Can a function with infinitely many holes in the domain still be continuous?

Question

If we have a function $f:A\rightarrow \mathbb{R}$ where $A$ has infinitely many holes and is subset of $\mathbb{R}$( i.e. $\mathbb{Q} \subset \mathbb{R}$). Then can $f$ be a continuous function still?

The textbook we are using defines a function $f:A\rightarrow \mathbb{R}$ is continuous at a point $c\in A$ if for $(\forall\epsilon > 0)(\exists \delta > 0)$ such that $|f(x) - f(c)|<\epsilon$ whenever $|x - c|<\delta$ and $c\in A$. This seems to imply that we can have a continuous function $f:\mathbb{Q}\rightarrow \mathbb{R}$ by this definition.

I ask because when I look at this proof : Prove that the inverse image of an open set is open. I can't shake this feeling about properly working with the holes of the preimage. Since even as we shrink the $\epsilon$ in the proof, the holes in the rationals should always exist in the set.

I am not sure if my understanding of the definitions is correct which may resolve this problem.

Answer 1

Yes. I think your issue is that you're assigning "blame" to the wrong mathematical object, and/or intuiting the effect backwards.

That $\mathbb{Q}$ is totally disconnected¹ is a fault of $\mathbb{Q}$, not of functions defined on $\mathbb{Q}$. This fault does not impose restrictions on what can be a continuous function. In fact, it's quite the opposite: each "hole" of $\mathbb{Q}$ can be seen as an opportunity for a function to "jump" while still being continuous. For example, the following is a continuous function $\mathbb{Q} \to \mathbb{R}$:

$$ f(x) = \begin{cases} 0 & x < \sqrt{2} \\ 1 & x > \sqrt{2} \end{cases} $$

_{1: This is a technical term; it means that every subset that has at least 2 points must be disconnected}

Answer 2

Continuousness is a topological notion. A function is continuous if the preimage of any open set is an open set.

This has actually something to do with the intuitive notion of drawing a curve without taking the pencil of the paper precisely only in the case where the domain is a connected subset of $\Bbb R $. "Connected" meaning "with no holes", in your words.

The morality is that what is wrong is the simplified notion of continuousness taught at elementary levels.

Note however that this condition can not always be non-trivially fulfilled: if $ A $ is a set with more than one element endowed with the trivial topology $\{A, \emptyset \} $ and if $ B $ is non-empty endowed with the discrete topology, then only constant maps $ A\to B $ are continuous.