Can a countable union of disjoint compact Cartesian products be compact?

Question

Suppose we are given a disjoint collection $(A_i)_{i \geq 1} \subset \mathbb{R}^d$, each $A_i$ beeing a Cartesian product of $d$ closed intervals. Assume that we can simplify the union of all those sets in the following way $$ M=\bigcup_{i \geq 1} A_i = \bigcup_{j=1}^{n} B_j, $$ where $B_1, \ldots, B_n$ are disjoint Cartesian products of $d$ intervals. (We do not know if the $B_j$ consist of open or closed or half opened intervals.)

We know that the LHS is a countable union of disjoint compact sets. I am interested if we can say something about $M$. Is $M$ compact again? I know that this is not true in general, but does the finite union on the RHS help? If we could show that the $B_j$ are necessarily closed (and hence compact), this would imply the compactness of $M$.

Answer 1

If you have no information about the $B_i's$ you cannot conclude anything substantial. Remember that by Heine-Borel, if $M$ is compact then it must be closed ad bounded. The boundedness doesn't seem to be an issue, the big problem is, is the union closed.

Answer 2

This question turns out to be vacuous - no countable disjoint union of nonempty closed rectangles (in $\mathbb{R}^n$) can be connected.

Note that in the situation you describe, at least one of the connected components of $\bigcup B_i$ must cover infinitely many $A_i$s by pigeonhole, and each of these are connected; this is why the point above is relevant to your question. Specifically: there is no countable set $A_i$ of disjoint closed boxes such that their union can be written as a finite union of boxes of any type.

This is a consequence of the Baire category theorem; see e.g. Does not exist cover of $\mathbb{R}^n$ by disjoint closed balls.

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To motivate this, let's look at the case $n=1$ in more detail.

Let's assume $\bigcup A_i$ is connected, where each $A_i$ is a nonempty closed interval; WLOG (why?) let's also assume that each $A_i$ is bounded. Then we define a subsequence $A_{i_k}$ as follows:

• $i_1=1, i_2=2$.

• For $n>1$, $i_{n+1}$ is the least number $>i_n$ such that $A_{i_{n+1}}$ lies between $A_{i_n}$ and $A_{i_{n-1}}$.

That is, we pick a subsequence of the $A_i$s that "zig-zags" appropriately.

Now let $I_n$ be the smallest closed interval containing both $A_{i_n}$ and $A_{i_{n+1}}$ (so, fill in the middle).

Then $\mathcal{I}=\bigcap I_n$ is an intersection of compact sets, so nonempty; but (exercise) any element of $\mathcal{I}$ is not an element of any $A_i$. But this is a problem - that means the $A_i$s don't fully cover the gap between $A_1$ and $A_2$, hence their union isn't connected.

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Note that this proof, in fact, shows (with minor modifications) that in $\mathbb{R}^n$ the disjoint union of countably many compact sets is never compact. Exercise: is this true in an arbitrary topological space? What about uncountably many?