Pancakes, anyone?

Open sets are not pancakes.

There are many reasons open sets are not pancakes. One is that pancakes are yummy.

But it’s helpful to imagine covering someone with pancakes… …when what you’re actually doing is covering a space with open sets.

So, in this chapter, I conflate “pancakes” and “open sets.”

When you see a reference to “pancakes”, feel free to mentally cross it out and replace it with “open sets,” and vice versa, depending on your preference.

(Thank you, Francis Su’s real analysis lectures, for the pancake analogy!)

How many ways can you cover someone with pancakes?

Given a particular friend…

…there are lots of ways to completely cover them with pancakes.

Similarly, given a set in space…

…there are a lot of ways to completely cover it with open sets. For example, a cover of $(0,1)$ could be….

Just one pancake. One bigger pancake. Two overlapping pancakes. Infinitely many overlapping pancakes.
The set $(0,1)$.

The set $(-1,2)$.

The sets $(-1,\frac{3}{4})$ and $(\frac{1}{4},2)$.

The infinite collection of sets $\{(\frac{1}{n},1): n \in \mathbb{N}\}$.

So, what exactly is a cover?

Let’s say a “cover” of your friend … …is a collection of pancakes… …that covers every inch of them.

And let’s say a “cover” of a set $S$… …is a collection of open sets… …that completely contains $S$.

Formally, a $\textit{cover}$ of $S$ is a collection $\mathcal{C}$ of open sets $U$ such that:

$$ S \subseteq \bigcup\limits_{U \in \mathcal{C}} U.$$

In other words, a cover $\mathcal{C}$ of the set $S$ is a collection of open sets $U$ (pancakes) such that if you take the union of all the open sets in the collection $\mathcal{C}$, it contains the set $S$.

How many pancakes did we use?

Pancakes don’t grow on trees.

So ideally, we only use a finite number of pancakes.

Note that some covers use finitely many pancakes. And some covers use infinitely many.

So, when I say a cover is finite, I mean the cover uses finitely many pancakes.

Formally, when I say that $\mathcal{C}$ is a finite cover, I mean that there are only finitely many open sets in $\mathcal{C}$. Of course, each of these open sets could contain infinitely many points.

Did you need all of those pancakes?

Sometimes, we used more pancakes than we needed to completely cover a person.

So once you have a cover…

…you might want to eat the pancakes you didn’t need. A subcover is any subcollection of those pancakes that still covers your friend.

If it doesn’t completely cover your friend, it isn’t a subcover.

And if it requires adding new pancakes that weren’t in the original cover, it isn’t a subcover either.

Informally, a subcover is a cover …

…. after you’ve eaten some of the pancakes in it.

Formally, a subcover is a subcollection of a cover, that is still a cover.

So if you consider this particular cover…

…this is a subcover of it. …this is not.

How many pancakes can we remove?

Sometimes, you’ll start with an infinite cover…
…and be able to remove enough pancakes to end up with a finite subcover.

But other times, you won’t be able to.

For example, we suppose we want to cover the set $(-1,1)$…

with the following collection…

$$\left\{\left(-\frac{1}{2},\frac{1}{2}\right), \left(-\frac{3}{4},\frac{3}{4}\right), \left(-\frac{4}{5},\frac{4}{5}\right), \cdots \right\}$$

…which we can write succinctly as…

$$\left\{ \left(-\frac{n}{n+1},\frac{n}{n+1} \right) : n \in \mathbb{N}\right\}.$$

There is no way you can take away so many pancakes that you leave a finite number of pancakes there, and it still covers the set $(-1, 1)$.

Informally, this is because when you take any finite subset of the cover, you end up exposing a bit of edge of the set. You’ll prove it formally in the puzzle at the end of this section!

How do we know which covers have spare pancakes?

Ok fine, so we know that when you look at covering $(-1,1)$…

There are some covers … …that have finite subcovers.
And there are other covers …. that have no finite subcover.

So what?

Here’s the big thing: there are some sets, which, given any cover of them, always reduce to a finite subcover. In those sets, any infinite cover will always have spare pancakes to munch on.

For example, look at $[-1,1]$.

You can cover it like this… …and get a finite subcover.
And you can cover it like this… …and get a finite subcover.

And in fact, it turns out, no matter how you cover it, you can reduce that to a finite subcover.

So what’s so great about this set?

It’s so low-maintenance!

Given an infinite amount of information about the set’s cover… …you can always just reduce that…

…to a finite amount of information.

So, let’s call any set like this a “compact set” as a thank-you to the conciseness of the information needed to describe any cover of the set.

Formally, let’s say a set is compact when every single cover of it (no matter how big) has a finite subcover.

So the set $[-1,1]$ is compact.

Because (we’ll prove later) every cover of it has a finite subcover.
But the set $(-1,1)$ is not compact.

Because there is a cover of it with no finite subcover.

What’s the point of compact sets?

“Compactness is the next best thing to finiteness,” says Prof. Francis Su.

It’s true. That is, a compact set may not be finite, but it’s the “next best thing” — any open cover of it can be reduced to a finite one.

This ability to reduce a lot of information to just a finite amount means compact sets are much easier to prove things about than other sets.

As a result, compact sets play a big role in differentiation and integration, as we’ll see later.


We know $\left\{ \left(-\frac{n}{n+1},\frac{n}{n+1} \right) : n \in \mathbb{N}\right\}$ covers $(-1,1)$. Prove that no finite subset of it covers $(-1,1)$.
Scroll if you'd like a hint...

If you had a finite subcover $\mathcal{F}$, then it has a “largest pancake.” That is, there is a largest value of $n$ such that the pancake $\left( -\frac{n}{n+1},\frac{n}{n+1} \right)$ is in $\mathcal{F}$.

Scroll if you'd like a hint...

Suppose the largest pancake in $\mathcal{F}$ is $\left( -\frac{N}{N+1},\frac{N}{N+1} \right)$.

Then, the interval $\left[\frac{N}{N+1}, 1\right)$ is uncovered…so $\mathcal{F}$ can’t be a cover.