# Quick question from someone who has never taken topology

## Summary:

Suppose I wanted to prove that the Cartesian product of two compact sets is also a compact set. Does it suffice to simply state that the Cartesian product of the finite subcovers for the two sets covers the Cartesian product of the two sets?

## Main Question or Discussion Point

I only took an introductory real analysis course, and that was during the spring of last year.

I apologize for the unnecessary and possibly stupid question, in any case.

## Answers and Replies

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fresh_42
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Yes, if you add that the Cartesian product is equipped with the product topology. But you start with a open cover of the product set. You have to break this down onto the two components, not the other way around.

PeroK
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Summary: Suppose I wanted to prove that the Cartesian product of two compact sets is also a compact set. Does it suffice to simply state that the Cartesian product of the finite subcovers for the two sets covers the Cartesian product of the two sets?
No. Because these sets may not be a sub cover of the original open cover.

For example, if both open covers have a set of which the compact set in question is a subset, then the Cartesian product of these may not be part of the original open cover.

For example, if both open covers have a set of which the compact set in question is a subset, then the Cartesian product of these may not be part of the original open cover.
I am trying to wrap my head around what you are saying. To this end, I tried drawing a picture of what you are describing.

Here is a Cartesian product of two compact sets in two-dimensional Euclidean space.

Then I drew two members of the open covers for both $A$ and $B$. I assumed that these two open covers cover the whole line containing their respective sets. I drew only the two members of the covers that are supersets of $A$ and $B$, for simplicity's sake. They are colored red.

And then I drew the Cartesian product of these two members of the open cover:

You said that the red rectangle was not going to be part of the original open cover, which by previous assumption:

"I assumed that these two open covers cover the whole line containing their respective sets."

...cover the whole two-space. I cannot quite understand what you mean when you say that the Cartesian product may not be a part of the original open cover. I assume you mean part of the original open cover of the rectangle.

Come to think of it, though, the Cartesian product of two open covers comprise only one class of open covers of a closed rectangle. That is to say, you can cover the rectangle with not just open rectangles, but infinitely more types of sets in two-space. For example, you can cover it with a combination of open circles, open triangles, or some crazy shapes that don't contain their borders.

The definition of compact was roughly this, I think: "If $A \times B$ is compact, then for all open covers of $A \times B$, there exists a finite subcover that covers $A \times B$." If an open cover of the rectangle consists entirely of the Cartesian product of two open covers, then my opening post states that this contains a finite subcover.

My guess is that you are saying that the Cartesian product of two finite subcovers, which would just be an open rectangle containing the closed rectangle, may not be contained inside an arbitrary open cover of the rectangle, which may not necessarily be a single open rectangle. I drew an arbitrary open cover of the rectangle in purple below.

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PeroK
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That's not a proof.

I don't see that your rectangles generalise to compact sets and open covers.

Your language is too loose. You don't have a Cartesian product of open covers, you have a Cartesian product of sets.

My notes should give you a clue on how to construct a counterexample to your claim.

Finally, you should aim for a proof. Then you should see it's not so simple as it looks at first.

fresh_42
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I have said where to start, too.
1. List what is given: 3 facts.
2. Look at the definitions: 2 basic ones.
3. Write down what you have to show, i.e. end up with: 1 statement.
4. See whether you can find the way from 1 and 2 to 3.
(... and check whether your proof works in @PeroK's case in post #3 as well!)

You must be careful with drawings when it comes to topology. They can help if you know what to do, but you could as well easily overlook the critical cases. Maybe you should play a bit with the concepts first. Take $\mathbb{R}^1$ as an example and try to understand what the difference is between $(\mathbb{R^2}=\mathbb{R}^1\times \mathbb{R}^1\, , \,\|\cdot\|)$ with the norm topology and $(\mathbb{R^2=\mathbb{R}^1\times \mathbb{R}^1}\, , \,|\cdot| \times |\cdot |)$ with the product topology.

I have said where to start, too.
I apologize. I should have asked for further clarification on your response.

Yes, if you add that the Cartesian product is equipped with the product topology. But you start with a open cover of the product set. You have to break this down onto the two components, not the other way around.
First, I am confused by the use of the term "product topology", and for that matter, "norm topology". Also, you told me that I started with an open cover of the product set, which I assume to be the Cartesian product of the finite covers of $A$ and $B$ and $A\times B$ itself, respectively. You said that I should break this and the "product topology" down into the "two components". I am confused as to what these two components are.

