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Lets consider some natural number x in binary. Let the least significant digit represent the inclusion or exclusion of 0, the next least significant represent 1, and so on upwards. For some examples:

$0$ ; $0_2$ ; {}
$7$ ; $111_2$ ; {0, 1, 2}
$8$ ; $1000_2$ ; {3}
$14$ ; $1110_2$ ; {1, 2, 3}

By cantors diagonalization argument I know that this can't be a bijection between the natural numbers and their powerset, but I'm having trouble proving that it's not. Can I prove that this is not a bijection without just quoting the cardinality of the sets (and preferably with a counterexample)?

matthew schallenkamp
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2 Answers2

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That's a bijection between $\mathbb N$ and the set of all finite subsets of $\mathbb N$. But $\mathbb N$ also has infinite subsets…

José Carlos Santos
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The proposed bijection only works for finite sets, as José Carlos Santos pointed out.

Even for finite sets, it doesn't work.

One issue is the multiplicity of assignments due to leading zeroes.

Consider the set $\{1\}$. This should be $01_2 = 1$, but $1_2=1$ has already been assigned to the set $\{0\}$.

Similarly, the set $\{3\}$ should also be $0001_2 = 1$.

Zubin Mukerjee
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  • I'm considering the bits in reverse order from how you are, so instead of ${3}$ being $0001_2$ it is $1000_2$. That should alleviate the leading $0$'s issue. – matthew schallenkamp Jan 15 '19 at 20:06