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We know that the Schroder-Bernstein Theorem is useful to show two sets are the same size. Assume $f: A \rightarrow B$ is one-to-one & $g: B \rightarrow A$ is also one-to-one.

Then $A$ and $B$ have the same cardinality; there is a one-to-one function h from A onto B.

My attempt at the proof:

Consider the function $f: (0,1) \rightarrow [0,1]$ defined by $f(x)=x$, $f$ would be one-to-one. The function $g: [0,1] \rightarrow (0,1)$, defined by $g(x) = \frac{1}{2}x + \frac{1}{8}$ is also one-to-one.

Then, there is a bijective function $h$ from $(0,1)$ onto $[0,1]$. Let $h(x)=\frac{1}{2}x + \frac{1}{8}$. Observe that $[0,1]=|[1/8, 5/8]|$. Since $[1/8, 5/8]⊂(0,1)$ it follows that $|[0,1]|=|[1/8, 5/8]|≤|(0,1)|$.

Thus, $|(0,1)|≤|[0,1]|$ and $|[0,1]≤|(0,1)|$, so $|(0,1)|=|[0,1]| $

Any suggestions? How can I make this more clear?

Miguelgondu
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user19059
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  • $x\in(0,1)\Longleftrightarrow0<x<1$ and $x\in[0,1]\Longleftrightarrow0\leq x\leq1$ – mjh Aug 09 '14 at 17:28
  • @mih Thanks so much. – user19059 Aug 09 '14 at 17:31
  • You're welcome. If you want an alternative way to look at it, note that the closed unit interval $[0,1]$ is just the open unit interval $(0,1)$ with the endpoints added, so $(0,1)\cup{0,1}=[0,1]$. Then it follows easily since $A\subseteq A\cup B$ for all sets $A$ and $B$: $(0,1)\subseteq(0,1)\cup{0,1}=[0,1]$. – mjh Aug 09 '14 at 17:35
  • First if all, the notation here is all wrong. What does it mean for on cardinality to be the subset of another? – Thomas Andrews Aug 09 '14 at 17:38
  • @ThomasAndrews The actual problem is: Use the Schroder-Bernstein Theorem to prove that |(0,1)|=|[0,1]|. My thoughts were to attempt to show that |(0,1)| and |[0,1]| are subsets of each other. – user19059 Aug 09 '14 at 17:41
  • $|X|$ is not a set, so it can't be a subset of anything. @user19059 You can say $|X|\leq |Y|$, you cannot say $|X|\subseteq |Y|$. – Thomas Andrews Aug 09 '14 at 17:44
  • There is a solution in the back of the text. I just want to make sense of it. The start of the proof given: Since (0,1) $\subseteq$ [0,1], the function i: (0,1) -> [0,1] defined by i(x)=x. @ThomasAndrews – user19059 Aug 09 '14 at 17:45
  • Yes, $(0,1)\subseteq [0,1]$, but that means $|(0,1)|\leq |[0,1]|$. – Thomas Andrews Aug 09 '14 at 18:14
  • I get it, thank you so much @ThomasAndrews – user19059 Aug 09 '14 at 19:12
  • Related posts: http://math.stackexchange.com/questions/300815/show-that-open-segment-a-b-close-segment-a-b-have-the-same-cardinality, http://math.stackexchange.com/questions/783020/mathbb-r-has-the-same-cardinality-of-any-interval (and many other posts) – Martin Sleziak Aug 16 '14 at 05:48

3 Answers3

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Choose an element in $(0,1)$. Can you identify it with a point in $[0,1]$?

For example, I might identify $\frac 12$ in $(0,1)$ with $\frac 12$ in $[0,1]$. You might say "That's obvious!" I'd say you're right, you're probably overthinking it (which happens all the time).

davidlowryduda
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We can define the following one-to-one function: \begin{align*} f&:[0,1]\to(0,1) \\ f&(x)=\left\{ \begin{array}{lcr} 1/2 && \mbox{if}\;x=0,\\ 1/3 && \mbox{if}\;x=1,\\ 1/(n+2) && \mbox{if}\;x=1/n,\,n=2,3,\cdots,\\ x && \mbox{if}\;x\neq0,1/n,\,n\in\mathbb{N}. \end{array} \right. \end{align*}

Mehdi Nadjafikhah
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I am not sure if you mean $(0,1) \subseteq [0,1]$ or that the cardinality of $(0,1)$ is smaller than $[0,1]$ (I would write $|(0,1)| \leq |[0,1]|$ or $(0,1) \preceq [0,1]$ for this).

About the proof: $x \in (0,1) \Leftrightarrow x <1 \wedge x>0 \Rightarrow x \leq 1 \wedge 0 \leq x \Leftrightarrow x \in [0,1]$

Martin Sleziak
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supinf
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