1. List what is given: 3 facts.
2. Look at the definitions: 2 basic ones.
3. Write down what you have to show, i.e. end up with: 1 statement.
I want to be sure if I am working with the right information here, so I'm listing here what I think the right information is:

1.
(a) $A$ is compact.
(b) $B$ is compact.
(c) If $\alpha$ and $\beta$ are finite covers for $A$ and $B$ respectively, then $\alpha \times \beta$ is a finite open cover for $A\times B$.

2.
(a) "A set $U$ is compact iff for all open covers containing $U$, there exists a finite subcover that also contains $U$."
(b) "An open cover for a set $U$ is a collection of open sets that contain $U$."

3. "For every open cover of $A\times B$, there is a finite subcover."

fresh_42
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First, I am confused by the use of the term "product topology",...
We have a compact set $A\subseteq X$ in one topological space, and another compact set $B\subseteq Y$ in another topological space. At least you haven't specified the situation any further. Now we consider $A\times B$ which is a subset of $X \times Y$. But $X \times Y$ isn't automatically a topological space. We have to define a topology on it. One possibility is the product topology. Without mentioning it, $A\times B$ is compact doesn't make sense. We need a topology. Which one?
... and for that matter, "norm topology".
In normed spaces as in my example the real Euclidean plane, we have a norm: the distance of a point from the origin, and a metric, i.e. a distance between two points. Basic open sets are then defined as all points with a distance properly smaller than a certain value, the so called open balls $B_r(x_0)=\{\,x \in \mathbb{R}^2\,|\,|x-x_0| < r\,\}$. So the norm and with it the distance defines openness. But this is a different topology than the one gained by the product $\mathbb{R}^1 \times \mathbb{R}^1\,.$
Also, you told me that I started with an open cover of the product set, which I assume to be the Cartesian product of the finite covers of $A$ and $B$ and $A\times B$ itself, respectively. You said that I should break this and the "product topology" down into the "two components". I am confused as to what these two components are.
Yes, it appeared to me that you had the right idea but confused the direction of prove. As I wasn't sure, I just said what to do.
I want to be sure if I am working with the right information here, so I'm listing here what I think the right information is:

1.
(a) $A$ is compact.
(b) $B$ is compact.
(c) If $\alpha$ and $\beta$ are finite covers for $A$ and $B$ respectively, then $\alpha \times \beta$ is a finite open cover for $A\times B$.
No, that is the end of the story, not it's beginning. What we have at the beginning is:
1. $X\times Y$ is the Cartesian product of two topological spaces equipped with the product topology. (see link above)
2. $A\subseteq X$ and $B\subseteq Y$ are compact sets.
3. $A\times B \subseteq \bigcup_{\iota \in I} U^\times_\iota$ is a (possibly infinite) open cover of the set $A\times B$.
2.
(a) "A set $U$ is compact iff for all open covers containing $U$, there exists a finite subcover that also contains $U$."
(b) "An open cover for a set $U$ is a collection of open sets that contain $U$."
Let's stay with the compact sets $A$ and $B$ and reserve $U$ for open sets. Yes, we know that if $A\subseteq \cup_{\iota \in I^{A}} U^{A}_\iota$ is a open cover, that is the $U^{A}_\iota \subseteq X$ are all open, then there exist a set $\{\,i_1,\ldots,i_n\,\} \subseteq I^{A}$ such that $A\subseteq U^{A}_{i_1}\cup \ldots \cup U^{A}_{i_n}$. Similar is true for $B \subseteq\cup_{\iota \in I^{B}} U^{B}_\iota$ is a open cover, then there are $\{\,j_1,\ldots,j_m\,\} \subseteq I^{B}$ such that $B\subseteq U^{B}_{j_1}\cup \ldots \cup U^{B}_{j_m}$.

This counts as one definition.
The other one I meant is that of the product topology: What are open sets in $X\times Y\,?$
3. "For every open cover of $A\times B$, there is a finite subcover."
Yes. So we may assume an open cover
$$A\times B \subseteq \bigcup_{\iota \in I} U^\times_\iota \subseteq X\times Y$$
This is at first an arbitrary open cover, i.e. the index set $I$ can be infinitely large and all we know is that the $U_\iota$ are open in the product topology. But we only have information about $A$ and $B$ separately. We do not have open sets $U^{A}_{i_k}$ and $U^{B}_{j_l}$ at this stage of the proof!

