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$f$ 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.

Myshkin
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    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 05:47
  • yes a point of rationals and irrationals I meant to say – Myshkin Dec 14 '12 at 05:47
  • $f(x)=x$ 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. – Gyu Eun Lee Dec 14 '12 at 05:47
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    @Kuttus Is $f$ a continuous function? –  Dec 14 '12 at 05:50
  • BenjaLim No, just an injective map – Myshkin Dec 14 '12 at 05:51
  • Who is "we" and why do you need to find this out? – Jonas Meyer Dec 14 '12 at 06:00
  • Here are 2 related questions: http://math.stackexchange.com/questions/11320/one-to-one-function-from-the-interval-0-1-to-mathbbr-setminus-mathbbq and http://math.stackexchange.com/questions/105990/showing-that-mathbbr-and-mathbbr-backslash-mathbbq-are-equinumerous?lq=1 – Jonas Meyer Dec 14 '12 at 06:09

3 Answers3

4

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.

Brian M. Scott
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  • 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 06:08
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    @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 06:16
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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 $\mathbb{Q}$.

Deven Ware
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  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.
WimC
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