How many real-valued functions with one point of discontinuity are there?

Question

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 $\mathbb{R}\setminus\lbrace x_0 \rbrace$?

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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?

Answer 1

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.