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I understand that theorem lets you prove the existence of a bijection from a set A to a set B just by proving that there is a one-to-one function that maps A to B has another one-to-one function that maps B to A.

Also since the question asks for same cardinality, proving that a bijection exists is sufficient. I don't understand, how to apply my knowledge to this situation ? Any help would be greatly appreciated.

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  • $\begingroup$ notice that the CSB theorem does not require an injective function with an inverse that is also injective. All that is requires are injections in both directions. One need not be related in any way to the other. $\endgroup$ – Ittay Weiss Nov 17 '13 at 2:56
  • $\begingroup$ I have written what I understand about the topic and question. I am lacking the ability apply what I understand and have no idea how to get started. Andres Caicedo, Serkan, dfeuer, Norbert, Asaf Karagila $\endgroup$ – Gdgames Gamers Nov 18 '13 at 2:13
  • $\begingroup$ One direction is clear since $(0,1)\subset[0,1]\Rightarrow|(0,1)|\le|[0,1]|$ Let $A=[0,1]\setminus\{0,1,\frac{1}{2},\frac{1}{3}...\}$ then; $(0,1)=\{\frac{1}{2},\frac{1}{3}...\}\cup A$. Consider $f:[0,1]\to(0,1)$ with $f(x)$=$\ \begin{cases} \frac{1}{2} &if\ x= 0 \\ \frac{1}{n+2} &if\ x=\frac{1}{n},n\in\mathbb N \\ x &x\in A \end{cases} \ $ , Note that $f$ is injective, $\Rightarrow|[0,1]|\le|(0,1)|$ and Therefore $|[0,1]|=|(0,1)|$ $\endgroup$ – derivative May 7 '14 at 20:44
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Construct an injective function $f:(a,b)\to [a,b]$ (which is very easy to do). Now construct an injective function $g:[a,b]\to (a,b)$ (which can be done in many ways, for inspiration, try to think geometrically - what can you do the the segment $[a,b]$ to get it inside the interval $(a,b)$?). Now conclude by the CSB theorem that there exists a bijection between the two given sets and thus that they have the same cardinality.

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  • $\begingroup$ I have not learnt about injective functions. Could explain it little bit more ? $\endgroup$ – Gdgames Gamers Nov 17 '13 at 2:57
  • $\begingroup$ injective means the same is one-to-one. A function is injective if $f(x)=f(y)$ implies $x=y$. $\endgroup$ – Ittay Weiss Nov 17 '13 at 2:59
  • $\begingroup$ Just construct $f: [a, b] \to [\frac{a}{2}, \frac{b}{2}]$? @IttayWeiss $\endgroup$ – Don Larynx Nov 17 '13 at 3:18
  • $\begingroup$ that will not always do the trick @DonLarynx $\endgroup$ – Ittay Weiss Nov 17 '13 at 5:07
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You should provide injective functions $(0,1)\hookrightarrow [0,1]$ and $[0,1]\hookrightarrow (0,1)$.

That would mean $|(0,1)|\le |[0,1]|\le |(0,1)|$ and Cantor-Shroder-Bernstein theorem asserts exactly that in a situation like $|A|\le|B|\le|A|$ we must have $|A|=|B|$.

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Shroder-Bernstein theorem:

If $(\alpha$ sm $\gamma)$ and $(\beta$ sm $\delta)$ and $(\gamma \subset \beta)$ and $(\delta \subset \alpha)$ then $\alpha$ sim $\beta$.

where "sm" stands for "is similar to" or " has one-to-one(bijective) relation to."

Let $\alpha = (0,1), \gamma = (0,\dfrac{1}{2})$, $\beta=[0,1], \delta = [0, \dfrac{1}{2}]$

Let $f$ be the relation of "$\times 2$", e.g. $ 1=f(\dfrac{1}{2})$, then $(\alpha$ sm $\gamma) $ and $(\beta$ sm $\delta)$.

$(0,\dfrac{1}{2})\subset [0,1] \rightarrow \gamma \subset \beta$

$[0,\frac{1}{2}] \subset (0,1) \rightarrow \delta \subset \alpha$.

In virtue of Shroeder-Bernstein theorem, $\alpha$ sm $\beta$, i.e. $(0,1)$ is similar to $[0,1]$.

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  • 1
    $\begingroup$ You are going to post essentially the same answer on every which one of these nearly identical questions? $\endgroup$ – Asaf Karagila May 7 '14 at 21:00
  • $\begingroup$ He was asking for specific constructs, but I'm going to stop right here. $\endgroup$ – George Chen May 7 '14 at 21:05
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    $\begingroup$ Duplicates are not necessarily the same question word for word, or even same objects per objects. These are questions whose answers are relatively close to easily extrapolate from one into the other. Are you seriously going to argue that this answer here, and this answer here are not the same answer with different variables? $\endgroup$ – Asaf Karagila May 7 '14 at 21:07
  • $\begingroup$ For people of your calibre, they are obviously the same. But if the op sees what you see, why does he ask them both in the first place? $\endgroup$ – George Chen May 7 '14 at 21:12
  • $\begingroup$ My caliber or not my caliber. These questions were made six months ago. The OP either moved on, or understood that. Yes, the two questions are not quite the same. One asks about any bijection, the other about using the Cantor-Bernstein theorem. If your answer is still relevant, six months later, certainly there's no need to post it twice. $\endgroup$ – Asaf Karagila May 7 '14 at 21:15

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