# When is the union of infinitely many closed sets, closed?

It is known that, in general, the union of infinitely many closed sets need not be closed. However, in the following case, apparently, the union is closed:

Suppose there is a large closed polygon $C$, inside which there is a square $S$ (green). Consider the set of all closed convex objects that contain $S$ and are contained in $C$. Then, apparently, the union of all these closed objects is closed.

My questions:

• Is the above claim true, and if so, how to prove it?
• In general, what are conditions for infinite set of closed sets to be closed, especially in $\mathbb{R}^2$?
• This doesn't apply in your situation, but a common sufficient condition for the union of a family of closed sets to be closed is that the family be locally finite. This means that for any point, there is a neighbourhood of that point that meets only finitely many of the closed sets. – user49640 Jun 13 '17 at 3:48
• Echoing the comment above, the result that the union of a locally finite family of closed sets is closed, is a result that holds in every topological space. – DanielWainfleet Jun 13 '17 at 6:33

The claim is true. Let $A$ be the set you define and which we wish to prove closed.

A point $x$ belongs to $A$ if and only if the convex hull of $S \cup \{x\}$ is contained in $C$. (In that case, its closure is also contained in $C$.) This convex hull consists in turn of the union $B_x$ of all segments joining $x$ to a point in $S$.

$B_x$ is convex because if $xz_1 z_2$ is a triangle with $z_1, z_2 \in S$, and the points $y_1$ and $y_2$ lie on sides $xz_1$ and $xz_2$, respectively, then for any point $w$ on segment $y_1 y_2$ the ray $xw$ meets side $z_1 z_2$, which is contained in the convex set $S$.

Thus a point $x$ belongs to $A$ if and only if the segment $xz$ is contained in $C$ for every point $z$ in $S$. Therefore $A$ is the intersection, for all $z \in S$, of the set $C_z$ which is the union of all segments with one endpoint at $z$ which are contained in $C$. Hence we need only prove that $C_z$ is closed.

After a translation, we may assume $z = 0$. The set $C_0$ is the intersection of the sets $(1/t)C$ for all $t \in (0,1]$, and all the sets $(1/t)C$ are obviously closed.

• (a) Why must the square $S$ be closed? (b) in the last sentence, why is the set $C_0$ the intersection of all sets $(1/t)C$? – Erel Segal-Halevi Jun 13 '17 at 5:09
• I don't know if $S$ must be a closed square to prove what you want, but I use the fact that it is closed when I show that $B_x$ is closed. Your requirements are that there be a closed convex set containing $x$ and $S$. For your second question, a point $x$ belongs to $C_0$ if and only if $tx \in C$ for all $t \in (0,1]$. – user49640 Jun 13 '17 at 5:40
• If S is not closed, can't you just replace S with its closure and get the same result? – Erel Segal-Halevi Jun 13 '17 at 11:23
• Yes, you're right. Just take the closure of the convex hull of $S \cup \{x\}$. It's contained in $C$ because $C$ is closed. So I didn't need to mention anything about compactness. I'm editing the answer now. – user49640 Jun 13 '17 at 11:39

(0). The closure bar denotes closure in $\mathbb R^2.$

(1). Exercise: If $Y$ is a closed bounded convex subet of $\mathbb R^2$ and $x\in \mathbb R^2$ then the convex hull of $\{x\}\cup Y$ is closed.

(2). Let $A$ be the family of closed convex subsets of $C$ that have $S$ as a subset. Let D be the set of $p\in C$ such that the convex hull of $\{p\}\cup S$ is not a subset of $C.$ We have $(\cup A )\cap D=\phi$ so $$\cup A\subset C \backslash D.$$ Take any $p\in D.$ There exists $q\in S$ such that the line segment joining $p$ to $q$ contains a point $r\not \in C.$ Take an open ball $B(r,d)$, of radius $d>0$, centered at $r$, such that $B(r,d)\cap C=\phi.$ If $s>0$ and $s$ is sufficiently small then for any $p'\in C\cap B(p,s),$ the line segment from $p'$ to $q$ will intersect $B(r,d).$ So $B(p,s)\cap C\subset D.$

So $D$ is open in the space $C.$

Therefore $C$ \ $D$ is closed in the space $C.$ And $C=\overline C$ so we have $$C \backslash D =\overline {C \backslash D}.$$ But for any $x\in C$ \ $D$ the convex hull of $\{x\}\cup S$ is closed, and is a subset of $C$. Therefore $$C\backslash D\subset \cup A.$$

We have now $\cup A\subset C \backslash D \subset \cup A,$ so $$\overline {\cup A}=\overline {C\backslash D}=C\backslash D=\cup A.$$