Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to show there does not exist a strictly increasing function: $f: \mathbb{Q} \rightarrow \mathbb{R}$ that is surjective.
I started by assuming that such a function exists. Then, this implies the function is continuous because it is strictly monotone with an interval as its image. Now, I think I should somehow use that to contradict that there are enough values in $\mathbb{R}$ for each value in $\mathbb{Q}$, but I'm stuck. Hints?

share|improve this question
2  
$\Bbb Q$ is countable, $\Bbb R$ is not? –  David Mitra Mar 2 at 20:43
1  
What do you understand "continuous" to mean when $\mathbb Q$ is involved? Topology with the rationals is weird. –  Ben Millwood Mar 2 at 20:43
    
I'm in a intro to real analysis class, so I'm only allowed to use basic definitions of continuity. I cannot use the cardinality of $\mathbb{Q}$ or $\mathbb{R}$ –  Paul Malinowski Mar 2 at 20:45
1  
Note that the strictly increasing condition implies that the function is also injective –  Mark Bennet Mar 2 at 20:45
    
For the sake of the problem, is there a way to do it only using the sequential or episolon-delta definition of continuity? –  Paul Malinowski Mar 2 at 20:47
show 2 more comments

2 Answers 2

up vote 2 down vote accepted

The proof that follows avoids using the fact that the cardinality of $\mathbb R$ is larger than the one of $\mathbb Q$.

Let $x_0\in \mathbb R\smallsetminus\mathbb Q$. Then as $f$ is (strictly) increasing, then the limit $$ a=\lim_{x\to x_0^-}f(x), $$ exists and it is a real number. In fact for every $x,y\in\mathbb Q$, with $x<x_0<y$, $$ f(x)<a<f(y), $$ as $f$ is strictly increasing.

This means that $a\not\in\mathrm{Ran}(f)$, and hence $f$ is not surjective.

Note. Let me explain better why for every $x,y\in\mathbb Q$, with $x<x_0<y$, we have that $f(x)<a<f(y)$. As $x<x_0<y$, there are $x_1,y_1\in\mathbb Q$, such that $x<x_1<x_0<y_1<y$, and since $f$ is strictly increasing, we have that $$ f(x)<f(x_1)\le \lim_{z\to x_0^-}f(z)\le f(y_1)<f(y). $$

share|improve this answer
    
The step where you note that $x_0 > x \in \mathbb Q$ implies $a > f(x)$ might need some unpacking, since it relies on some non-trivial properties of $\mathbb Q$ (specifically, that it is densely ordered); in particular, it would not hold if we replaced $\mathbb Q$ with $\mathbb Z$ (although the conclusion would obviously still be true). –  Ilmari Karonen Mar 2 at 21:44
    
is there a way to do this without using left/right limits? –  Paul Malinowski Mar 7 at 21:32
add comment

Let $p$ be an irrational number. Show that the supremum of $f(x)$ for $x \in \mathbb Q$ with $x < p$ and the infimum of $f(x)$ for $x \in \mathbb Q$ with $x > p$ are be equal. If this is $y$, then there is no rational $x$ with $f(x) = y$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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