User: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.