Re: less-than-uncool preschool
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Reply #40 –

Now we can give each natural number its conventional name.

We already have 1, from P1.

We know from P2 that 1* exists, and from P3 that 1* ≠ 1. So let's give 1* a different name. Let's define 2 = 1*.

We know from P2 that 2* exists, and from P3 that 2* ≠ 1, and from P4 that 2* ≠ 2 (because 2 = 1*, and 2 ≠ 1). So let's give 2* a different name. Let's define 3 = 2*.

We know from P2 that 3* exists, and from P3 that 3* ≠ 1, and from P4 that 3* ≠ 2 (because 2 = 1*, and 3 ≠ 1), and from P4 again that 3* ≠ 3 (because 3 = 2*, and 3 ≠ 2). So let's give 3* a different name. Let's define 4 = 3*.

Carrying on in this way - giving the successor of each newly-named number its own name - we are sure never to name the same number twice.

Let's define 5 = 4*, 6 = 5*, 7 = 6*, 8 = 7*, 9 = 8*, and 10 = 9*.

Our notation "10" consists of two symbols ("digits") strung together, but should be treated as a single symbol. Similarly, the conventional name for every natural number other than those in {1, 2, 3, 4, 5, 6, 7, 8, 9} will also consist of two or more digits strung together, but each such multi-digit name should be treated as a single symbol. Each digit will be one of {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9"}, but the leftmost digit will not be "0".

For convenience, let's define a function *t* : ℕ \ {1, 2, 3, 4, 5, 6, 7, 8, 9} → ℕ, where for any *n* ∈ dom *t*, the number *t*(*n*) is the number whose conventional name is given by taking the conventional name of *n* and deleting the rightmost digit.

Let *n* ∈ dom *t*. Then we have the following three cases, to derive the conventional name for *n** from the conventional name for *n*:

Case 1: If the rightmost digit of the conventional name for *n* is "0", then the conventional name for *n** is given by appending the digit "1" to the right of the conventional name for *t*(*n*).

Case 2: If the rightmost digit of the conventional name for *n* is one of {"1", "2", "3", "4", "5", "6", "7", "8"}, then let *m* be the number in {1, 2, 3, 4, 5, 6, 7, 8} whose name is the same as the rightmost digit of the conventional name for *n*. In this case, the conventional name for *n** is given by appending the digit corresponding to the conventional name for *m** to the right of the conventional name for *t*(*n*).

Case 3: If the rightmost digit of the conventional name for *n* is "9", then the conventional name for *n** is given by appending the digit "0" to the right of the conventional name for *t*(*n*)*.

Case 3 is recursive, because the conventional name for *t*(*n*)* will itself fall into one of those three cases, unless it consists of only one digit, in which case the recursion is terminated.

So for every natural number that is named by the process above, the successor of that number is also named.

Therefore, if we let *A* = {*n* ∈ ℕ | *n* is given a name according to the process in this post}, then P5 tells us that *A* = ℕ. So every natural number is given a name.