Proof Verification: Show that if Y is compact, then the projection is a closed map

Question

I searched the site for other proofs that may be similar to mine and couldn't find one. I was hoping someone could review my variation in particular for correctness. Thanks in advance.

Problem:

Show that if Y is compact, then the projection $\pi$$_1$: X $\times$ Y -> X is a closed map.

Proof:

Let $\pi$$_1$: X $\times$ Y -> X be the projection function.
Let Y be compact. Let A be a closed set of X $\times$ Y.

Then A can be written as: X $\times$ Y - C where C is an open set in X $\times$ Y.

Note: C can be written as the union of basis elements. We can then just consider some open set of the form U $\times$ V for U open in X and V open in Y.

Then consider the infinite union of open sets: $\bigcup$$_{i=0}^\infty$U$_i$ $\times$ V$_i$.

Then two cases arise here:

case 1: $\bigcup$$_{\in I}V_i$ does not cover Y.

case 2: $\bigcup$$_{\in I}V_i$ does form an open covering of Y.

In case 1 we are left with: $\pi$$_1$(X $\times$ Y - $\bigcup$$_{i \in I}U_i$ $\times$ V$_i$) = X since the RHS of the difference does not contain all the x value combinations with Y for any particular x. Hence since X is closed in X, $\pi$$_1$ is a closed map in this case.

In case 2 we can take a finite covering Y of Y by compactness = $\bigcup$$_{i=0}^n$V$_i$. If we take the corresponding intersection:

D = $\bigcap$$_{i=0}^n$U$_i$ where each i corresponds to a V$_i$ that we chosen in our finite cover. Then in D we have the points in X that are removed completely in the calculation: X $\times$ Y - C. Note: removed completely refers to the fact that D represents the points that are of the form ($x_1$,Y) for $x_1$ $\in$ X and thus are completely removed from the operation X $\times$ Y - C. So those particular x coordinates from D will not be present in any tuple in X $\times$ Y - C.

We can repeat this procedure for EVERY open covering of Y that the set V$_i$ may hold and then compute the intersection of the associated U$_i$'s. Each result is an open set of points to be removed from X in the calculation X $\times$ Y - C. Since each $\bigcap$$_{i=0}^n$U$_i$ (the intersection result) is open, then the union of all the intersection results is also open. Call this final open set B.

Finally, we have that $\pi$$_1$(X $\times$ Y - $\bigcup$$_{i \in I}U_i$ $\times$ V$_i$) = X - B since we are only interested in the x values of the resulting set = (X $\times$ Y - $\bigcup$$_{i \in I}U_i$ $\times$ V$_i$) we must only consider x values on the RHS that attain all the points in ($x_1$, Y) for $x_1$ $\in$ X. For any $x_2$ that does not have all possible combinations with Y as $x_1$ does, the compliment operation will not fully remove that $x_2$ coordinate from the space X $\times$ Y. Thus the $x_2$ coordinate will thus still be in the projection result. Hence to compute the projection result we have X - B since only B values are fully removed (i.e represented in a form of ($x_1$,Y).

So in case two we have a closed map as well.

The case where we have a finite union of open sets in X $\times$ Y is just an easier case of the above. Hence $\pi$$_1$ is a closed map.

Answer 1

I would not accept this as a proof: it's way too sloppy. Some critique points:

• You suggest that an infinite union is countable (it could well be much bigger).

• You don't give a proof for the claim the projection of the set equals $X$ when the $V_i$ don't cover $Y$. Just some handwaving.

• You don't remove "isolated points" in the computation. Introduce notation. Explicitly use the compactness of $Y$. Why do you suddenly talk about all covers of $Y$ that contain some $V_i$?

• Define the set $B$ properly, and prove the equality (if it holds at all).

There are much better written proofs available on this site, e.g. in this thread and the simplest ones work from the assumption that $\pi[A]$ is not closed and use one cover of $Y$ and its finite subcover to get a contradiction.