# Strongly additive open cover

Call an open cover $\mathscr{U}$ of a metric space $M$ strongly additive if whenever $U,V\in\mathscr{U}$ and $U\cap V\ne\emptyset$, then $U\cup V\in\mathscr{U}$. Prove that $M$ is compact and connected if and only if every strongly additive open cover of $M$ contains $M$.

My proof:

(1) If $M$ is not compact, there exists a strongly additive open cover of $M$ that doesn't contain $M$.

Suppose $M$ is not compact. Then it has an open cover $\mathscr{U}$ with no finite subcover. Consider this cover, appended by all finite unions of sets in this cover. It is strongly additive, and it does not contain $M$.

(2) If $M$ is not connected, there exists a strongly additive open cover of $M$ that doesn't contain $M$.

Suppose $M$ is not connected. Then it can be partitioned into two disjoint, nonempty open subsets $A,B$. Take this as a cover. It is strongly additive, and it doesn't contain $M$.

(3) If there exists a strongly additive open cover of $M$ that doesn't contain $M$, then $M$ is not compact or not connected.

Suppose there exists a strongly additive open cover $\mathscr{U}$ of $M$ that doesn't contain $M$. Consider the collection $\mathscr{V}\in\mathscr{U}$ of sets in $\mathscr{U}$ that are not a proper subset of any other set in $\mathscr{U}$. Note that $\mathscr{V}$ still covers $M$, because any set in $\mathscr{U}$ outside $\mathscr{V}$ is a subset of some set in $\mathscr{V}$. Also, any two sets in $\mathscr{V}$ are disjoint, because otherwise their union belongs to $\mathscr{U}$, and each of them is a proper subset of their union. So the open cover $\mathscr{V}$ consists of disjoint open sets. Since $M\not\in\mathscr{V}$, $\mathscr{V}$ consists of at least two elements. Let $A\in\mathscr{V}$. Let $B$ be the union of all sets in $\mathscr{V}$ beside $A$. $B$ is open as a union of open sets. Then $A,B$ are disjoint, nonempty open subsets of $M$ whose union is $M$. So $M$ is not connected.

In (3), I didn't have to use the condition "or compact". Where did I go wrong?

EDIT: Second try on (3) -- used the compactness following Tim's comment

Suppose there exists a strongly additive open cover $\mathscr{U}$ of $M$ that doesn't contain $M$, and that $M$ is compact. We'll show that $M$ is not connected. Since $M$ is compact, $\mathscr{U}$ has a finite subcover $\mathscr{U}'$. Take all unions of sets in $\mathscr{U}'$ and add it to $\mathscr{U}'$ -- note that $\mathscr{U}'$ still doesn't contain $M$, and is now strongly additive. Consider the collection $\mathscr{V}\in\mathscr{U}$ of sets in $\mathscr{U}'$ that are not a proper subset of any other set in $\mathscr{U}'$. Note that $\mathscr{V}$ still covers $M$, because any set in $\mathscr{U}'$ outside $\mathscr{V}$ is a subset of some set in $\mathscr{V}$. Proceed as in the old step (3).

• I'm not sure if this is where you went wrong but the sentence "any set in $\mathscr{U}$ outside $\mathscr{V}$ is a subset of some set in $\mathscr{V}$" isn't obviously true to me. – Tim kinsella Jun 20 '13 at 3:00
• @Timkinsella Ah I see now that my reasoning here was wrong. – PJ Miller Jun 20 '13 at 3:07
• So the finite cover obtained from compactness is important. – PJ Miller Jun 20 '13 at 3:18
• Yeah, that probably works. I think once you have $\mathscr{U}'$ I would use a "downward induction" (or whatever it's called) argument to get the cardinality of $\mathscr{U}'$ down to 1: If M is connected then there exist two elements of $\mathscr{U}'$ that overlap. Replace both of them with their union (which is not M since the union is in $\mathscr{U}'$) to get a cover of smaller size. Continue until you get a contradiction. This is probably equivalent to what you're doing. – Tim kinsella Jun 20 '13 at 3:36