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$A=\{x^2 :0<x<1\}$, $B=\{x^3: 1<x<2\}$ 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 onto.

well, at a glance I have drawn the picture of $x^2$ and $x^3$ and they intersects at $(0,0)$ and $(1,0)$ so in between I can map each point of one curve to one point of another, so I feel there may be a bijection from $A$ to $B$. Will be pleased if you help.

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You could start by writing down the sets $A$ and $B$ in simpler forms. They are both open intervals. –  Thomas E. Jun 11 '12 at 12:27
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Ah yes, $A=(0,1)$ $B=(1,8)$ so homeomorphic, there exist a continous bijection map. am I right? –  Une Femme Douce Jun 11 '12 at 12:29
    
That is correct. –  Thomas E. Jun 11 '12 at 12:29

1 Answer 1

up vote 1 down vote accepted

I see you've already figured out $1$ (and hence, $3$ and $4$).

For $2$, Cantor's theorem about DLOs (dense linear orders) tells us that there is an order-preserving bijection $f:\Bbb Q\cap A\to\Bbb Q\cap B$. Note that if there is a continuous function $g:A\to B$ such that $f=g\restriction(\Bbb Q\cap A)$, then this function is unique (since $f$ is a continuous function $\Bbb Q\cap A\to B$, $\Bbb Q\cap A$ is dense in $A$, and $B$ is Hausdorff). It shouldn't be difficult to justify that there is such an extension $g$ of $f$--using methods similar to the definition of $\Bbb R$ by either Cauchy sequences or Dedekind cuts of $\Bbb Q$--nor should it be difficult to show that this extension is a bijection.

Alternately, $x\mapsto(x+1)^3$ is bijective and takes rationals to rationals, but if you want something that takes only rationals to rationals, then you'll need an approach as above.

Edit: As KReiser points out in the comments, I was far too eager to bring the big guns to the table when it wasn't necessary (that's a weakness of mine). Ignore everything I said above (though it is all accurate, apart from the claim that you need the DLO approach for a function that takes only rationals to rationals). The function $x\mapsto 1+7x$ is a homeomorphism $A\to B$ that maps $\Bbb Q\cap A$ onto $\Bbb Q\cap B$. H/T to KReiser, and apologies for the misleading answer.

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One does not need something so fancy as Cantor's theorem on dense linear orders to show that there exists a homeomorphism between $A$ and $B$ which restricts to an order-preserving homeomorphism on $\mathbb{Q}\cap A$ and $\mathbb{Q}\cap B$: indeed, $f:A\to B$ given by $x\mapsto 7x+1$ does the trick. –  KReiser Jun 11 '12 at 18:49
    
Right you are! Making it far too complicated.... –  Cameron Buie Jun 11 '12 at 18:50

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