4
$\begingroup$

How can I show that the cardinality of this set: $A=\{f: \mathbb{R} \rightarrow \mathbb{R} \text{ continuous} : f(\mathbb{Q})\subseteq\mathbb{Q}\}$ is $2^{\aleph_{0}}$?

I know that $A\subseteq \{f: \mathbb{R} \rightarrow \mathbb{R} \text{ continuous} \}$ so #$(A)\leq2^{\aleph_{0}}.$ But I don't know how to show the other inequality. Thanks a lot for your help! I know there is a post with the same question, but I don't understand the answer :(

Cardinality of $A=\{f: \mathbb R \to \mathbb R , f \text{ is continuous and} f(\mathbb Q) \subset \mathbb Q\}$

$\endgroup$
4
  • $\begingroup$ Can you please link to the post you've mentioned for context? $\endgroup$
    – user61527
    Dec 6 '13 at 4:33
  • $\begingroup$ Here is the link math.stackexchange.com/questions/594915/… $\endgroup$
    – Maxi
    Dec 6 '13 at 4:36
  • $\begingroup$ it would help if you specify what is unclear about the other answer you mention. $\endgroup$ Dec 6 '13 at 5:11
  • $\begingroup$ @Maxi: let me know if my answer seems unclear. I constructed an uncountable collection of continuous maps from $\mathbb Q \rightarrow \mathbb Q$ that extend continuously into continuous maps $\mathbb R \rightarrow \mathbb R$ $\endgroup$
    – user99680
    Dec 6 '13 at 5:20
3
$\begingroup$

Idea: we construct , on each interval $[n,n+1]$ , a countably-infinite collection of linear functions (linear functions $ax+b$ , with both a,b in $\mathbb Q$) mapping $\mathbb Q$ to itself, that extend to continuous functions $f: \mathbb R \rightarrow \mathbb R$. Linear functions extend because they are uniformly-continuous on a dense subset, and uniform continuity is sufficient to extend a function from a dense subset into the whole space. Then we extend each of these functions from $[n,n+1]$ to $[n+1,n+2]$ continuously. Then, each of the $|\mathbb Q|$ functions on $[n,n+1]$ can be (linearly)extended to $[n+1,n+2]$ in $|\mathbb Q|$ ways, so that the total cardinality is $|\mathbb Q| \times |\mathbb Q| \times....$ $|\mathbb Q|$ times.

Consider the integers $ \mathbb Z$ . We construct an uncountable collection of linear maps from $\mathbb Q $ to $\mathbb Q$, and we use the fact that linear maps, being uniformly-continuous on the dense subset $\mathbb Q$ of $\mathbb R$, extend to a continuous map $f: \mathbb R \rightarrow \mathbb R$ .Start at, say $0$. Then, following the idea of the link, any line thru the point $0$ with rational slope maps $\mathbb Q$ to $\mathbb Q$: take $px+q$ , with $p,q$ both in $\mathbb Q$, since Rationals are closed under multiplication, then $px$ is Rational as a product of Rationals, and when we add $b$ to it we have a sum of Rationals, which is Rational. Notice this choice of line can be made in $|\mathbb Q|= \aleph_0$ ways . Now, extend the function at $x=1$ , starting at the image $a(1)+b$ , and then extend the same way from $x=2 $ to $x=3$ , i.e., you defined $a'x+b'$ in $[1,2]$ to be $a'x+b'$ , with both $a',b'$ Rational. This means that each of the $|\mathbb Q|$ choices in each of the interval $[n,n+1]$ can be combined with $\mathbb Q$ choices in $[n+1, n+2]$ , for all integers $n$. So you have a total of $|\mathbb Q|\times |\mathbb Q |\times...|.....$ , all of this $|\mathbb Q|$ times, which gives you an uncountable collection of functions $f: \mathbb Q \rightarrow \mathbb Q$ , that extend to continuous functions $F: \mathbb R \rightarrow \mathbb R$, and you're done.

$\endgroup$
2
  • $\begingroup$ Thanks a lot! Now I understand! $\endgroup$
    – Maxi
    Dec 6 '13 at 13:29
  • $\begingroup$ I did another edit, I think it clarified a few things. $\endgroup$
    – user99680
    Dec 6 '13 at 20:06
2
$\begingroup$

Define $T(x) = \max\{0,1-|2x|\}$, so that $T(x)$ is a continuous "tent function" supported on the interval $({-}\frac12,\frac12)$. For any subset $S\subseteq\mathbb Z$ of the integers, the function $$ F_S(x) = \sum_{n\in S} T(x-n) $$ is a continuous function with a "tent" of width $1$ at every integer in $S$ and flat everywhere else. There are $2^{\aleph_0}$ subsets $S$ of $\mathbb Z$, hence $2^{\aleph_0}$ such continuous functions (they're all different, by checking their values on integers); and they all map rational numbers to rational numbers.

$\endgroup$

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.