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Consider the sets

$$ \ A=\{x^{2}: 0<x<1 \} \text{ and }B=\{x^{3}: 1<x<2 \} \ $$

Show that there is a one-one function and onto map between them.

Let $ \ f : A \rightarrow B $ be a map defined by

$$ \ f(x)= x^{\frac{3}{2}} , \ \ \forall \ x \in (0,1) $$

Then this map is clearly one-one. But how to show it is onto? Also is the map well defined? Any help would be appreciated ?

Did
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MAS
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    Just use the fact that $A=(0,1)$ and $B=(1,8)$. You should be able to find an explcit bijection between these two intervals. – TonyK Jun 02 '17 at 16:03
  • is that the reason , that any two open interval are equivalent ? – MAS Jun 02 '17 at 16:05
  • I don't understand. Is what the reason? – TonyK Jun 02 '17 at 16:05
  • I mean there exist one-one and onto map. This is due to the fact that any two open intervals are equivalent or bijective . – MAS Jun 02 '17 at 16:07
  • Ah, I see. Yes, that is the reason. – TonyK Jun 02 '17 at 16:07
  • ok thank you very much. – MAS Jun 02 '17 at 16:08
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    In order to use that argument, you must show that $x^2$ and $x^3$ are both continuous, monotonic increasing functions, or at least continuous functions with the given $\sup$ and $\inf$ for each respective function. – Alex Jones Jun 02 '17 at 16:11
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    The idea here is probably to suggest the mapping $$f(x)=(\sqrt{x}+1)^3$$ but, indeed, the exercise is so bizarre that one almost feels it is a moral duty to answer $$f(x)=1+7x$$ instead. Anyway, $f(x)=x^{3/2}$ does not send $A=(0,1)$ to $B=(1,8)$. – Did Jun 02 '17 at 16:29
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    No, $B \neq (1,8)$. The correct is $B = (1, 2^{\frac13})$. – md2perpe Jun 02 '17 at 17:08
  • @md2perpe: $B$ is definitely equal to $(1,8)$. You must be thinking of ${x:1<x^3<2}$. – TonyK Jun 02 '17 at 19:16
  • What's wrong with my brain? shameful – md2perpe Jun 02 '17 at 19:31
  • yes you are right . $ B=(1,2^{1/3}) $ – MAS Jun 03 '17 at 10:07
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    Funny, nobody believes that $B=(1,2^{1/3})$ anymore (not even the OP who suggested it), except you? Of course, reading carefully the question would be an option to help you see the light... – Did Jun 03 '17 at 11:38

1 Answers1

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In general any two open intervals (However small) are equivalent.

Here's the general idea:

Define $f:(a,b)\rightarrow (c,d)$ by

$$f(x)=c+\Bigl(\frac{d-c}{b-a}\Bigr)(x-a)$$

This make $f$ as a bijection (check it!)

In case, your function is $f(x)=1+7x$ and it is your require bijection

Chinnapparaj R
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