Hello: I have what may be a silly question because did not find it in the books I looked for it and is this: Is the product of closed sets closed in the product topology?
The answer seems to me to be "yes".
Hello: I have what may be a silly question because did not find it in the books I looked for it and is this: Is the product of closed sets closed in the product topology?
The answer seems to me to be "yes".
Yes, the product of two closed sets is closed in the product topology, just as the product of two open sets is open.
But be careful! just as not every set in X x Y is the product of sets in x and y, so closed (or open) sets in X x Y may not be the product of closed (or open) sets in X and Y.
For example, the singleton set {(x, y)) in X x Y is the product of the two singleton sets {x} and {y} but {(x,y), (x, z)} in X x Y (so both y and z are in Y) is NOT the product of two sets in X and Y.
Theorem: let $\displaystyle \left\{X_j\right\}_{j\in\mathcal{J}}$ be a non-empty class of topological spaces and let $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ be under the product topology. Then, given any class of subsets $\displaystyle \left\{E_j\right\}_{j\in\mathcal{J}}$ such that $\displaystyle E_j\subseteq X_j,\text{ }\forall j\in\mathcal{J}$ we have that $\displaystyle \prod_{j\in\mathcal{J}}\overline{E_j}=\overline{\p rod_{j\in\mathcal{J}}E_j}$
Proof: Let $\displaystyle \bold{x}\in\prod_{j\in\mathcal{J}}\overline{E_j}$. Let $\displaystyle N$ be any neighborhood of $\displaystyle \bold{x}$. Clearly we may find a basic open set $\displaystyle B$ (in the defining open base) such that $\displaystyle \bold{x}\in B\subseteq N$. Since $\displaystyle \pi_j(\bold{x})$ is an adherent point of $\displaystyle E_j$ we may pick some $\displaystyle \alpha_j\in E_j\cap \pi_j(B)$. Clearly then $\displaystyle \prod_{j\in\mathcal{J}}\{\alpha_j\}\in B\cap \prod_{j\in\mathcal{J}}E_j$. Since $\displaystyle N$ was arbitrary it follows that $\displaystyle \bold{x}$ is an adherent point for $\displaystyle \prod_{j\in\mathcal{J}}E_j$ and thus $\displaystyle \bold{x}\in\overline{\prod_{j\in\mathcal{J}}E_j}$
Conversely, let $\displaystyle \bold{x}\in\overline{\prod_{j\in\mathcal{J}}E_j}$. Let $\displaystyle j\in\mathcal{J}$ be arbitrary and consider any neighborhood $\displaystyle N_j$ of $\displaystyle \pi_j(\bold{x})$. Since $\displaystyle \pi_j^{-1}(N_j)$ is open in the product topology we see that it contains a point $\displaystyle \bold{y}\in\prod_{j\in\mathcal{J}}E_j$. It follows then that $\displaystyle \pi_j(\bold{y})\in E_j\cap N_j$. In other words, $\displaystyle \pi_j(\bold{x})$ is an adherent point for $\displaystyle E_j$ and thus $\displaystyle \pi_j(\bold{x})\in\overline{E_j}$. It follows that $\displaystyle \bold{x}\in\prod_{j\in\mathcal{J}}\overline{E_j}$.
The conclusion follows. $\displaystyle \blacksquare$
There are two nice corollaries
Corollary:Let $\displaystyle \left\{E_j\right\}_{j\in\mathcal{J}}$ be a collection of closed sets such that $\displaystyle E_j\subseteq X_j$. Then $\displaystyle \overline{\prod_{j\in\mathcal{J}}E_j}=\prod_{j\in\ mathcal{J}}\overline{E_j}=\prod_{j\in\mathcal{J}}E _j$ from where it follows that $\displaystyle \prod_{j\in\mathcal{J}}E_j$ is closed.
Corollary: Let $\displaystyle \left\{\mathfrak{D}_j\right\}_{j\in\mathcal{J}}$ be a collection of dense subsets of $\displaystyle X_j$. Then, $\displaystyle \overline{\prod_{j\in\mathcal{J}}\mathfrak{D}_j}=\ prod_{j\in\mathcal{J}}\overline{\mathfrak{D}_j}=\p rod_{j\in\mathcal{J}}X_j=X$. From where it follows that $\displaystyle \prod_{j\in\mathcal{J}}\mathfrak{D}_j$ is dense in $\displaystyle X$