Is $(\Bbb{N}, \tau)$ a topological monoid with topology induced by an additive arithmetic function?

by Roll up and smoke Adjoint   Last Updated November 10, 2018 00:20 AM

For any function $f : \Bbb{N} \to \Bbb{N}$ we have that $d(m,n) = |f(m) - f(n)|$ is a pseudometric so it generates a topology.

If $f(mn) = f(m) + f(n)$ for all $m, n \in \Bbb{N}$ then we say that $f$ is completely additive.

The definition of continuity here is that for any open ball $B \subset (\Bbb{N}, d)$ and any $g(x) \in B$ there is an open ball $B' \ni x$ in the same space such that $g(B') \subset B$. Then this should extend naturally to all open sets.

So let $B = B_r(n) = \{ m : |f(m) - f(n)| \lt r\}$. Let $B \ni y = g(x) = mx$. Then $r \gt |f(y) - f(n)| = |f(mx) - f(n)| = |f(m) + f(x) - f(n)|$. I thought something would happen here but it didn't.

Is there a metrized topological monoid given an additive function?

Let $g(x) = ax$ (change of notation).

Define $B' = B_r(mn) = \{ z \in \Bbb{N} : |f(z) - f(m) - f(n)| \lt r\}$. Let $z \in B'$. Then $|f(mz) - \dots$

Working on it. I think this should work intuitively.

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