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".

Printable View

- Mar 1st 2010, 04:54 AMfacenianQuestion on topology relating to product spaces
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". - Mar 1st 2010, 07:10 AMHallsofIvy
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. - Mar 1st 2010, 10:16 AMfacenian
Thank you but my question was more general in the sense that I asked whether the product of any number of closed sets y general product space not yust a finite product space

- Mar 1st 2010, 01:07 PMDrexel28
**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 $\displaystyle X$ - Mar 2nd 2010, 04:44 AMmabrukaQuote:

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.

{(x,y), (x, z)} = $\displaystyle \{x\}\times \{y,z\}$ ?

Sorry i did not understand your example :) - Mar 2nd 2010, 03:26 PMDrexel28