Andrewdodd13

Let us define a set N₁ to be the set of all natural numbers greater than 0.

$N_{1}=\{f\in N:f>0:f\}$

We can define an enumeration function from the naturals to this set N₁ like so:

$f(x)=x+1$

Let us define another set which we shall call T (even though it can be easily seen that T = N), which we shall base on the union of our set N₁ with the set containing only the number 0.

$T=\{0\}\cup N_{1}$

We can enumerate this function with a basic function

$g(x)=x$

Since we have an enumeration function with an unbounded upper value for x for both N₁ and T, both sets have the same cardinality of that of the natural numbers. However, surely this cannot be right as |N₁ = |T - 1 (as N₁ is the same as T less 1 element).

So we either have:

$|N|=|N|+1$

or

$0\in N_{1}$

Wat.