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


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 $\left|[0,1]\right|=\left|(0,1)\right|$.

  • $\begingroup$ 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
  • 2
    $\begingroup$ @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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