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I was thinking about functions with removable discontinuities and a train of thought led to this. My understanding is that for most nicely-behaved functions $g:\mathbb{R}\to\mathbb{R}$, a function $f:\mathbb{R}\to\mathbb{R}$ where $$ f(x)=g(x)\frac{x-c}{x-c} $$ has a removable discontinuity at $x=c$ and $\lim_{x\to c}f(x)=g(c)$.

If we redefine $f$ as the product of $g$ and n-many factors of $\frac{x-c_{i}}{x-c_{i}}$, all we're accomplishing is adding n-many removable discontinuities to $f$, located at whatever $c_i\in\mathbb{R}$ values we choose. $\lim_{x\to c}f(x)=g(c)$ is still true for all $c\in\mathbb{R}$.

Say that we keep going and we add a discontinuity for every $c\in\mathbb{R}$.

$$f(x) = g(x)\prod_{c \in \mathbb{R}} \frac{x-c}{x-c}$$

What happens to the limit of $f$? Does $\lim_{x\to c}f(x)$ fail to exist for all $c\in\mathbb{R}$?

This seems a fundamentally different problem than discussed in this question because the infinitely-many removable discontinuities there are are discretely spaced, not continuous over an interval. I suppose we could constrain our adding-of-removable-discontinuities to some interval $(a,b)$ within $\mathbb{R}$ and define $f$ like so $$ f(x) = g(x)\prod_{c \in (a,b)} \frac{x-c}{x-c} $$ but I'm still wondering about the limit in that interval and near its boundaries at $x=a$ and $x=b$.

I'm also curious about the machinery for removing a removable discontinuity by piecewise-redefining a function. In my mind, redefining a function piecewise to take care of a removable discontinuity is normally like patching a hole in a ship. We can patch one hole, or many finite holes. But if the ship is "all holes" and has infinitely many removable discontinuities over an interval, does this technique work to patch all of them or are they no longer nicely removable?

DoodleVib
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    How precisely are you defining a product indexed by $\mathbb{R}$? – Noah Schweber Aug 08 '23 at 20:29
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    Right off the bat, it's worth noting that definitions via ambiguous notation are not quite definitions. It seems to me that $f:x\mapsto g(x)\cdot\frac{x-c}{x-c}$ either fails to be a definition at all (if we seek a function $\Bbb R\to\Bbb R$) or a trivial definition of a function $f$ which exactly equals $g$. Both options kinda ruin the question, it seems to me. – FShrike Aug 08 '23 at 20:32
  • You seem to be confusing removable discontinuities and removable singularities (in both cases the function can be made continuous by changing its definition, but in the first the function has the wrong value at the point of discontinuity, while in the second the function is not defined at the point of singularity). Either way, you can always find a continuous extension regardless of how many bad points there are (at least if you assume that the set of bad points is dense: the requirements on the extension become a bit unclear if you have an open set of bad points). – Rob Arthan Aug 08 '23 at 21:02

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