4.6: Compact Sets
( \newcommand{\kernel}{\mathrm{null}\,}\)
We now pause to consider a very important kind of sets. In Chapter 3, §16, we showed that every sequence
A set
If all of
(a) Each closed interval in
(a') However, nonclosed intervals, and
For example, the sequence
(b) Any finite set
(c) The empty set is "vacuously" compact (it contains no sequences).
(d)
Other examples can be derived from the theorems that follow.
If a set
- Proof
-
We must show that each sequence
clusters at some . However, as is also in so by the compactness of it clusters at some Thus it remains to show that as well.Now by Theorem 1 of Chapter
has a subsequence . As and is closed, this implies (Theorem 4 in Chapter
Every compact set
- Proof
-
Given that
is compact, we must show (by Theorem 4 in Chapter 3, §16) that contains the limit of each convergent sequence .Thus let
As is compact, the sequence clusters at some i.e., has a subsequence However, the limit of the subsequence must be the same as that of the entire sequence. Thus ; i.e., is in as required.
Every compact set
- Proof
-
By Problem 3 in Chapter 3, §13, it suffices to show that
is contained in some finite union of globes. Thus we fix some arbitrary radius and, seeking a contradiction, assume that cannot by any finite number of globes of that radius.Then if
the globe does not cover so there is a point such thatBy our assumption,
is not even covered by Thus there is a point withAgain,
is not covered by so there is a point not in that union; its distances from and must therefore be .Since
is never covered by any finite number of -globes, we can continue this process indefinitely (by induction) and thus select an infinite sequence with all its terms at least -apart from each other.Now as
is compact, this sequence must have a convergent subsequence which is then certainly Cauchy (by Theorem 1 of Chapter 3, §17). This is impossible, however, since its terms are at distances from each other, contrary to Definition 1 in Chapter 3, §17. This contradiction completes the proof.
Note 1. We have actually proved more than was required, namely, that no matter how small
Note 2. Thus all compact sets are closed and bounded. The converse fails in metric spaces in general (see Problem 2 below). In
In
- Proof
-
In fact, if a set
is bounded, then by the Bolzano-Weierstrass theorem, each sequence has a convergent subsequence If is also closed, the limit point must belong to itself.Thus each sequence
clusters at some in so is compact.The converse is obvious.
Note 3. In particular, every closed globe in
The converse is obvious.
(Cantor's principle of nested closed sets). Every contracting sequence of nonvoid compact sets
in a metric space
For complete sets
- Proof
-
We prove the theorem for complete sets first.
As
we can pick a point from each to obtain a sequence since it is easy to see that is a Cauchy sequence. (The details are left to the reader.) Moreover,Thus
is a Cauchy sequence in a complete set (by assumption).Therefore, by the definition of completeness (Chapter 3, §17),
has a limit This limit remains the same if we drop a finite number of terms, say, the first of them. Then we are left with the sequence which, by construction, is entirely contained in (why?), with the same limit P. Then, however, the completeness of implies that as well. As is arbitrary here, it follows that i.e.,The proof for compact sets is analogous and even simpler. Here
need not be a Cauchy sequence. Instead, using the compactness of we select from a subsequence and then proceed as above.Note 4. In particular, in
we may let the sets be closed intervals (since they are compact). Then Theorem 5 yields the principle of nested intervals: Every contracting sequence of closed intervals in has a nonempty intersection. (For an independent proof, see Problem 8 below.)