TITLE: When is the union of infinitely many closed sets, closed?
QUESTION [5 upvotes]: 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 $\mathbb{R}^2$?

REPLY [2 votes]: (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 $$\overline {\cup A}=\overline {C\backslash D}=C\backslash D=\cup A.$$
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