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I was given the following task:

Find the cardinality of the set $C$ of real-valued functions with exactly one point of discontinuity.

What I've found so far is that $$ \lvert \mathbb{R} \rvert \le \lvert C \rvert $$ because the map $f \colon \mathbb{R} \to C$ defined as $$ f(x_0) := x \mapsto \frac{1}{x-x_0} $$ is injective.

Now, can I prove that $\lvert C \rvert \le \lvert \mathbb{R} \rvert$?

I'd like to apply the Schroeder-Bernstein theorem. However, I don't know how to proceed. Should I take the same approach as in finding the cardinality of the set of continuous real-valued functions, and find a dense subset of $\mathbb{R}\setminus\lbrace x_0 \rbrace$?

P.s.: After reading this question, and confronting it with the answer given by @Arthur, I'm trying to do the following:

  1. Choose a point of discontinuity $x_0$;
  2. Choosing two continuous functions on $(-\infty,x_0)$ and $(x_0,+\infty)$, which can be done by taking their restrictions on the dense subsets $\mathbb{Q} \cap (-\infty,x_0)$ and $\mathbb{Q} \cap (x_0,+\infty)$;
  3. Choose a function value for $x_0$, i.e. assigning a function on the set of points of discontinuity.

Then, we can conclude that $$ \lvert C \rvert \le \lvert \mathbb{R} \rvert \cdot \lvert \mathbb{R} \rvert \cdot \lvert \mathbb{R} \rvert \cdot \lvert \mathbb{R} \rvert = \lvert \mathbb{R} \rvert. $$

Am I right here?

Community
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  • Simmons, in "Introduction To Topology And Modern Analysis" calls it the Schroeder-Bernstein theorem. Engelking, in "General Topolgy" (in a hint to an exercise) calls it Cantor-Bernstein. Elsewhere I have seen it called C-S-B. – DanielWainfleet Feb 07 '17 at 14:19
  • @user254665 I added a link to the Wikipedia entry. Thank you! –  Feb 07 '17 at 14:27

1 Answers1

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You need to choose four things:

  1. A point of discontinuity $x_0$
  2. A continuous function on $(-\infty, x_0)$
  3. A continuous function on $(x_0,\infty)$
  4. A function value at $x_0$

For each of these, you have $|\Bbb R|$ options to choose from.

Arthur
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  • Thank you for the answer! Should I motivate points (2) and (3) using the same argument I did use with continuous real-valued functions? (i.e. equality on rationals implies equality on $\mathbb{R}$? –  Feb 07 '17 at 14:14
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    @FilippoDeBortoli You could do that. You could also give an explicit bijection $C(-\infty, x_0) \to C(\Bbb R)$ by using some homeomorphism between $(-\infty, x_0)$ and $\Bbb R$, for instance $\log(x_0-x)$. – Arthur Feb 07 '17 at 16:00