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:

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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 $\mathbb{R}^2$?
  • 3
    $\begingroup$ 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. $\endgroup$ – user49640 Jun 13 '17 at 3:48
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ (a) Why must the square $S$ be closed? (b) in the last sentence, why is the set $C_0$ the intersection of all sets $(1/t)C$? $\endgroup$ – Erel Segal-Halevi Jun 13 '17 at 5:09
  • $\begingroup$ 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 $t \in (0,1]$. $\endgroup$ – user49640 Jun 13 '17 at 5:40
  • $\begingroup$ If S is not closed, can't you just replace S with its closure and get the same result? $\endgroup$ – Erel Segal-Halevi Jun 13 '17 at 11:23
  • $\begingroup$ 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. $\endgroup$ – 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 $$\overline {\cup A}=\overline {C\backslash D}=C\backslash D=\cup A.$$


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