So how do we get a finite index set $\{\,i_1,\ldots,i_n\,\}\subseteq I$ with $A\times B \subseteq U^\times_{i_1} \cup \ldots\cup U^\times_{i_n}\,?$ We have to find those open sets!

WWGD
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One issue to consider is that not all product open sets are of the form $U_a \times U_b ; U_a \in \tau _A, U_b \in \tau_B$. These are just a basis for the product topology. There is no easy way I am aware of of characterizing the open sets of the product topology.

mathwonk
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Does this work?
Assume A and B compact (in Heine Borel sense). Take any open cover of AxB.

EDIT:
we want to find a finite subcover, i.e. a cover by a finite number of sets that are among those in the original cover. notice that it suffices to find a finite refinement, i.e. a finite cover by sets each one of which is contained in one set of the original cover, since then we can go back and replace the finitely many sets of the refinement by the finitely many larger sets of the original cover that contain them.

1) since product sets are a basis, we can reduce to a refinement by product open sets of form UxV.

2) for each point a of A, the subspace {a}xB is homeomorphic to B hence compact, so we can take a finite subcover of {a} x B by some of those product sets. next we want to take a further refiement by making all the ( finitely many) product sets covering {a}xB have the same base set U.

3) for each point a, choose a refining cover of {a}xB by product sets with all first factors equal, say to Ua, by intersecting all the first factors in the given finite subcover of {a} x B. then {a} xB is contained in
Ua x Va,1 + Ua x Va,2 +...+Ua x Va,n. In particular B = Va,1 +...+Va,n, (where + denotes union.)

4) Since the sets Ua cover A, and A is compact, choose a finite subcover of A by some of the sets Ua. say A = Ua1 + Ua2 +...+Uam.

(not essential, but simplifies notation: By taking n = max of all n's that occur, may assume there are exactly n V's for every one of the finitely many a's remaining, i.e. can just repeat some of the sets.)

5) now you are done.
I.e. AxB = (Ua1 x Va1,1 + Ua1 x Va1,2 +...+ Ua1 x Va1,n) +.....+ (Uam x Vam,1 +....+Uam x Vam,n)
= Ua1 x B +....+ UamxB = AxB.

I.e. we have a finite refinement of the original cover, which can then be replaced by a finite subcover.

this was fun, since i never saw clearly how to do this as a student. it was a phd prelim problem when i was a student but the grader picked on my "solution" as i recall, as too complicated. maybe it still is.

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WWGD
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Edit:I wonder if this works, since it seems too simple. Use that compactness is equivalent to every net having a convergent subnet. Then the net $(x_{\alpha}, y_{\beta}) \in A \times B$ has the convergent subnet $(x_{\alpha'},y_{\beta'})$ where each is convergent in $A,B$ respectively using the projection maps $\pi_1, \pi_2$ respectively. But then again maybe we can't just assume that compactness and every net has a convergent subnet are equivalent and we should prove it.

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WWGD
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Does this work?
Assume A and B compact (in Heine Borel sense). Take any open cover of AxB.
1) reduce to a subcover by product open sets of form UxV.

2) for each point a of A take a finite subcover of {a} x B by some of those product sets.

3) for each point a, choose a subcover of {a}xB by product sets with all first factors equal, say to Ua, by intersecting all the first factors in the given finite subcover of {a} x B. then {a} xB is contained in
UaxVa,1 + UaxVa,2 +...+UaxVa,n. In particular B = Va,1 +...+Va,n.

4) Choose a finite subcover of A by some of the sets Ua. say A = Ua1 + Ua2 +...+Uam.

(not essential: By taking n = max of all n's that occur, may assume there are exactly n V's for every one of the finitely many a's remaining.)

5) now you are done.
I.e. AxB = (Ua1 x Va1,1 + Ua1 x Va1,2 +...+ Ua1 x Va1,n) +.....+ (Uam x Vam,1 +....+Uam x Vam,n)
= Ua1 x B +....+ UamxB = AxB.

this was fun, since i never saw clearly how to do this as a student. it was a phd prelim problem when i was a student but the grader picked on my "solution" as i recall, as too complicated. maybe it still is.
What do you mean by the + sign in a general topological space?

mathwonk
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sorry, i meant union when I wrote +. and i used the wrong language, meaning refinement instead of subcover.

