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Georg Cantor postulated a theorem that states that for any set (even if it's an infinite set) $A$, the power set of A ($\mathbb{P}(A)$) has cardinality greater than $A$.

Could this theorem also be extended to the idea that:

For any set $A$, if $A \subset B$ but $B \not\subset A$, then cardinality of B $>$ cardinality of A.

I am trying to see if uncountable infinite sets $A$ and $B$. Note: $A$ = set of elements in the open range $(0, 1)$, and $B$ = set of elements in the open range $(0, \infty)$

If the mentioned idea holds I can easily prove that $A \subset B$ but $B \not\subset A$ and thus $|B| > |A|$.

If this the right intuition or did I misinterpret something at some point?

Asaf Karagila
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Joel
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    Let $B$ be the natural numbers and $A$ be the even numbers. We got $A \subset B$ but $B \not\subset A$. But the cardinality of $A$ and that of $B$ are the same. You misinterpret something here. Cantor proved that $|\mathcal{P}(A)| > |A|$ for all set $A$, but the idea that you suggest is not true and it surely does not extend the idea of Cantor's theorem. – M.G Feb 05 '16 at 20:52
  • Cool. That's a clear counter example for A and B being countable sets. Could we also find one for uncountable sets? – Joel Feb 05 '16 at 20:54
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    You provided one ! You should be able to find a bijection between $(0,1)$ and $(0,\infty)$, thus showing that their cardinality is the same. Such a bijection is $f(x) := \frac{1}{x} - 1$, defined on $(0,1)$. – M.G Feb 05 '16 at 20:54
  • Indeed, one definition of "infinite set" is to have the same cardinality as a proper subset. Given an infinite set, $X$, find a countable subset ${x_1,x_2,\dots,x_n,\dots}$. Then define $f(x)=x_{i+1})$ when $x=x_i$ and $f(x)=x$ otherwise. Then $f$ is a bijection between $X$ and $X\setminus {x_1}$. – Thomas Andrews Feb 05 '16 at 20:55
  • A simpler example for uncountably infinite is the bijection $[0,1]\to [0,2]$ defined as $x\mapsto 2x$. – Thomas Andrews Feb 05 '16 at 20:59
  • The theorem clearly cannot be extended this way because of the many counterexamples that existed before Cantor was even born. – Asaf Karagila Feb 05 '16 at 21:00
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    Basically, abandon your conception that being a proper subset means being "strictly smaller" in the sense of cardinality. That conception works for finite sets, but is 100% wrong for infinite sets - ever infinite set has proper subsets of the same cardinality. – Thomas Andrews Feb 05 '16 at 21:02
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    See also: http://math.stackexchange.com/q/182236 – Asaf Karagila Feb 05 '16 at 21:07
  • So, for uncountable infinite sets $A$ and $B$ to have the same cardinality, it suffices to show that there's a one way bijection. In this case, it is from $A \to B$. – Joel Feb 05 '16 at 21:17
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    Well, any bijection is invertible, so it is not really one way. If we have $f:A\rightarrow B$ bijective, then we have $f^{-1}:B\rightarrow A$ also bijective.

    But we actually define cardinality of infinite sets this way. The infinite set $A$ has the same cardinality as $B$ iff there exists a bijection between $A$ and $B$.

     – kagof Feb 05 '16 at 21:19

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Your intuition holds up for finite sets, but infinities tend to defy intuition sometimes.

Consider the function $f:(0,1)\rightarrow (0,\infty)$, defined by $f(x)=\tan(\frac{\pi x}{2})$. This is a bijection between the two sets, thus they have the same cardinality.

kagof
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