$f: [0,1]\rightarrow \mathbb{R}$ be an injective function, then :

Question

Question is :

$f: [0,1]\rightarrow \mathbb{R}$ be a one one function, then which of the following statements are true?

$(a)$ $f$ must be onto

$(b)$ range of $f$ contains a rational number

$(c)$ range of $f$ contains an irrational number

$(d)$ range of $f$ contains both rational numbers and irrational numbers

I can see that $f$ need not be onto function by considering :

$f : [0,1]\rightarrow \mathbb{R}$ with $f(x)=x$ is one-one , but not onto.

Now, I should not take a continuous function to show a contradiction for $(b)/(c)/(d)$ because, any continuous function takes intervals to intervals and thus, Range of $f$ will have both rational and irrational numbers.

I recalled all functions i have thought could be discontinuous but those are not helping me at all.

I would be thankful if some one can help me to crack this problem and similar kind of problems.

THank You.

Since $[0,1]$ is uncountable, the range would have to contain an irrational number. Not sure what else can be said though. — Prahlad Vaidyanathan, Oct 05 '13 at 09:30
Continuity is not given and is irrelevant - this question is all about the cardinalities of the sets involved. — Anthony Carapetis, Oct 05 '13 at 09:31
@PrahladVaidyanathan : yes.. This does makes lots of sense to me :) — , Oct 05 '13 at 09:32
@PrahladVaidyanathan : can we say anything about the rational numbers presence?? — , Oct 05 '13 at 09:34
I would guess that you can avoid all rational numbers, but I would suspect that constructing such a function would involve the axiom of choice. — Prahlad Vaidyanathan, Oct 05 '13 at 09:37

score 9 · Accepted Answer · answered Oct 05 '13 at 09:44

9

Consider $f : [0, 1] \to \mathbb R$ defined as: $$ f(x) = \begin{cases} x &: x \text{ irrational} \\ x + \sqrt{2} &: x \text{ rational} \end{cases} $$

It is easy to show that this is an injective function from $[0, 1]$ into $\mathbb R - \mathbb Q$.

answered Oct 05 '13 at 09:44

Ayman Hourieh

39,603

$f: [0,1]\rightarrow \mathbb{R}$ be an injective function, then :

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