0
$\begingroup$

I wan to show that there is no continuous real-valued function of a real variable that sends rationals to irrationals and irrationals to rationals by using only $1)$ and $2)$:

$1)$ We can use the result that if $f$ is continuous then it is sequentally continuous, that is, for every sequence $x_n$ that tends to $c$ the associated sequence $f(x_n)$ tends to $f(c)$ and also the result that if $f$ is continuous then $\lim_{x \to c}f(x)=f(c)$.

$2)$ We can use the result that between every different two numbers on the real line there is an infinite number of irrationals between them and an infinite number of rationals between them.

We should not use anything else, no countability/uncountability arguments, no intermediate value theorem, no that if $A$ is connected then $f(A)$ is connected etc...

Surely $1)$ and $2)$ seem to be too weak to deduce the result but even if they are how to prove it? Can someone explain to me why this can (or cannot) be done by only using $1)$ and $2)$?

If $1)$ and $2)$ are indeed too weak then use also:

$3)$ $\bigcap_{i=1}^{\infty}A_i$ is non-empty if $A_{j+1} \subset A_j$ for every $j \in \mathbb N$ and $A_j$ is non-empty compact set for every $j \in \mathbb N$ and we know that compact means closed and bounded.

$\endgroup$
2
  • $\begingroup$ It's completely unclear to me why you want to only allow a small selection of tools and ban everything else, but if you just want to prove the result, recall that $\mathbb R\setminus \mathbb Q$ is a $G_{\delta}$ set whose preimage would have to be a $G_{\delta}$ also. $\endgroup$ Sep 23, 2017 at 0:57
  • $\begingroup$ @ChristianRemling Use of those sets highly depends on the use of countability, so we should not mention them here. Christian, I just wanted to see is it possible to deduce the result by using only 1), 2) and 3). I like minimal assumptions and alike things. $\endgroup$
    – user114642
    Sep 23, 2017 at 1:03

1 Answer 1

2
$\begingroup$

The two or three facts you mention will not rule out continuous functions which take rationals to rationals and irrationals to irrationals (since the are plenty of such). So you probably need something else to rule out what you want to rule out.

Here is a better example:

You have two sets of reals $A=\mathbb{Q}$ and $B=\mathbb{R} \setminus \mathbb{Q}.$ You wish to show that there is no continuous map $f$ with $f(a) \in B$ for every $a \in A$ and $f(b) \in A$ for every $b \in B.$ You want to do this using three facts. 1) and 3) do not depend on $A$ and $B$ while 2) is that in every open interval $(s,t)$ there are infinitely many members of $A$ and infinitely many members of $B.$ Although you do not use it, there is the fourth property that $A$ and $B$ are disjoint with $A \cup B =\mathbb{R}.$

I will now give an example with all four properties where the function $f(x)=x+1$ has the desired behavior. This means that to prove the thing you want, something else about your $A$ and $B$ needs to be used.

First let

$$S= \cdots [-4,-3) \cup [-2,-1) \cup [0,1) \cup [2,3)\cdots$$ and $$T= \cdots [-3,-2) \cup [-1,0) \cup [1,2) \cup [3,4)\cdots.$$

Then let $A$ be the rational members of $S$ together with the irrational members of $T$ while $B$ is the irrational members of $S$ together with the rational members of $T.$ Then 1) 2) 3) still hold along with the fact that $A,B$ partition the reals. However, as claimed, the function $f(x)=x+1$ has the desired property.

$\endgroup$
2
  • $\begingroup$ But you do not use anywhere sequential continuesness which could help us to approach points in various ways in any way desired? I mean, with the help of that property of continuous functions could it be that we can conjecture the existence of continuous f that sends rationals to irrationals and irrationals to irrationals and then approach some point (or points) with some special sequences to show that such an f cannot exist? What result(s) from analysis do you think we need to prove that such an f cannot exist, and please, choose some result(s) that is(are) as weak as possible. $\endgroup$
    – user114642
    Sep 23, 2017 at 15:16
  • $\begingroup$ I don’t use any of 1,2,3,4 . I just pointed out that they apply equally well for my $A,B$ and yours so anything they imply in your case also applies in mine. I would say that your 3) is more advanced than countable and uncountable. Perhaps it is the case that if $B$ is uncountable and dense in the reals and $f$ is continuous then either $\{f(b)\}$ is uncountable or $f$ is a constant function. $\endgroup$ Sep 23, 2017 at 16:40