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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.

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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 5:47
yes a point of rationals and irrationals I meant to say – Un Chien Andalou 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. – Gyu Eun Lee Dec 14 '12 at 5:47
@Kuttus Is $f$ a continuous function? – user38268 Dec 14 '12 at 5:50
BenjaLim No, just an injective map – Un Chien Andalou Dec 14 '12 at 5:51
up vote 2 down vote accepted
  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.
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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.

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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 6:08
@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


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 $\mathbb{Q}$.

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This doesn't answer the question. – user38268 Dec 14 '12 at 5:49
@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
@BenjaLim can you elaborate? this seems to answer the question to me – Deven Ware Dec 14 '12 at 5:53

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