Let $f: \mathbb Q \to \mathbb R$ be a continuous function.


An *extension* of $f$ is a function $\tilde f: \mathbb R \to \mathbb R$ such that $\tilde f = f$ on $\mathbb Q$.


We say an extension $\tilde f$ of $f$ is *maximally continuous* if for any other extension $g$ of $f$, we have that if $g$ is continuous at $x \in \mathbb R$, then so is $\tilde f$.



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> **Question:** For any continuous function $f: \mathbb Q \to \mathbb R$, does there exist a maximally continuous extension of $f$?
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**Remark:** One can always obtain an extension that is continuous at every point in $\mathbb Q$, see for example, the answer in [this post](https://mathoverflow.net/questions/394565/extending-continuous-functions-from-mathbb-q-to-mathbb-r) by Fedor Petrov.



Maximally continuous extension of continuous functions from $\mathbb Q$ to $\mathbb R$
There even exists a largest set $X$ to which $f$ can be continuously extended.


The trick is the following result (which I state here in more generality, to point out which topological assumptions one needs):


**Theorem.** Let $X,Y$ be topological spaces, where $Y$ is $T\_3$, and let $D \subseteq X$ be dense. Let $f: D \to Y$ be continuous and assume that the following property is satisfied for each point $x \in X \setminus D$ (equivalently, each point $x \in X$):


$(\*)$ There exists $y\_x \in Y$ such that, for each net $(x\_j)$ in $D$ that converges to $x$, the net $(f(x\_j))$ converges to $y\_j$.


Then $f$ has a (obviously, unique) continuous extension to $X$.


(I gave a proof of this in a topology course a year ago, but unfortunately I don't know a reference - maybe somebody else can help out with a reference?)


**Application to your situation.** For $D = \mathbb{Q}$, let $X$ be the set of all $x \in \mathbb{R}$ which satisfy property $(\*)$. Then $f$ can be continuously extended to $X$, and this extension is clearly the largest one.


