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Is the function defined by

$$\begin{align} \mathbb{Q} &\rightarrow \mathbb{R}\\ q &\mapsto \frac{1}{q-e} \end{align}$$

a continuous function?

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Did you mean $\;x\to\frac1{x-e}\;$ , or else what's the relation between $\;q\,,\,\,x\;$ ? – Joanpemo Feb 25 at 10:31

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 $\;\Bbb R\;$ ).

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Yes, of course. Even it is continuous as a function from R{e} (that is R without e) to R.

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I disagree. The domain $\mathbb{Q}$ is not continuous in the first place. Therefore the image will have some gaps in it and so not continuous. – Question Asker Feb 25 at 10:32
@QuestionAsker: The continuity of the domain is not necessary for all definitions of continuous to apply. I require it on my own courses for freshmen/sophomore, but on more advanced course we use a different definition. Either in terms of a metric or, later, in terms of a topology. With either of those concepts used in the definition of continuity this answer makes perfect sense (and is also correct). – Jyrki Lahtonen Feb 25 at 11:08
Ok thank you. I understand now. – Question Asker Feb 25 at 11:10
Another way to see this is that it is a composition of the inclusion function $i: \mathbb{Q} \rightarrow \mathbb{R} - \{e\}$, $i(x) = x$, with the function $f: \mathbb{R} - \{e\} \rightarrow \mathbb{R}$, $f(x) = \frac{1}{x - e}$, both of which are continuous. – filipos Feb 25 at 13:49

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 $q\in\mathbb Q$.

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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.

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