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Showing Cardinality between sets.
(a)$\left( {a,\infty } \right) \sim R$
(b)$\left[ {0,1} \right) \sim \left( {0,1} \right)$
(c)$\left( {a,b} \right) \sim R$
For (a) since $\left( {a,\infty } \right) \subseteq R$, I can use $f\left( x \right) = {e^x}$ which has domain $R$ and range $\left( {0,\infty } \right)$, ans show $\left( {0,\infty } \right) \sim R$, and I know two open intervals are cardinal $\left( {a,\infty } \right) \sim \left( {0,\infty } \right)$ therefore $\left( {a,\infty } \right) \sim R$. Correct?
For (b) I can use $f\left( n \right) = {1 \over {n + 1}}$ and shift $1$ of $\left( {0,1} \right)$ to ${1 \over 2}$ and show the function is one to one. How do I show it is onto?
For (c) $\left( {a,b} \right) \subseteq R$ => $\left( {a,b} \right) \to R$ is one to one. Now, since $\left( {a,b} \right) \to \left( {c,d} \right)$ is onto for some $\left( {c,d} \right)$ in $R$ so is onto? So, $\left( {a,b} \right) \sim R$ ? Correct?

  • (c) isn't onto. – Nosrati Jan 24 '17 at 12:35
  • @MyGlasses, for (b) for what value of $n$ in $\left( {0,1} \right)$ I will have $0$ in $\left[ {0,1} \right)$. I mean is it onto, $f\left( n \right) = {1 \over {n + 1}}$.. I guess that's not definition of onto? –  Jan 24 '17 at 12:49
  • What about other points.? and $\frac12$ also.? – Nosrati Jan 24 '17 at 12:51
  • @MyGlasses for $1/2$ I have $1$ which shouldn't be. Hmm...I have to find a function $f(x)$ which exhibits asymptotic behavior at $x=0$. What about just $n/n+1$ –  Jan 24 '17 at 13:10
  • http://math.stackexchange.com/questions/1233238/construct-an-explicit-bijection-f0-1-to-0-1-where-0-1-is-the-closed?rq=1 – Nosrati Jan 24 '17 at 13:15
  • http://math.stackexchange.com/questions/160738/how-to-define-a-bijection-between-0-1-and-0-1 – Nosrati Jan 24 '17 at 13:16
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    http://math.stackexchange.com/questions/1136573/constructing-a-bijection-between-a-b-to-0-1?rq=1 – Nosrati Jan 24 '17 at 13:17

1 Answers1

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For (c). Take $\tan x:(-\dfrac{\pi}{2},\dfrac{\pi}{2})\to(-\infty,\infty)$ is bijective, so $\tan\dfrac{\pi}{2}x:(-1,1)\to\mathbb{R}$ and with $g(x)=\dfrac{2}{b-a}x+\dfrac{a+b}{a-b}:(a,b)\to(-1,1)$ thus $\tan\dfrac{2}{\pi}g(x):(a,b)\to\mathbb{R}$ is bijective.

Nosrati
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