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.


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

| cite | improve this answer | |
  • $\begingroup$ Then with the proof that I linked in the post. As we shrink the $\epsilon$ band on the image of the function, does the continuity of $f$ affect the elements which "fill" the holes? Or could I think of the $\epsilon$ band having two parts on the image, those within the band and a subset of that subset which are the image of the preimage? Does that make sense? $\endgroup$ – Andrew Shedlock Aug 13 '18 at 23:47
  • $\begingroup$ @AndrewShedlock: I'm not entirely sure what you mean by "fill the holes"; I don't see anything I would recognize as meeting that description. I have a guess as to your mistake, though. When considering $\mathbb{Q}$ as a metric space, the ball $B(x,e)$ consists of all rational numbers whose distance from $x$ is less than $e$. There are no irrational numbers in this ball, because there are no irrational numbers in $\mathbb{Q}$. $\endgroup$ – user14972 Aug 13 '18 at 23:51
  • $\begingroup$ If we had the identity function $f(x) = x$ on $f:\mathbb{Q}\rightarrow \mathbb{R}$ and looked at the interval $(0,1)\in\mathbb{R}$. Any open subinterval of $(0,1)$ will contain elements which are not mapped to it by the domain. Or do these elements which are not included not considered part of the set? $\endgroup$ – Andrew Shedlock Aug 14 '18 at 0:19
  • $\begingroup$ I bring this up because of the proof for the preimage of an open set of a continuous function must be an open set, but we cannot have an open subinterval of the range from the preimage because of the irrational numbers $\endgroup$ – Andrew Shedlock Aug 14 '18 at 0:42
  • $\begingroup$ @AndrewShedlock: There are indeed elements of $(0,1) \subseteq \mathbb{R}$ that are not in the image of $f$. $f$ is not required to be a surjective function to be continuous. $\endgroup$ – user14972 Aug 14 '18 at 1:14

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.

| cite | improve this answer | |
  • $\begingroup$ Well, the constant map is continuous... $\endgroup$ – Jair Taylor Aug 13 '18 at 23:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.