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.

$f$ is an injective map from $[0,1]\rightarrow\mathbb{R}$, then we need to find out which of the followings is true

  1. $f$ must be onto

  2. range of $f$ must contain a point of $\mathbb{Q}$

  3. range of $f$ must contain a point of $\mathbb{Q}^c$

  4. range of $f$ must contain points of both $\mathbb{Q}$ and $\mathbb{Q}^c$

I am not getting counter examples or examples for 2,3, I am guessing answer 1 or 4, I am not getting how to apply the injectiveness.

share|improve this question
1  
What do you mean by ‘a $\Bbb Q$’? There is only one $\Bbb Q$. Do you mean ‘a rational’? (Similarly for the complement.) –  Brian M. Scott Dec 14 '12 at 5:47
    
yes a point of rationals and irrationals I meant to say –  Une Femme Douce Dec 14 '12 at 5:47
    
$f(x)=x$ is an injective map that satisfies none of these properties if restricted to $[0,1]$. But this answer seems trivial, so maybe there's more to this question. –  neuguy Dec 14 '12 at 5:47
1  
@Kuttus Is $f$ a continuous function? –  user38268 Dec 14 '12 at 5:50
    
BenjaLim No, just an injective map –  Une Femme Douce Dec 14 '12 at 5:51

3 Answers 3

If the map is required to be continuous, its image will be a closed interval in $\Bbb R$, so the answer to (2)-(4) will be yes. If there are no restrictions on it besides injectivity, then the answer is entirely a question of cardinality; the range of the map must have cardinality $|[0,1]|=2^\omega=\mathfrak c$. Thus, it can’t be a countable set. From this you can answer (2)-(4) easily.

share|improve this answer
    
This is just a side remark on the phrase "entirely a question of cardinality": For parts (2) and (4) that is to some extent a matter of perspective, because they can be answered by algebraic means without any knowledge of a notion of cardinality, as in WimC's answer. (That method would work with $\mathbb Q$ replaced by an arbitrary proper subgroup of $\mathbb R$.) –  Jonas Meyer Dec 14 '12 at 6:08
1  
@Jonas: True, though I think that this is an accidental side effect of the fact that $\Bbb Q$ and its complement are by far the most familiar sets that would work here and were therefore used in the problem. I’m still content to say that at bottom it’s a matter of cardinality. –  Brian M. Scott Dec 14 '12 at 6:16
  1. $x \mapsto x$
  2. $f(x) = \sqrt{2}+x$ if $x \in [0,1] \cap \mathbb{Q}$ and $f(x) = x$ otherwise.
  3. $[0,1]$ is not countable
  4. See 2.
share|improve this answer

HINT:

The cardinality of $[0,1]$ is the same as $\mathbb{R}$, which is also the same as $\mathbb{Q}^{c}$. But all of these are bigger than $\mathbb{Q}$.

share|improve this answer
1  
This doesn't answer the question. –  user38268 Dec 14 '12 at 5:49
1  
@BenjaLim: It’s a perfectly good hint. Unfortunately, I can cancel only one of the downvotes. –  Brian M. Scott Dec 14 '12 at 5:52
1  
@BenjaLim can you elaborate? this seems to answer the question to me –  Deven Ware Dec 14 '12 at 5:53

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.