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".

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- 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 be a non-empty class of topological spaces and let be under the product topology. Then, given any class of subsets such that we have that

**Proof:**Let . Let be any neighborhood of . Clearly we may find a basic open set (in the defining open base) such that . Since is an adherent point of we may pick some . Clearly then . Since was arbitrary it follows that is an adherent point for and thus

Conversely, let . Let be arbitrary and consider any neighborhood of . Since is open in the product topology we see that it contains a point . It follows then that . In other words, is an adherent point for and thus . It follows that .

The conclusion follows.

There are two nice corollaries

**Corollary:**Let be a collection of closed sets such that . Then from where it follows that is closed.

**Corollary:**Let be a collection of dense subsets of . Then, . From where it follows that is dense in - 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)} = ?

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