If $x\in\mathbb{R}\setminus\mathbb{Z}$ and if $x_n \to x $ then $\lfloor x_n \rfloor \to \lfloor x \rfloor$ I need to prove that if $(x_n)$ is a real sequence converging to some $x\in\mathbb{R}\setminus\mathbb{Z}$ then $\lfloor x_n \rfloor \to \lfloor x \rfloor$.
I'll be using the continuity of $f(x) = \lfloor x \rfloor$ on $\mathbb{R}\setminus\mathbb{Z}$ to prove this.  For any $c \in \mathbb{R}\setminus\mathbb{Z}$, we have $\lim_{x \to c} \lfloor x \rfloor = \lfloor c \rfloor$, hence, $f$ is continuous on $\mathbb{R}\setminus\mathbb{Z}$.
Now let $(x_n)$ be any real sequence converging to some $x \in \mathbb{R}\setminus\mathbb{Z}$. Then by definition of continuity, we have $\lim_{n\to\infty} f(x_n) =\lim_{n\to\infty} \lfloor x_n \rfloor = \lfloor x \rfloor$.
Is my claim and proof correct?
 A: Here is a possible argument: since $x$ is not an integer, it is contained in some interval $(k, k+1)$ with $k \in \mathbb{Z}$. This is not hard to prove, try making a drawing and play with the floor function.
Now, since $x_n \to x$, there exists $n_0 \in \mathbb{N}$ such that $x_n \in (k,k+1)$ for $n \geq n_0$. To prove this, take $\varepsilon > 0$ so that $(x - \varepsilon, x + \varepsilon) \subseteq (k,k+1)$ by for example considering the minimum distance from $x$ to $k$ and $k+1$, and use the definition of convergence of sequences for this $\varepsilon$.
Finally, since $x_n \in (k,k+1)$ for $n \geq n_0$, by definition of the floor function we have that $[x_n] = k = [x]$ and so $|[x_n] - [x]|  = 0$ for all $n \geq n_0$. This implies that $[x_n] \to [x]$.
A: You do not have the continuity of $f(x)$, but you DO have, for each $x \in \mathbb{R} \setminus \mathbb{Z}$, continuity of $f(x)$ in a small enough ball $B_x$ around $x$ [I think as far as the exercise it suffices for the student to note this fact?]. That is all you need: For each such $x$ there is an $n_0$ such that $x_{n'} \in B_x$ for all $n' \ge n_0$. And then you can finish as you did on the sequence $\lfloor x_{n'} \rfloor; n' \ge n_0$
