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Let $A=\{x^2:0<x<1\}$ and $B=\{x^3:1<x<2\}$. Which of the following statements is true?

  1. There is a one to one, onto function from $A$ to $B$.
  2. There is no one to one, onto function from $A$ to $B$ taking rationals to rationals.
  3. There is no one to one function from $A$ to $B$ which is onto.
  4. There is no onto function from $A$ to $B$ which is one to one.

I have been trying to solve the problem. Could someone point me in the right direction? Thanks in advance for your time.

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What have you tried? A bijection for (1) is readily written down, thus (3) and (4) are falsified immediately. And it's hard to believe that an example you quickly find for (1) is not also a counterexample for (2). –  Hagen von Eitzen Dec 19 '12 at 19:05
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Thanks Stefan Hansen for makiung the image readable. –  Hagen von Eitzen Dec 19 '12 at 19:08
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@HagenvonEitzen Actually, it's not that hard to believe. I think the first example you're "expected" to find (and the one that leads you into the trap here) is the map that takes $x^2$ to $(x+1)^3$, which is a bijection that doesn't preserve rationality –  Zimul8r Dec 19 '12 at 19:30
    
@Zimul8r Oh, I wasn't even able to fall for that. Then again, all we need is that $A,B$ are uncountable and $A\cap \mathbb Q,B\cap \mathbb Q$ are infinite. –  Hagen von Eitzen Dec 19 '12 at 21:19

2 Answers 2

A hint: The sets $A$ and $B$ are described in a somewhat cumbersome way. Find really simple descriptions of these two sets.

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Allow me to point you into the direction of the function $t\mapsto 7t+1$.

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Sir, do you mean that '$t$'is taken from $A$ and $7t+1$ is from the set $B$? –  user52976 Dec 19 '12 at 19:31
    
@user33640: Yes. you are right. He noted the right map from $(0,1)\to (1, 8)$. Check it. –  Babak S. Dec 19 '12 at 19:37
    
sir,why are you taking elements of the form $(a,b)$ from the sets $A$ and $B$..If you clarify it,i will be grateful. –  user52976 Jan 10 '13 at 10:14
    
These $(a,b)$ in Babak's comment are not to be interpreted as elements (ordered pair with components $a$ and $b$) but as open intervals (i.e. $(a,b)=\{x\in\mathbb R\mid a<x<b\}$). Note that $A=(0,1)$ and $B=(1,8)$. –  Hagen von Eitzen Jan 10 '13 at 13:15

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