1}{n}$.

Then $\lim_{n \rightarrow \infty}a_n=a$, but $f(a_n)=a-1$ for all $n \in \mathbb{N}$ and therefore $\lim_{n \rightarrow \infty} f(a_n)=a-1 \neq a= f(a)$

Now take $a \in \mathbb{R}-\mathbb{Z}$ such that $n-1 < a <n$, $n \in \mathbb{Z}$. Now if we restrict to a neighborhood $U$ of $a$ such that $U \subseteq B_{(n-a)/2}(a)$ we can see that $g(x)=n-1$ for all $x \in U$, that it's continuous