How does my bijection between the natural numbers and the powerset of natural numbers break down?

Question

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)?

Answer 1

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

Answer 2

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

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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 $0001_2 = 1$.