# 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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