Topology induced by uniformity (open and closed)

Question

If V is an open (closed) in product topology X$\times$X that induced by uniformity , then $V(x) $ is open (closed) in $X$ ?

Answer 1

For $x\in X$ and let define the function \begin{align} &\sigma_x:X\to X\times X& &y\mapsto (x,y) \end{align} Note that $\sigma_x$ is continuous, for if $\pi_1,\pi_2:X\times X\rightrightarrows X$ denote the canonical projections, then \begin{align} &(\pi_1\circ\sigma_x)(y)=x& &(\pi_2\circ\sigma_x)(y)=y \end{align} are both continuous functions, hence $\sigma_x$ is continuous by universal property of product topology. Then $$V(x)=\{y\in X:(x,y)\in V\}=\sigma_x^{-1}[V]$$ which is open (closed) as $V$ is open (closed) in $X\times X$, by continuity of $\sigma_x$. Note that this holds for every topology on $X$ (not necessarily induced by an uniformity).