# 16.2 Compact Sets

A set of real numbers \(S\) is said to be **covered by a collection \(O\) of open sets**, when
**every element of \(S\) is contained in at least one member of \(O\).** (The members of \(O\) can
contain numbers outside of \(S\) as well as those in \(S\).)

\(S\) is said to **compact**, if, **for every covering \(O\) of \(S\) by open sets, \(S\) is
covered by some finite set of members of \(O\).** A significant fact about a covering by open intervals
is: **if a point \(x\) lies in an open set \(Q\) it lies in an open interval in \(Q\) and is a positive
distance from the boundary points of that interval.**

We will now prove, just for fun, that **a bounded closed set of real numbers is compact.** The
argument does not depend on how distance is defined between real numbers as long as it makes sense as a
distance.

Open sets of real numbers are each unions of disjoint open intervals on the real line. We can consider a covering \(O\) of \(S\) by open sets to be a covering by their open intervals. Each open set in the covering containing the number \(x\) has an open interval in it containing \(x\). Thus a covering of \(S\) by open sets is actually a covering by open intervals as well.

Any interval in the covering of \(S\) by \(O\) that is contained in the union of other intervals in \(O\), is redundant in \(O\) and can be removed from it and the rest of \(O\) will still be a cover.

Every closed set of real numbers is a collection of disjoint closed intervals. For example, the collection \(S\) of intervals \([\frac{1}{2n+1}, \frac{1}{2n}]\) and \([\frac{-1}{2n}, \frac{-1}{2n+1}]\) for all positive \(n\) and the number \(0\), is a closed set. The minimum number in \(S\) is \(-\frac{1}{2}\) and the maximum is \(\frac{1}{2}\). Here the number \(0\) must be in it because it is a limit point of sequences of other of its numbers.

We can provide an explicit cover for \(S\) by the following infinite collection of open intervals:

\[\left\{\ \frac{1}{2n+1} - [\frac{1}{(2n+1)^2}, \frac{1}{2n} + [\frac{1}{(2n)^2} \right\}\] \[\left\{\ -\frac{1}{2n} - [\frac{1}{(2n)^2}, \frac{1}{2n+1} + [\frac{1}{(2n+1)^2} \right\}\] \[\text{and} \medspace \{-a,a\} \medspace \text{for some (small)} \medspace a.\]

To prove this claim we make use of several facts: first, *if a sequence of numbers in \(S\) is infinite, it
must have at least one limit point \(s\) in \(S\) since \(S\) is closed and bounded.* Second, *if a
number \(x\) is covered by an open set \(U\), then \(U\) contains numbers both smaller and larger than
\(x\).* Finally, *an open set containing a limit point of a sequence of distinct numbers must contain
an infinite number of those numbers.*

We can prove this result by actually constructing a finite set of open intervals that cover \(S\). To do so we set \(A_0\) to be the smallest number in \(S\), and let \(A_{j+1}\) be the smallest number in \(S\) greater than \(A_j\) that is in no member of \(O\) that contains \(A_j\).

\(A_{j+1}\) can be a lower boundary point of \(S\) (as will happen for all but one \(j\) encountered in the example above,) or it can be in the middle of or on the upper boundary of a closed interval of \(S\).

We will define an open interval \(O_j\) containing \(A_j\) inductively starting with the largest value of \(A_{j+1}\) as follows:

If \(A_{j+1}\) is the smallest number in its interval in \(S\) then let \(B_{j}\) be the largest number in \(S\) smaller than \(A_{j+1}\), and let \(O_J\) be any open interval in \(O\) containing both \(A_j\) and \(B_j\).

Otherwise, let \(B_j\) be any number in \(O_{j+1}\) that is smaller than (and not equal to) \(A_{j+1}\). By definition of \(A_{j+1}\) there is an open interval in \(O\) containing both \(A_j\) and \(B_j\), and let any such open interval be \(O_j\).

By construction, \(O_j\) contains only one of the \(A\)'s, namely \(A_j\). Thus it can contain no limit point of \(A\)'s by our third fact above. This means that the \(A\)’s, and \(O\)'s are finite in number. Also the \(O\)'s cover \(S\). Thus the \(O\)'s are a finite sizes covering of \(S\) by open intervals.

