1
$\begingroup$

Suppose that $\mathbf J$ is a set such that each element $J$ in the set $\mathbf J$ is a convex subset of the plane of real numbers $\mathbb{R}^2$. Then the intersection of all $J$ in $\mathbf J$ is a convex subset of the plane of real numbers $\mathbb{R}^2$.

  • 0
    Why isn't the intersection of all $J$ in **J** the empty set? The circle centered at $(0, 0)$ with radius 1 is in **J**, and the circle centered at $(2, 3)$ is in **J**, and their intersection is empty.So the intersection of every $J$ in **J** is also empty, no? – 2017-02-27
  • 0
    @Ovr: If you meant "disk" of radius 1 (not "circle") then, yes, the intersection of those two disks is the empty set. But the empty is set is convex (it vacuously satisfies the definition), so there's no issue. In any case, the set $J$ was not specified -- it need not be the collection of all convex subsets of $\mathbb{R}^2$ – 2017-02-27
  • 0
    Well, the problem is, you're not five. But the relevant question is whether you realize this site isn't for five year-old (and even if it were a five-year-old asking, it would be apparent.) You, on the other hand want to be spoon-fed...whaaaa, whaaaaa, help me, I can't feed myself! whaaaaa! – 2017-03-01

3 Answers 3

1

To say that $E \subseteq \mathbf{R}^2$ is convex is to say that for any $v, w \in E$, and any real number $0 \leq t \leq 1$, the element $tv + (1-t)w$ is in $E$.

Apparently, you want to show that the intersection of any collection of convex subsets of $\mathbf{R}^2$ is a convex subset of $\mathbf{R}^2$. Let $E$ be the intersection of all the sets $J \in \mathbf{J}$. Let $v, w \in E$, and let $0 \leq t \leq 1$. You need to show that $tv + (1-t)w$ is in $E$.

Since $v$ and $w$ are in $E$, they are in each set $J \in \mathbf{J}$. Since each $J$ is convex, $tv + (1-t)w$ is in $J$. This shows that $tv + (1-t)w$ is in every set $J \in \mathbf{J}$, which means that $tv + (1-t)w$ is in $E$.

  • 0
    how are you getting tv+(1-t)w? – 2017-02-27
  • 0
    That's what I'm using as the *definition* of a convex set. Is that different from the definition you're using? – 2017-02-27
0

Let $H$ be the intersection of all $J \in \mathbf J$.

Let $a,b \in H$, and let $ab$ denote the line segment with endpoints $a,b$.

Then \begin{align*} &a,b \in H\\[6pt] \implies\; &a,b \in J,\text{ for all }J \in \mathbf{J}\\[6pt] \implies\; &ab \subseteq J,\text{ for all }J \in \mathbf{J}\\[6pt] \implies\; &ab \subseteq H \end{align*}

It follows that $H$ is convex.

0

The explanation "like you're five" won't be a formal proof and will be full of intuitive and inexact words, of course, but I'll give my best shot nonetheless:

Imagine a collection of convex sets on the Cartesian coordinate system. Remember that a convex set on the Cartesian coordinate system means that you can connect the line segment between any two chosen points in the set without going outside of the set. For instance, you can connect any two points inside a circle without having to go outside, so the inside of a circle is convex - however, the outside of a circle is not convex, since you can imagine choosing two points on a line containing a diameter of the circle, with the points on this line chosen on opposite sides of the center, and the segment passing through them will contain the diameter and therefore exit your set.

Many other sets are convex: the interior of squares, the region "above" the graph of $y=e^x,$ and the empty set itself (since there are no two points in the empty set failing the convex definition).

Picture a whole bunch of these convex sets all at once and pretend that they all overlap together to form some set (else, the set of where all overlap is the empty set, which is convex as explained above).

Now imagine any two points on that overlap. Well, we know that these two points belong to each of the convex sets we imagined in this overlap. Because these two points in the overlap belong to every one of these convex sets, the segment between them must belong to every one of these convex sets, for if this segment didn't fit into to one of these sets that set would fail to be convex by definition, contradiction our initial assumption.

Since the points we imagined were not specified, this above description applies to any two points we choose in the intersection set. But if any two points in a set also have their linking segment entirely within the set as well, that means this intersection set is itself convex. The proof is complete.

Take special note that this is, in a sense, the "largest" convex set that is contained within every convex set in your collection!