products of uniform space

Question

$X$ is a topological space(or even a nonempty set), and $(Y,Φ)$ is a uniform space, then $Y^X$ is a uniform space, too.

$\widetilde{U}=\{(f,g):$for any $x\in X, (f(x),g(x))\in U\}$, $\Psi=\{\widetilde{U}:U \in Φ\}$ is just the uniform structure on $Y^X$.

There are two questions which bother me.

First, $\Psi$ is a filter, how to prove that $\Psi$ is an upper set, or upward closed?

Second, the product of uniform spaces $(X_t,\mathfrak{A}_t),\,t\in T$, is the uniform space $(\prod X_t,\prod\mathfrak{A}_t)$. This is another general definition of the product of uniform space. The two definitions of uniform structure of $Y^X$ must coincide?

Thanks a lot.

Answer 1

It seems the following.

$\Psi$ is a filter, how to prove that $\Psi$ is an upper set, or upward closed?

I think that the family $\Psi$ is not a uniformity, but only a base of a uniformity. For instance, let $X=\{0,1\}$, $|Y|\ge 2$ and $\Phi$ be a uniformity on the set $Y$ consisting of all entourages of the diagonal $\Delta_Y$. Then the diagonal of the space $Y^X$ belongs to the family $\Psi$. So if $\Psi$ were uniformity then each entourage of the diagonal of the space $Y^X$ belongs to the family $\Psi$. But an entourage $\mathcal U=\{(f,g)\in Y^X\times Y^X: f(0)=g(0)\}$ does not belong to the family $\Psi$, because if $\mathcal U\in\Psi$ then $\mathcal U=\widetilde{\Delta_Y }$, which is not true.

the product of uniform spaces $(X_t,\mathfrak{A}_t),\,t\in T$, is the uniform space $(\prod X_t,\prod\mathfrak{A}_t)$. This is another general definition of the product of uniform space. The two definitions of uniform structure of $Y^X$ must coincide?

It depends on the definition of the entourages on the $\prod X_t$. The approach with $\widetilde{U}$ is like a box product topology on $Y^X$.