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The property of being a bounded set in a metric space is not preserved by homeomorphism. For example, the interval (0, 1) and the whole of R are homeomorphic under the usual topology. So to generalise theorems in Real analysis like "a continuous function on a closed bounded interval is bounded" we need a new concept.
This is the idea of compactness. We will give a definition which applies to metric spaces later, but meanwhile, phrased purely in terms of open sets we have:
Definitions
A topological space is compact if every open covering has a finite sub-covering.
An open covering of a space X is a collection {U_{i}} of open sets with U_{i} = X and this has a finite sub-covering if a finite number of the U_{i}'s can be chosen which still cover X.
The most important thing is what this means for R with its usual metric.
Theorem
The interval [0, 1] is compact under the usual metric on R.
Proof
Let {U_{i}} be an open covering of [0, 1]. The trick is to consider the set A = {x [0, 1] | [0, x] can be covered by finitely many of the U_{i}'s}. Then use the Completeness property of R to take to be the least upper bound of A.
Suppose < 1. Then is contained in some open set U_{i0} and so lies in an -neighbourhood lying in U_{i0} .
But now [0, - /2] is covered by finitely many of the U_{i}'s and so this collection, together with U_{i0} covers [0, + /2] which contradicts the definition of .
A similar proof shows that any closed bounded interval of R is compact. We will see later that in fact any closed bounded subset of R (with its usual metric) is compact.
Theorem
A compact subset of R with its usual metric is closed and bounded.
Proof
If a set A R is not closed then there is a limit point p A. Then cover A by complements of closed -neighbourhoods of p for p = 1, ^{1}/_{2} , ^{1}/_{3} , ... .
For example If A = (0, 1) and p = 0 then (0, 1) = (^{1}/_{2} ,1) (^{1}/_{3} ,1) (^{1}/_{4} ,1) ...
We cannot take a finite subcover to cover A.
A similar proof shows that an unbounded set is not compact.
Properties of compactness
Proof
Let {U_{i}} be an open cover of f(C). Then {f ^{-1}(U_{i})} is an open cover of C and can therefore be reduced to a finite subcover. The corresponding collection of U_{i}'s will be a finite sub-cover of f(C).
Corollary
If X is compact and ~ is any equivalence relation then X/~ is compact.
Proof
The natural map p: X X/~ is continuous and onto.
Proof
If {U_{i}} is an open cover of A C then each U_{i} = V_{i} A with V_{i} oopen in C. Then the collection {V_{i}} together with the open set C - A cover C and hence have a finite subcover. The corresponding U_{i}'s then cover A.
Corollary (The Heine-Borel theorem)
Any closed bounded subset of R with its usual metric is compact.
Proof
Any such subset is a closed subset of a closed bounded interval which we saw above is compact.
Remarks
Proof
The closed bounded interval is compact and hence its image is compact and hence is also a closed bounded subset which is in fact an interval also, by connectedness. Thus the function is bounded and its image is an interval [p, q]. It attains its bounds at points mapped to p and q.
Proof
Suppose C X is compact. To show that X - C is open we take x X - C and try and show that x is in an open subset of X - C.
For each y C we can find disjoint open sets U_{y} and V_{y} separating x and y: x U_{y} y V_{y} . The set U_{y} where the intersection is over all y C does not meet C and hence is in X - C. Unfortunately, it is not necessarily open since a topology is not closed under infinite intersections. However, since C is compact, we may discard all but finitely many of the V_{y}'s and the intersection of the corresponding U_{y}'s will be the open set we need.
Proof
This result is known as Tychonoff's theorem after Andrei Tychonoff (1906 to 1993) who proved it for a product of infinitely many spaces. Even for two spaces the proof is surprisingly tricky.
Corollary
The closed unit square [0, 1] [0, 1] is compact.
Hence too are spaces like the Möbius band, Real projective palne, torus, sphere, ... made from it by identification.
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