Continuous function rational domain. Is the function defined by
$$\begin{align}
\mathbb{Q} &\rightarrow \mathbb{R}\\
q &\mapsto \frac{1}{q-e}
\end{align}$$
a continuous function?
 A: Let $\;(a,b)\subset\Bbb R\;$ , and assume for simplicity $\;0<a<b\;$ , with the other cases being similar. Denote by $\;f\;$ your function (assuming $x=q\;$ in your question):$${}$$
$$f^{-1}(a,b)=\left\{\,q\in\Bbb Q\;;\;\;\frac1{q-e}\in (a,b)\,\right\}=\left\{\,q\in\Bbb Q\;;\;\;a<\frac1{q-e}<b\,\right\}=$$$${}$$
$$=\begin{cases}\left\{\,q\in\Bbb Q\;;\;\;q<\frac1a+e\;,\;\;\frac1b+e<q\in (a,b)\,\right\}=\left(\frac1b+e,\,\frac1a+e\right)\cap\Bbb Q,&q\ge e\\{}\\
\left\{\,q\in\Bbb Q\;;\;\;q>\frac1a+e\;,\;\;\frac1b+e>q\in (a,b)\,\right\}=\left(\frac1a+e,\,\frac1b+e\right)\cap\Bbb Q,&q< e\end{cases}$$$${}$$
and in both cases the set is open in the relative topology of $\;\Bbb Q\;$ in $\;\Bbb R\;$ , so the function is continuous (assuming the usual, Euclidean topology on $\;\Bbb R\;$ ).
A: Yes, of course. Even it is continuous as a function from $\mathbb{R} \setminus \{e\}$ (that is $\mathbb{R}$ without $e$) to $\mathbb{R}$.
A: You can also argue by sequential continuity: If $q\in\mathbb Q$ then $q\neq e$. Let $q_n\to q$, then because the function $f(x)=\frac{1}{x-e},\,f:\mathbb R\to\mathbb R$ is continuous at each $x\neq e$, then you have that $f(q_n)\to f(q)$ and so $f$ is continuous at each $q\in\mathbb Q$.
A: Note that if a function $g:\mathbb{Q}\rightarrow \mathbb{R}$, such that $g(x)=x-e$ is continuous since $x\neq e$ for all $x\in\mathbb{Q}$ since for any $\epsilon'>0$, $|g(x)-g(y)|=|x-y|<\delta:=\epsilon'$ whenever $|x-y|<\delta$.
Now let $f(x)=\frac{1}{g(x)}$, and let $\epsilon>0$ be given. Let $y\in\mathbb{Q}$. As $g(y)\neq 0$, choose $\epsilon'=\frac{\frac{\epsilon}{2}(g(y))^2}{1+\frac{\epsilon}{2}|g(y)|}$ so that $|g(y)|-\epsilon'>0$.  
Now consider the following for all $x\in(y-\delta,y+\delta)$ 
\begin{equation}
|f(x)-f(y)|=|\frac{1}{g(x)}-\frac{1}{g(y)}|=\frac{|g(x)-g(y)|}{|g(x)||g(y)|},
\end{equation}
which gives
\begin{equation}
\begin{split}
|f(x)-f(y)|&=\frac{|g(x)-g(y)|}{|g(y)|}|\frac{1}{g(x)}-\frac{1}{g(y)}+\frac{1}{g(y)}|,\\
&\leq \frac{\epsilon'}{|g(y)|}(|f(x)-f(y)|+\frac{1}{|g(y)|})\\
\end{split}
\end{equation}
which gives
\begin{equation}
(1-\epsilon'/|g(y)|)|f(x)-f(y)|\leq\frac{\epsilon'}{|g(y)|^2},
\end{equation}
or \begin{equation}
|f(x)-f(y)|\leq\frac{\epsilon'}{|g(y)|^2(1-\epsilon'/|g(y)|)}=\frac{\epsilon}{2}<\epsilon,
\end{equation}
which proves the continuity of $f$ at $y$.
Remark: The above steps essentially prove that
for any real valued continuous function $f$ on a subset $A$ of $\mathbb{R}^k$, such that $f(x)\neq 0$, the function $g(x)=\frac{1}{f(x)}$ is also continuous.
