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

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Homework Help Overview

The discussion revolves around a proof concerning a function f that maps from the set of real numbers, \(\mathbb{R}\), to the set of rational numbers, \(\mathbb{Q}\). The objective is to show that there exists a rational number \(q\) such that the preimage \(f^{-1}(q)\) is an infinite set within \(\mathbb{R}\).

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of \(\mathbb{R}\) being uncountable and how this relates to the finiteness of the sets \(f^{-1}(q)\). There are attempts to clarify the relationship between the elements of \(\mathbb{R}\) and the preimages under the function f.

Discussion Status

The conversation is ongoing, with participants exploring the implications of assuming all \(f^{-1}(q)\) are finite. Some have suggested that this leads to a contradiction given the uncountability of \(\mathbb{R}\), while others are seeking to clarify the reasoning behind these assertions.

Contextual Notes

Participants are grappling with the definitions and implications of countability and uncountability in the context of the function f and its preimages. There is a focus on understanding how the union of all \(f^{-1}(q)\) relates to the entirety of \(\mathbb{R}\).

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


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}...?
 


burak100 said:
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?
 


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


burak100 said:
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?
 


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?
 


burak100 said:
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)??
 


Dick said:
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) "
 
  • #10


burak100 said:
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.
 
Last edited:
  • #11


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?
 
  • #12


burak100 said:
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.
 

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