I like the idea to use nets since it occurred to me the proof should be easy for metric spaces using sequences. But I also agree that nets may seem a little unintuitive and/or abstract and don't really prove what is asked without further argument, which is perhaps longer than doing it directly. But I like the observation; I did "learn" nets once sometime in the 60's from J.L. Kelley's general topology.

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WWGD
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Nice trivia: Examples of sets open in $\mathbb R^2$ product topology that are not of the form:
$U_a \times U_b ; U_a \in \tau_A ; U_b \in \tau_B: \{(x,y): x \neq y \} , \{(x,y): x^2+y^2 >1 \}$

Source is http://at.yorku.ca/cgi-bin/bbqa , a heaven for anyone into pointset topology by Henno Brandsma from the famed Dutch point set topology school.

WWGD
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sorry, i meant union when I wrote +. and i used the wrong language, meaning refinement instead of subcover.

I like the idea to use nets since it occurred to me the proof should be easy for metric spaces using sequences. But I also agree that nets may seem a little unintuitive and/or abstract and don't really prove what is asked without further argument, which is perhaps longer than doing it directly. But I like the observation; I did "learn" nets once sometime in the 60's from J.L. Kelley's general topology.
I agree it is a bit unintuitive. I came to understand it better when I had to use it to show (net) convergence of the Riemann integral. I was curious about what sense the R-integral converged. I researched it and I had to use all this material in net convergence.

fresh_42
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I thought that product topology and norm topology would be different. Looks as if they are not. But knowing this should make the proof easier.

mathwonk
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there are many norms on R^2, euclidean norm, with circular "balls", max norm, with square "balls", and sum norm with diamond shaped balls. the balls are always a basis. all norms on R^2 are equivalent, i.e. define same topology.

WWGD
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Interesting result ( not originally mine) on product spaces: $\mathbb R^3$ " Is not a square root" , meaning it is not homeomorphic to a product $Z \times Z$ for some $\mathbb Z$. Assume it is, so that $h: Z^2 \rightarrow \mathbb R^3$ is one homeomorphism. Use then the fact that homeomorphisms $h$have a degree $\pm 1$ assigned to them and degree map D is multiplicative, meaning $D(f \circ g)=D(f)D(g)$D so that $h \circ h=1$. Consider again the assumed homeo $h: X^2 \rightarrow \mathbb R^3$. We then use that it follows that $X^4 , \mathbb R^6$ are also homeo . We now construct a linear homeo $L: X^4 \rightarrow X^4$ so that $D(L \circ L)=1$ but its equivalent map $L': \mathbb R^6 \rightarrow \mathbb R ^6$ has the property $D( L' \circ L') =-1$ , a contradiction. Will finish ASAP as they are closing my coffee shop. Feel free to finish it for me.

Math_QED
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I thought that product topology and norm topology would be different. Looks as if they are not. But knowing this should make the proof easier.
Haha yes, the product $\prod_{i \in I} X_i$ is metrizable iff $I$ is at most countable and $X_i$ is metrizable for all $i \in I$.

If there are only finitely many factors, it is easily seen that the maximumnorm induces the product topology, so norm topology and product topology definitely coincide.

WWGD
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Any two norms in a finite-dimensional space coincide, as Wonk said. Moreover, a useful result is that many properties , including compactness here, can be tested by using the basis: If every cover by basic open sets admits a finite basic subcover, then the space is compact.

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fresh_42
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If there are only finitely many factors, it is easily seen that the maximumnorm induces the product topology
This had been the step I hadn't made (likely late at night). I only saw an open square and an open ball at my inner eye. That the square is actually the maximumsnorm ... Well, happens.

WWGD
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None of which sheds any light in proving compactness of the product.

Math_QED
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None of which sheds any light in proving compactness of the product.
I think the context of OP's question was an Euclidean space. Product of two compacts is then closed and bounded. Done.

In metric spaces just take a sequence in the product and extract a convergent subsequence.

In arbitrary topological spaces use the cover definition as above explained. The proof I saw was Tychonoff' theorem which used convergence of ultrafilters,but this was for a product with any number of factors (countable or uncountable) and definitely overkill for finitely many factors.

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