# Quesion on topology relating to produc spaces

• March 1st 2010, 04:54 AM
facenian
Question 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".
• March 1st 2010, 07:10 AM
HallsofIvy
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.
• March 1st 2010, 10:16 AM
facenian
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
• March 1st 2010, 01:07 PM
Drexel28
Quote:

Originally Posted by facenian
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

Theorem: let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be a non-empty class of topological spaces and let $X=\prod_{j\in\mathcal{J}}X_j$ be under the product topology. Then, given any class of subsets $\left\{E_j\right\}_{j\in\mathcal{J}}$ such that $E_j\subseteq X_j,\text{ }\forall j\in\mathcal{J}$ we have that $\prod_{j\in\mathcal{J}}\overline{E_j}=\overline{\p rod_{j\in\mathcal{J}}E_j}$

Proof: Let $\bold{x}\in\prod_{j\in\mathcal{J}}\overline{E_j}$. Let $N$ be any neighborhood of $\bold{x}$. Clearly we may find a basic open set $B$ (in the defining open base) such that $\bold{x}\in B\subseteq N$. Since $\pi_j(\bold{x})$ is an adherent point of $E_j$ we may pick some $\alpha_j\in E_j\cap \pi_j(B)$. Clearly then $\prod_{j\in\mathcal{J}}\{\alpha_j\}\in B\cap \prod_{j\in\mathcal{J}}E_j$. Since $N$ was arbitrary it follows that $\bold{x}$ is an adherent point for $\prod_{j\in\mathcal{J}}E_j$ and thus $\bold{x}\in\overline{\prod_{j\in\mathcal{J}}E_j}$

Conversely, let $\bold{x}\in\overline{\prod_{j\in\mathcal{J}}E_j}$. Let $j\in\mathcal{J}$ be arbitrary and consider any neighborhood $N_j$ of $\pi_j(\bold{x})$. Since $\pi_j^{-1}(N_j)$ is open in the product topology we see that it contains a point $\bold{y}\in\prod_{j\in\mathcal{J}}E_j$. It follows then that $\pi_j(\bold{y})\in E_j\cap N_j$. In other words, $\pi_j(\bold{x})$ is an adherent point for $E_j$ and thus $\pi_j(\bold{x})\in\overline{E_j}$. It follows that $\bold{x}\in\prod_{j\in\mathcal{J}}\overline{E_j}$.

The conclusion follows. $\blacksquare$

There are two nice corollaries

Corollary:Let $\left\{E_j\right\}_{j\in\mathcal{J}}$ be a collection of closed sets such that $E_j\subseteq X_j$. Then $\overline{\prod_{j\in\mathcal{J}}E_j}=\prod_{j\in\ mathcal{J}}\overline{E_j}=\prod_{j\in\mathcal{J}}E _j$ from where it follows that $\prod_{j\in\mathcal{J}}E_j$ is closed.

Corollary: Let $\left\{\mathfrak{D}_j\right\}_{j\in\mathcal{J}}$ be a collection of dense subsets of $X_j$. Then, $\overline{\prod_{j\in\mathcal{J}}\mathfrak{D}_j}=\ prod_{j\in\mathcal{J}}\overline{\mathfrak{D}_j}=\p rod_{j\in\mathcal{J}}X_j=X$. From where it follows that $\prod_{j\in\mathcal{J}}\mathfrak{D}_j$ is dense in $X$
• March 2nd 2010, 04:44 AM
mabruka
Quote:

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.
But drexel isnt

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

Sorry i did not understand your example :)
• March 2nd 2010, 03:26 PM
Drexel28
Quote:

Originally Posted by mabruka
But drexel isnt

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

Sorry i did not understand your example :)

I didn't say that.