Proof that f maps an infinite set from Real Numbers to Rational Numbers

[burak100]

infinite set

Homework Statement

$f: \mathbb{R} \rightarrow \mathbb{Q}$,

show that there is a $q \in \mathbb{Q}$ st. $f^{-1}(q)$ is infinite set in $\mathbb{R}$.

Homework Equations

The Attempt at a Solution

how can we show that is true?

 

[Dick]

Do you know R is uncountable? That would be a clue.

 

[burak100]

infinite set means ; we can't find 1-1 correspondence between the set {1,...,n} and $f^{-1}(q)$ , here $f^{-1}(q)$ has n elements.
So, can we say $f^{-1}(q)$ is uncountable then we can't find 1-1 correspondence between the set {1,...,n}...?

 

[Dick]

infinite set means ; we can't find 1-1 correspondence between the set {1,...,n} and $f^{-1}(q)$ , here $f^{-1}(q)$ has n elements.
So, can we say $f^{-1}(q)$ is uncountable then we can't find 1-1 correspondence between the set {1,...,n}...?

Try to get the idea of your proof before you start writing symbols. Assume the opposite, that all of the f^(-1)(q) are finite. And you didn't answer my question. Do you know R is uncountable, and do you know what that means?

 

[burak100]

yeah R is uncountable but I can't find a relation to this question..

 

[Dick]

yeah R is uncountable but I can't find a relation to this question..

Every element if R is in one of the f^(-1)(q) sets, right? Suppose they are all finite?

 

[burak100]

I didn't understand clearly but ;

for all $q \in \mathbb{Q}$, we suppose all the sets, $f^{-1}(q)$ are finite. That means, every element of $\mathbb{R}$ is one of these sets, $f^{-1}(q)$ ...right?

 

[Dick]

I didn't understand clearly but ;

for all $q \in \mathbb{Q}$, we suppose all the sets, $f^{-1}(q)$ are finite. That means, every element of $\mathbb{R}$ is one of these sets, $f^{-1}(q)$ ...right?

Every element of R is in one of the f^(-1)(q) whether those sets are finite or not, just because f:R->Q. Pick an element x of R, for which q in Q is x in f^(-1)(q)??

 

[burak100]

Every element of R is in one of the f^(-1)(q) whether those sets are finite or not, just because f:R->Q. Pick an element x of R, for which q in Q is x in f^(-1)(q)??

I think that is right because just now we said
" every element of R is in one of the set f^{-1}(q) "

 

[Dick]

I think that is right because just now we said
" every element of R is in one of the set f^{-1}(q) "

Yes, but can you tell me why?? Just because we said it doesn't prove it. You are going to need this as part of your proof.

 

[burak100]

Ok. I think; we have $\mathbb{R}= \bigcup\limits_{q \in \mathbb{Q}} f^{-1}(q)$, and we suppose that for all q in Q , $f^{-1}(q)$ finite, then union would be countable but R is uncountable so contradiction... right?

 

[Dick]

Ok. I think; we have $\mathbb{R}= \bigcup\limits_{q \in \mathbb{Q}} f^{-1}(q)$, and we suppose that for all q in Q , $f^{-1}(q)$ finite, then union would be countable but R is uncountable so contradiction... right?

Right, if you are clear on why the union of all the f^(-1)(q) is R.