# My first entry! “Helly’s Theorem”

I love compactness. I really do. It turns infinite things into (almost) finite things. I could gush about how great it is, but instead let me tell you about one problem where compact sets act as the delimiter.

Here is one way to characterize compactness:

A space X is compact if and only if any family of closed sets with the Finite Intersection Property (FIP) has a common point.

[Remember that a collection has the FIP if every finite subcollection has a common point (i.e. has non-empty intersection).]

This has a pretty clear connection to filters, as filters are collections of sets with the FIP (and the intersection is in the filter!) and closed under supersets. One example of a filter is the collection of all subsets of the real line that contain a closed interval around 0.

A closely related notion is that of being 2-linked. A collection A is 2-linked if any two sets in A have non-empty intersection. For example the collection of real intervals $\{(n,n+2): n \in \mathbb{Z}\}$ is 2-linked. Another example is the set of sides of a polygon triangle. (Why not a square?)

Then of course we can talk about being 3-linked which means that any 3 sets have non-empty intersection (we will now say that this is called ‘meeting’). Obviously, $\{(n,n+2): n \in \mathbb{Z}\}$ is 2-linked, but not 3-linked. (edit: Yeah, so not only is this not obvious, but it is not true! I address this here.)

Then we could go on to define n-linked for an arbitrary natural number n.

Question 1: How is the FIP related to being n-linked?
Question 2: Can you find, for each n, an example of a collection that is n-linked but not n+1 linked?
Question 3: How is n-linkedness related to the dimension of the real line?

I’ll get to these later. But you should think about them. 1 and 2 are not hard. 3 takes some thought, but just try to come up with a conjecture.

So we know that if a collection is 2-linked then it doesn’t have to have the FIP. But maybe if the sets are particularly nice this might work. One thing that would is if the sets were all close together. Maybe we could make the sets so close to each other that they had no choice but to be really friendly.

Attempt 1: The three sides of a triangle. Oh right, we saw this one already and it isn’t 3-linked. This isn’t good news. The sides of a triangle are already “pretty close” (i.e. they are contained in a bounded set).

Ok, so maybe we need 3-linked to get going.

Attempt 2: On the circle, the angles $[0,\pi], [\pi,2\pi], (0,2\pi), [0,\pi)\cup(\pi,2\pi]$, form a 3-linked family that isn’t 4-linked. (That example took me an embarrassingly long time to come up with).

All is not lost though, the reason I had such a hard time with that example is because of the following fact, called Helly’s Theorem which says:

If a family of closed convex sets in the plane are 3-linked, then they have the FIP.

Well that isn’t quite what Helley’s Theorem says, but the idea is there. It says that adding convex to 3-linked pushes us all the way to n-linked for all n.

Here is the full statement:

Helly’s Theorem: If a (countable) family of closed convex sets (at least one of which is bounded) in the plane are 3-linked, then they have a point in common, as they have the FIP.

The single bounded set here plays an interesting role; it guarantees that ‘most of the action takes place near the origin’. The one bounded set and the 3-linked property ensures that ‘everything is taking place inside of that bounded set’.

What do I mean? Let me illustrate with an example where there is no bounded set.

Example 1: Let $F_n =\{(x,y) : n\leq x\}$, i.e. the plane to the right of the line $x=n$. This is a family of closed convex sets, none of which is bounded. This family is 3-linked and in fact has the FIP, but the family has no point common to all of them.

Well, we seem to be pushing everything ‘off the map’ to avoid having a common point. How does this change when there is a bounded set present? (Oh yeah, closed and bounded is compact in the plane, so from now on, I will say compact instead of bounded.)

Example 2: Let $F_0$ = Unit square = $[0,1]\times[0,1]$ and let $F_n = \{(x,y): y\leq 1/n\}$. These are the planes below the lines $y=1/n$. Here the family has a bunch of common points, namely: $\{0\}\times[0,1]$.

Here we had our sets pushing all the way to the edge of the square, but that was fine.

Next week I will look at a couple of generalizations of Helley’s Theorem, which I think are not true! In the mean time here is something for you to chew on: We know that 2-linked + convex doesn’t give us the FIP, but 3-linked + convex does. What kind of weaker ‘meeting’ properties would still give us the FIP? Maybe we give up on trying to get the FIP, but still want that there are numerous possible (almost) common points?

## 6 thoughts on “My first entry! “Helly’s Theorem””

1. Peter Krautzberger says:

Welcome, Mike! Nice post. Looking forward to reading more.

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2. François says:

Hi Mike, welcome to Boole’s Rings! I look forward to meeting you someday…

Somehow this post didn’t show up on the main Booles’ Rings page. You may want to have that fixed. Also, I think the correct spelling is Helly.

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1. mpawliuk says:

Hi François, I look forward to meeting you too! You’re right it is “Helly”.

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3. Ivan Khatchatourian says:

Am I totally missing something about your example of a 2-linked collection? It seems to me that (2, 4) and (100, 102) are pretty disjoint. Certainly in that collection of intervals, any two adjacent members have nonempty intersection, but that’s also true of the sides of a square…

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1. Ivan Khatchatourian says:

Nevermind, I just saw that you noticed this in your second post. Carry on.

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