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Hi guys I am having trouble understanding cardinality. I am given this practice question.

1) Use Cantor-Schroder-Bernstein Theorem to prove that the intervals $(0,1)$ and $[0,1]$ have the same cardinality?

My attempt :

I know the CSB theorem is

Let $A, B$ be sets and if $f: A \rightarrow B$ , and $g: B \rightarrow A$ are both injections, then there exists a bijection from $A$ to $B$

I have to show that $|(0,1)| = |[0,1]| $ so we need $f: (0,1) \rightarrow [0,1]$ and we can set $f(x) = x$ and also we need $g: [0,1] \rightarrow (0,1)$ and we can let $g(x) = \frac{1}{2}x + \frac{1}{4}$

but after I get stuck I dont know what to do i know have to set these equations up but get confused after.

Please help out any help or hints will be greatly appreciated

Thank you

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You’re done as soon as you verify that the maps $f$ and $g$ are injections, that the range of $f$ is a subset of $[0,1]$, and that the range of $g$ is a subset of $(0,1)$: at that point you have an injection $f:(0,1)\to[0,1]$ and an injection $g:[0,1]\to(0,1)$, and the CSB theorem tells you outright that there exists a bijection $h:[0,1]\to(0,1)$ and hence that $\left|[0,1]\right|=\left|(0,1)\right|$.

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  • $\begingroup$ but thats where im confused from what I know atleast to show something is injective that implies that if f(x) = f(y) then x=y. Do i just use my 2 functions that I made up or do I have to show by (0,1) --> [0,1] i dont get that. like how do i do it by that if you know what I mean $\endgroup$ – MathGeek Apr 15 '13 at 11:28
  • $\begingroup$ @MathGeek: All you have to prove is that your functions are injective (one-to-one). Is it true that if $f(x)=f(y)$, then $x=y$? Very obviously yes, since $f(x)=x$ and $f(y)=y$. Is it true that if $g(x)=g(y)$, then $x=y$? Again the answer is yes, though this time it takes just a little algebra to verify the fact. $\endgroup$ – Brian M. Scott Apr 15 '13 at 11:33
  • $\begingroup$ OOO awesome, alright so I have to use my functions that I set up, I thought there was some trick behind it. Thanks a lot Brian M. Scott :) $\endgroup$ – MathGeek Apr 15 '13 at 11:38
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    $\begingroup$ @MathGeek: No, all of the hard work was done in proving the CSB theorem; that’s what makes it so nice. You’re welcome! $\endgroup$ – Brian M. Scott Apr 15 '13 at 11:39

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