Find the cardinality of the set of all continuous real-valued functions of one variable.

Trying to solve this problem I stumbled upon one casualty whichI've been struggling for a while with. This is my solution:

The cardinality of this set is bounded below by the cardinality of $\mathbb R$, because $f: \mathbb R \to \mathbb C(\mathbb R)$ given by $f(x) = \mbox{function y = x}$ is an injection. Now, I tried to bound this set from above:
$$|\mathbb R| = |\mathbb R^{\aleph_0}| = |\mathbb R^\mathbb Q|$$
And now I tried to make an injection $$I: \mathbb C(\mathbb R) \to \mathbb R^\mathbb Q$$
given by $$I(x) = x \upharpoonright \mathbb Q$$
However, in order for this solution to be acceptable, I'd have to prove this:
$$ x\upharpoonright \mathbb Q = y \upharpoonright \mathbb Q \Longrightarrow x=y$$
And this is where I got stuck. How can I tackle this proof?

$$\mathbb C(\mathbb R) - \mbox{Continuous real-valued functions}$$

  • 1
    $\begingroup$ could you define your notations? $\endgroup$ – ziggurism Feb 5 '18 at 14:05

$|C(\mathbb{R})|$ is at least $|\mathbb{R}|$ because for every $c \in \mathbb{R}$, the constant function $f(x) = c$ belongs to $|C(\mathbb{R})|$. It is at most $|\mathbb{R}^\mathbb{Q}|$, because $\mathbb{Q}$ is dense in $\mathbb{R}$ and so a continuous function on $\mathbb{R}$ is determined by its values on $\mathbb{Q}$. And as you note, $|\mathbb{R}| = |\mathbb{R}^\mathbb{Q}|$.

  • $\begingroup$ But the irrationals are dense in $\mathbb R$ as well. Could you, please, add some more explanations? $\endgroup$ – Aemilius Feb 5 '18 at 14:12
  • $\begingroup$ If you know the values of a continuous function $f$ on every rational number, you know its values on every real number. If $x$ is real, then take a sequence $q_1, q_2, \ldots$ of rationals whose limit is $x$; then, by continuity, $f(x)$ is the limit of the sequence $f(q_1), f(q_2), \ldots$. There's nothing special about the rationals here: any dense subset of $\mathbb{R}$ would do. $\endgroup$ – Connor Harris Feb 5 '18 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.