In the example above. the \(A\)'s are numbers of the form \(-\frac{1}{2n}\) from \(n=1\) until a value for \(n\) for which \(2n\) is approximately \(\frac{1}{a}\), with a similar number of positive \(A\)'s, for a total of roughly \(\frac{1}{a}\) \(A\)'s all together.

**
Exercises:
1. Prove the claim above that the \(O\)'s defined cover \(S\).
2. Show that the finite set of open intervals chosen from the members of \(D\) by the construction above
contains the fewest open intervals possible in a cover of \(S\) by open intervals.
**

We provide the example and construction above to give you intuition about what this result means. The usual simple proof involves breaking any closed set \(S\) in half choosing a number in S, and repeating these actions on a half of \(S\) which requires an infinite number of members of \(O\) to be covered. At each stage the size of the new \(S\) is half of that of the old \(S\), and at least one of the halves must need an infinite number of members of \(O\) to be covered if \(S\) does. The sequence of numbers chosen must, if infinite, have a limit point \(x\), and with \((a,b)\) any open interval covering \(x\), \((a,b)\) will contain all points in our sequence whose interval has length less than the smaller of \(b-x\) and \(x-a\); and that will be larger than the length of our interval \(O_j\) at some stage, and all subsequent members of our sequence will be inside \((a,b)\) and covered by it. (This is exactly what happens at \(0\) in the example.) This means that the later \(S\)'s require only one member of \(O\) to be covered. And one is very finite. An advantage of this proof is that it works as well for an \(n\) dimensional space whose elements are \(n\)-tuples of real numbers, exactly as it works for real numbers. (This argument is repeated in detail below.)

We have been discussing these various concepts in the context of real numbers, but they can also be defined in many other contexts. The definition of limit requires a definition of distance, but given such a definition, the concepts of closed, open, sequentially compact, complete and compact are also defined. Sets of points in which a distance between any pair of them is defined is said to be metric.

The concepts open closed and compact mentioned here can also be defined when there is no metric, by specification of which subsets of the whole set are open.

In a metric space \(S\) for any distance d we define the d neighborhood of a point \(x\) to consist of all elements of \(S\) whose distance from \(x\) is strictly less than \(d\). Any open set \(O\) in \(S\) is a union of neighborhoods of its elements each intersected with \(O\).

Suppose a closed bounded subset \(C\) of a finite dimensional space S is covered by open sets. Then it is covered by neighborhoods contained in those sets as well. We will argue that it must be covered by a finite collection of those neighborhoods, hence by a finite number of those open sets.

If \(S\) is \(n\) dimensional , we can cut \(C\) into a finite size grid of n-cubes (by which we mean sets of points whose length in any direction is at most some constant, and if \(O\) is necessarily infinite the number of members of \(O\) that are needed to cover least one of the \(n\)-cubes must be infinite as well. We choose a point in any such \(n\)-cube, reduce attention to that \(n\)-cube and repeat these cut and choose steps. The resulting sequence of elements must converge to some point \(x\), whose cover must still require an infinite number of neighborhoods. But a single point can be covered by one neighborhood, which, by this argument tells us that an infinite number of neighborhoods never were required.

An argument like that proves that a closed and bounded set in \(S\) is compact for any finite dimensional space defined over the real numbers.

When there is no metric strange things can happen. Suppose we have the integers, or rational numbers or real numbers, (with no definition of distance among them) and the closed sets consist of all finite sets. That means the open sets are all sets of elements that lack only a finite number of elements.

In any such space of points and definition of open sets, all sets are compact!

Given any set \(Q\), and any cover of \(Q\) by open sets, and any open set \(O\) in that cover, \(O\) can miss only a closed set which means a finite set, say \(d\), of elements of \(Q\). These can be covered by at most \(d\) open sets in the cover of \(Q\), which means that there is a cover of \(Q\) by at most \(d+1\) open sets that are in the original cover.

Thus compact sets need not, in general, be closed or bounded with these definitions.

A definition of open sets in a set of points is called a topology.

The subject considered above, called point set topology, was studied extensively in the \(19^{th}\) century in an effort to make calculus rigorous. It contains many interesting results of which what is above is a tiny and random sample.