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I am considering functions $f:\mathbb{R}\rightarrow \mathbb{R}$ with property that $\forall_{r\in\mathbb{R}}$ exists a limit $\lim_{x\rightarrow r}f(x)$ (it doesn't have to be equal to $f(r)$). I have two problems.
The first with showing that such function $f$ has countable number of discontinuity points. I have found only that monotonic functions have such property that it doesn't help me a lot.
Second problem is connected with evaluating the power of set of such functions. We know of course that the power of set of the continuous functions is equal to continuum which is a hint, but still it doesn't help me much.
I appreciate any help, because I am thinking about it from several days.

Martin Sleziak
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1 Answers1

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Whatever you indicate, it can be either a function with removable discontinuity or a continuous function.
Whenever $f$ have removable discontinuity, then $f$ can't be a monotone function, For example, $f(x) = \begin{cases} 2, & \text{if x=1} \\ x, & \text{otherwise} \end{cases} $

And, how many are there such type functions? Then it is uncountable, you can see that easily. As, we can creat, for each $a\in\mathbb{R} $, $f(x) = \begin{cases} a, & \text{if x=a+1} \\ x, & \text{otherwise} \end{cases} $

And, so, clearly cardinality of power set of such functions is equal to the cardinality of $\mathbb{R}^{\mathbb{R}} $.

Novice
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A learner
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    Then it is uncountable --- Clearly, the cardinality of such functions is greater than or equal to $2^{\aleph_0} = c,$ so a more interesting question is whether the cardinality is greater than $c$ (which is possible for a set of functions, since the set of all functions from the reals to the reals has cardinality $2^c).$ One can show that each such function has at most countably many discontinuities, and with some more work, one can show that the larger set of functions, each of which has a countable (= finite or countably infinite) number of discontinuities, has cardinality $c.$ – Dave L. Renfro Jul 29 '20 at 15:51
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    Indeed, the still larger set of functions that are pointwise limits of continuous functions (these are the Baire one functions) has cardinality $c.$ In fact, the still (much) larger collection of Baire functions (= Borel measurable functions) also has cardinality $c,$ and appropriate searches in Mathematics Stack Exchange will give questions dealing with this. Finally, there is a property intermediate between the OP's property and having a countable number of discontinuities that might be of interest, the set of regulated functions. – Dave L. Renfro Jul 29 '20 at 15:54
  • @Subhajit I think you have made a mistake in a definition of function, there should be $a+2$ instead of $a$ when $x=a+1$. Anyway, thank you very much for help. But still I don't see explicitly how to show that such functions have countable number of discontinuity points. –  Jul 29 '20 at 17:45
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    @Beginner there is no mistake. To show removable discontinuity at $a+1$, we can take any real number other than $a+1$ as the value of $f(a+1)$. – A learner Jul 29 '20 at 17:50
  • @Subhaijt yes, you are right. I catched the idea. I was thinking that there was connection with previous definition of function. But indeed it does not make difference. –  Jul 29 '20 at 17:53
  • @Dave L. Renfro could you tell me how to show that each such function has at most countably many discontinuities? Maybe it is easy, but anyway I don't see it. –  Jul 29 '20 at 18:17
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    @Beginner: See the answers to Is the set of discontinuity of $f$ countable? For the early history about this result (indeed, for much stronger results), see this answer. – Dave L. Renfro Jul 29 '20 at 19:02
  • @DaveL.Renfro thank you very much for help. –  Jul 30 '20 at 16:22