closure of dual group in pointwise convergence.

Question

I have something which seems kind of trivial but I can't seem to prove it. Let G be an abelian topological group and let T be the circle group. Denote by G* the group of all continuous homomorphisms from G to T. I need to show that G* is a p-closed subset of C(G,T) (the set of all continuous functions from G to T). p-closed means, closed in the pointwise convergence topology. Thanks for the help

Answer 1

Let $f ∈ C(G, T) \setminus G^*$. There are some $x, y ∈ G$ such that $f(x + y) ≠ f(x) + f(y)$. Take $U_{x + y}$, $V$ disjoint open in $T$ such that $f(x + y) ∈ U_{x, y}$ and $f(x) + f(y) ∈ V$. By continuity of $+$ there are some open $U_x ∋ f(x)$, $U_y ∋ f(y)$ such that $x' + y' ∈ V$ for every $x' ∈ U_x$, $y' ∈ U_y$. Now, $\{g ∈ C(G, T): g(x) ∈ U_x,\ g(y) ∈ U_y,\ g(x + y) ∈ U_{x + y}\}$ is an open neighborhood of $f$ disjoint with $G^*$.

That was maybe too elementary. We may observe, that $f \mapsto f(x + y)$ and $f \mapsto f(x) + f(y)$ are continuous functions so $\{f: f(x + y) = f(x) + f(y)\}$ is an equalizer of two continuous functions going into a Hausdorff space, which is closed.