I read that, if we work with a language $L$ which only has function symbols and if we consider two $L$-structures $\mathbb{A}$ and $\mathbb{B}$ (with domains $A$ and $\lbrace 0,1 \rbrace$ respectively), the set of all homomorphisms (functions from $A$ to $\lbrace 0,1 \rbrace$ which interract as we'd expect with the functions in both structures) from $\mathbb{A}$ to $\mathbb{B}$ is a closed set of $\lbrace 0,1 \rbrace ^A$ (with the product topology).
My question is: does this result have any direct applications in the domain of model theory? But also... I can't prove this, and I think this is mainly because I cannot see how I would prove anything is closed in that topology...
I have tried to look at the power set of $A$ rather the functions, it hasn't helped much, the open sets still are quite complicated, and I remain stuck at that point.
Any kind of help would really be appreciated,
Thank you!
If $\varphi:A\to\{0,1\}$ is not a homomorphism, there is is a function symbol $f\in L$ such that $\varphi$ doesn’t interact properly with $f$. If $f$ is $n$-ary, this means that there are $a_1,\ldots,a_n\in A$ such that
$$\varphi\big(f^{\Bbb A}(a_1,\ldots,a_n)\big)\ne f^{\Bbb B}\big(\varphi(a_1),\ldots,\varphi(a_n)\big)\;.$$
Let $a_0=f^{\Bbb A}(a_1,\ldots,a_n)\in A$, and let
$$U=\left\{\psi\in\{0,1\}^A:\psi(a_k)=\varphi(a_k)\text{ for }k=0,\ldots,n\right\}\;;$$
then $U$ is an open nbhd of $\varphi$ in $\{0,1\}^A$, and
$$\psi\big(f^{\Bbb A}(a_1,\ldots,a_n)\big)\ne f^{\Bbb B}\big(\psi(a_1),\ldots,\psi(a_n)\big)$$
for each $\psi\in U$, so no member of $U$ is a homomorphism from $\Bbb A$ to $\Bbb B$.