Existence of Uncomputable Mappings/Proof/Overview

Overview
every mapping is computable.

Hence for each mapping there exists a computer program or algorithm which computes it.

This computer program is a finite string of symbols from some alphabet.

Hence the set of all computer programs $\CC$ is countably infinite.

Consider now the set $S$ of mappings which maps the set of integers $\Z$ to the Boolean domain $\set {0, 1}$.

$S$ is countably infinite.

Let $S$ be placed in one-to-one correspondence with $\Z$.

Let $f_i$ be the mapping in $S$ which corresponds to the $i$th integer.

Now consider the mapping $f: \Z \to \set {0, 1}$ defined as follows:
 * $\forall n \in \Z: \map f n = \begin {cases} = 0 & : \map {f_n} n = 1 \\ 1 & : \text {otherwise} \end {cases}$

Then $\map f n$ cannot correspond to any integer.

This contradicts our supposition that $S$ is countably infinite.

Hence it cannot be the case that $\CC$ is countably infinite.

So $\CC$ is uncountable.

But this contradicts our deduction that $\CC$ is countably infinite.

Hence by Proof by Contradiction it cannot be the case that every mapping is computable.