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Suppose we take a function $f:E \to F$, where $E = \mathbb{Q}$, and $F$ is an arbitrary set.

Is it possible for $f$ to be continuous (and thus have a chance at being differentiable)? Or would it rather be considered to be defined on singular isolated points?

If its continuity isn't out of the question, then what is the exact criteria placed on $E$ in order for $f$ to even have a chance at being continuous? Could, say, a function defined on $\{\frac{a}{b}$ $|$ $a,b\in \mathbb{Z}, $ $b$ odd$\}$ be continuous?

virtualcode
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  • https://math.stackexchange.com/q/2524230/42969 – Martin R Nov 22 '23 at 20:32
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    Every function can be continuous or not, depending on the topology you consider. – TheSilverDoe Nov 22 '23 at 20:36
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    If $F$ is an arbitrary set, then what would it even mean for $f:E\to F$ to be differentiable? – Joe Nov 22 '23 at 20:40
  • @TheSilverDoe Unfortunately, I don't know what that means. We haven't studied topology yet. – virtualcode Nov 22 '23 at 20:41
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    If we assume that $F\subseteq\mathbb R$, then the idea of $f$ being differentiable at some $q\in\mathbb Q$ makes more sense. However, typically one requires that for $f$ to be differentiable at $x$, it must be the case that there is an open set $U$ of $\mathbb R$ such that $x\in U$ and $U\subseteq\operatorname{dom}(f)$. However, it is possible to define the derivative in a more general case – see here for instance. Whether that definition is useful is open to debate... – Joe Nov 22 '23 at 20:46
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    @TheSilverDoe a constant function is always continuous – Andrew Nov 22 '23 at 20:46
  • @Joe I see, so you mean that $f$ has to be defined on some neighbourhood of $x$ for us to even start thinking about differentiability. That makes sense. – virtualcode Nov 22 '23 at 20:51
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    @virtualcode: Yes, in what I would say is the most common definition of differentiability, we need $f$ to be defined in a neighbourhood of $x$. (I just didn't use the word neighbourhood in my comment, in case you weren't familiar with it.) It is possible to define the derivative more generally, but this "generalised derivative" does not have very nice properties – see this answer for example. – Joe Nov 22 '23 at 20:55
  • @Joe Thank you. Your answer definitely helped. – virtualcode Nov 22 '23 at 20:59
  • @virtualcode: No problem. I’m glad I could help. – Joe Nov 23 '23 at 08:04
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    I agree with the -at the moment 3- closing votes for lack of clarity and focus, so I shall answer only in comments. To follow the rules on this site and make it easier to answer, your post should restrict to one question (here: continuity, forget about differentiability), and bring some context. 1/3 "$F$ is an arbitrary set": For your question about continuity to make sense, it cannot be just a set, it must be a topological space. 2/3 "would it rather be considered to be defined on singular isolated points?": the points of $\Bbb Q$ are not isolated, and "singular" makes no sense here. – Anne Bauval Nov 23 '23 at 10:16
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    3/3 (your main question) "Could a function defined on $\Bbb Q$ or on ${\frac ab\mid a,b\in\Bbb Z,b\text{ odd}}$ be continuous?", "what is the exact criteria placed on $E$ in order for $f:E\to F$ to even have a chance at being continuous?" : what is your definition of continuity? a usual one is: $f:E\to F$ is continuous at $a\in E$ if $$\lim_{x\to a}f(x)=f(a).$$ – Anne Bauval Nov 23 '23 at 10:16
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    And there is no restriction on the topological space $E$ in this definition. If for instance $E$ is a subspace of $\Bbb R$, $\lim_{x\to a}f(x)=b$ if for every neighborhood $V\subset F$ of $b$, there exists some $\delta>0$ such that $$\forall x\in E\quad(0<|x-a|<\delta\implies f(x)\in V).$$ – Anne Bauval Nov 23 '23 at 10:16

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