Prove that for every computable function $f \colon \mathbb N \to \mathbb N$, there exists a computable function $g \colon \mathbb N \to \mathbb N$ such that for every $n$ we have $f(n) ≤ g(n)$.
Can anyone please explain me, why my proof below is not correct?
Lets assume, for the sake of arriving at a contradiction, that there exists a computable function $f$ such that for every computable function $g$, there
exists an $n$ such that $f(n) > g(n)$.
For each computable function $g$, let $n(g)$ be the smallest $n$ such that $f(n) > g(n)$. Let $n_0$ be the maximum of $n(g)$ over all computable functions $g$, and assume that this maximum is achieved for $g_0$, that is, $n(g_0) = n_0$.
Define a function $g_1$ that is equal to $g_0$ for all values except for $n_0$, and for $n_0$ we have $g_1(n_0) = f(n_0)$. This function $g_1$ is computable because $g_0$ is computable and $g_1$ differs from $g_0$ in only one point. By definition, $n(g_1) ≥ n(g_0) + 1$, and this contradicts the fact that $n(g)$ was maximized at $g_0$.