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I've recently started learning about topology, and I was able to prove that any function which is undefined on a countable subset of $\mathbb R$ is also not continuous. However, I am struggling to prove the same claim for a function which is undefined on an uncountable subset of $\mathbb R$. Here's my proof for the countable case:

Denote the set of points upon which $f$ is undefined as $A$. Then, the preimage of the empty set is $f^{-1}(\emptyset) = A$. If $f$ is continuous, then $A$ should be open (as $\emptyset$ is open). However, the singleton set is not open with respect to the usual topology over $\mathbb R$, and since $A$ is a countable union of singleton sets it must also not be open. Hence, the preimage of an open set is not necessarily open, meaning $f$ is not continuous over $\mathbb R$.

This proof breaks down for me, because if $A$ is uncountable it can possibly be open. Taking the function defined by $f(x) = \exp\Big(-\frac{1}{\sqrt{x^2-1}}\Big)$ as an example, the corresponding set $A$ for this $f$ is $A = (-1, 1)$. In this case, $f^{-1}(\emptyset) = (-1, 1)$ is open, so that means $f$ does take open sets to open sets... and so $f$ is continuous.

Can anyone please explain the flaw in my logic?

Baylee V
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    A function that is undefined at some points on $\mathbb{R}$ is, by the two modern definitions of function, not considered a function. – plop May 15 '21 at 11:49
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    If $A$ is the set where $f$ is undefined no, it absolutely does not follow that $A=f^{-1}(\emptyset)$. (That would say $f(t)\in\emptyset$ for every $t\in A$...) – David C. Ullrich May 15 '21 at 11:57
  • @DavidC.Ullrich Ahhh, I understand that was not a valid step. Thanks for pointing that out. – Baylee V May 15 '21 at 12:01
  • @plop I don't know if that makes sense. Wouldn't that imply that any map with removable discontinuities isn't a function? Or that the natural logarithm isn't a function? – Baylee V May 15 '21 at 12:05
  • @BayleeV the point is that the natural log is not considered a function on $\mathbb{R}$ but it is a function on the positive reals. Similarly, you can't speak of functions defined on $\mathbb{R}$ but aren't defined on all of $\mathbb{R}$ - this just doesn't make much sense. The idea is that they are defined on $\mathbb{R}$ "minus some points" but we can still discuss continuity for such functions. – Isky Mathews May 15 '21 at 12:13
  • @IskyMathews I see. So how would one formalize that $f(x) = x/x$ is discontinuous at $x = 0$, for example? The preimage (by this $f$) of any open set would be either $\mathbb R - {0}$ or $\emptyset$, depending on whether or not 1 is in the set, and both of these are open, implying that $f$ is continuous. – Baylee V May 15 '21 at 12:26
  • The function you have mentioned IS continuous at every point in its domain. It is NEITHER continuous nor discontinuous at $x=0$ as it is undefined at that point. See: https://math.stackexchange.com/questions/1431796/if-a-function-is-undefined-at-a-point-is-it-also-discontinuous-at-that-point – Isky Mathews May 15 '21 at 12:33
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    To be clear, continuity of $f:X\to Y$ relates the value of $f(x_0)$ to the value of $f(x)$ where "$x$ is close to $x_0$" (where the specific relation depends on the definition). Both $x$ and $x_0$ are members of $X$ - this is, more than anything, deeply important. This means we can have continuous functions $f:\mathbb{Q}\to \mathbb{R}$ which is a very different thing to continouous functions $f:\mathbb{R}\to\mathbb{Q}$ etc. What if I asked you whether or not $x^2$ was continuous at $x=elephant$? This doesn't make sense as elephant is not in the function's domain. – Isky Mathews May 15 '21 at 12:37

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It seems that your understanding of the concept of function is not the standard one. A function $f : X \to Y$ assigns to each input $x \in X$ a unique output $f(x) \in Y$.

Therefore the notation $f : \mathbb R \to \mathbb R$ implies that $f$ is defined on all of $\mathbb R$. There is nothing like a function $f : \mathbb R \to \mathbb R$ which is undefined on a nonempty subset of $\mathbb R$. What you have in mind is sometimes called a partial function. See here and here. But if you consider a partial function $f : X \to Y$ between topological spaces $X,Y$, you must be aware that it is actually a function $f : X ' \to Y$ defined on some $X' \subset X$. The continuity of $f$ has nothing to do with the points of $X \setminus X'$.

However, denoting certain partial functions as functions is well accepted in some contexts. In complex analysis (which does probably not belong to the field of your experience at the moment) one uses the concept of meromorphic functions on open subsets of the complex plane. Meromorphic functions are in fact partial functions in the strict sense.

Paul Frost
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