Proof. Suppose C is an open cover covering A . Then 0 Î U for some U in C. Then since 1/ n converges to 0 as n tends to infinity, there exists an integer N such that for all n > N, 1/n Î U. Now for n =1,¼, N, 1/n Î Un . Hence {U1, ¼ , UN , U} is a finite subfamily that covers A too.
The next notion is the notion of boundedness. A subset A of a metric space (M, d) is said to be bounded if and only if there exists a real positive number k such that d( x, y) < k for all x, y in A.
Theorem 5 (Heine-Borel). A subset A of R is compact if and only if A is closed and bounded.
Before we proceed with the proof. The following results will contribute to it and are important and useful on their own merits
Theorem 6. A compact subset A of a metric space (M, d) is bounded.
Proof. We are going to use an open cover of A by open balls. A typical open ball centred at x in A and of radius d > 0 is the set B(x, d) = {y Î M: d( y, x) < d}. For each a in A, let U(a) = B(a, 1) the unit ball centred at a.. Then C = {U(a) : a Î A} is an open cover for A. Since A is compact C has a finite subcover, say { U(ai) :i =1, ¼, n}. Let k = max {d( a i , a j ): 1£ i, j £ n }. Therefore, for any x, y in A, x Î a i and y Î a j for some 1£ i, j £ n, d( x, y) £ d( x , a i ) + d( a i , a j ) + d( a j , y ) < 2 + k and so A is bounded.
Theorem 7. Any compact subset A of a metric (Hausdorff) space is closed.
Proof. The proof uses the fact that any two distinct points x, y in a metric space can be separated in the sense that there are two disjoint open sets U and V with x Î U and y ÎV. We can take for instance, U = B(x, d(x, y)/2) and V =B(y, d(x, y)/2). This is the concept of a Hausdorff space. Let us fix an element y not in A. Then for each a in A, we have an open set U(a) and and an open set V(a) such that a Î U(a), y Î U(a) and U(a) Ç V(a) = Æ. Then C = {U(a) : a Î A} is an open cover for A. Since A is compact C has a finite subcover, say {U(ai) :i =1, ¼, n}. Then if we let U = È { U(ai) :i =1, ¼, n} and V = Ç {V(ai) :i =1, ¼, n}. Then U is a finite union of open sets and is therefore open and V is a finite intersection of open sets and is also open. Also A Í U and U Ç V = Æ. This is because U Ç V Í È {U(ai)ÇV :i =1, ¼, n}Í È {U(ai)Ç V(ai) :i =1, ¼, n}= Æ. Hence V is an open set containing y and V Í complement of A since V Ç A Í U Ç V = Æ. Hence each point y in the complement of y has an open set contained entirely in the complement of A, therefore the complement of A is a union of open sets and so is open. Therefore, A is closed. This completes the proof.
Proof of Theorem 5.
(Þ) Suppose A is a compact subset of R. Then by Theorem 6, A is bounded and is closed by Theorem 7.
(Ü) Suppose A is a closed and bounded subset of R. Then A Í [a, b] for some closed and bounded interval [a, b]. If we can show that [a, b] is compact, then A being a closed subspace of a compact space is therefore compact. (This is because any open cover for A together with the complement of A constitute an open cover for [a, b] and if [a, b] is compact there will be a finite subcover for A. ) Now let C be open cover for [a, b]. Define c = sup { x Î [a, b] : a finite subfamily of C covers [a, x]}. Obviously the set { x Î [a, b] : a finite subfamily of C covers [a, x]} is not empty since a belongs to it and is clearly bounded above by b. Therefore, by the completeness property of R, c exists. Then c > a. Why? a Î open set U in C since C is an open cover for [a, b]. Therefore, there exists a d > 0 such that (a - d, a + d ) Í U. Thus for any a < y < a + d. [a, y] Í U and so y Î { x Î [a, b] : a finite subfamily of C covers [a, x]}. Therefore, by the definition of supremum c ³ y > a. We shall show next that c = b. Now we have a < c £ b. Thus there exists an open set V in C such that c Î open set U. Then there exists d > 0 such that (c - d, c + d ) Í U . Take any d such that c - d < d < c. Then [d, c] Í U. Now since d < c, by the definition of supremum, there exists a point z in { x Î [a, b] : a finite subfamily of C covers [a, x]} such that d < z £ c . Hence there is a finite subfamily of C covering [a, z] and since [a, z] È [d, c] = [a, c] and [d, c] Í U, this subfamily together with U constitute a finite subfamily covering [a, c]. Hence c Î { x Î [a, b] : a finite subfamily of C covers [a, x]}. Hence c = b. This is because if c < b, then as above we can take a point e this time in (c , b)Ç(c - d, c + d ) Í U. Thus c < e < b and [c, e] Í U, and so since there is a finite subfamily of C covering [a, c] and U Î C , this subfamily and U constitute a finite subfamily covering [a, e]. Thus e Î { x Î [a, b] : a finite subfamily of C covers [a, x]}. Therefore, c = sup{ x Î [a, b] : a finite subfamily of C covers [a, x]} ³ e contradicting c < e . Hence c = b and so there is a finite subfamily covering [a, b] (Why? Reason as above.) and so [a, b] is compact. This completes the proof.
Theorem 8 (Bolzano-Weierstrass). Any bounded sequence in R has a convergent subsequence.
We shall give a proof of this theorem that can be adapted to a proof for a bounded sequence in R n .
Proof. By the Heine -Borel Theorem (Theorem 5), A bounded sequence {a n} in R lies inside a compact set, a large closed interval [c, d] Let us use the following notation for the sequence. Consider {a n} as the image of a function a : N ® R, where a(n) = a n .
If the image A = a (N) is finite, then there must exist an element y
in a (N) such that a -1 (y) is infinite. Therefore
{aj : j Î a -1 (y)}
is a convergent constant subsequence. We now consider the case A is infinite. Then
of course A is contained in [c, d]. Consider now the set of accumulation
point A' of A in R. A point x in R, is an accumulation
point of A, if any open set containing x contains a point of A
distinct from x. Claim that A' ¹ Æ . Suppose A' = Æ.
That means each point x in [c, d] has an open set U x such
that U x Ç A is finite.
Then the family of open sets {U x : x Î [c,
d] } covers [c, d]. Since [c, d] is compact by the
Heine-Borel Theorem, this family has a finite sub family {Ui , i
= 1, ¼ , n} such that [c,
d] Í U1 ÈU2È ¼ È Un
. Therefore, A Í A Ç [c, d] Í (U1Ç A) È(U2Ç A)È
¼ È (UnÇ A). But (U1Ç A) È(U2Ç A)È
¼ È (UnÇ A) is a union of finite set
and so is finite. Hence A being a subset of a finite set must be finite. We have
thus arrived at a contradiction since we have started with an infinite A. Take a
point x in A' . Then we shall construct a sequence {x j}
in A such that xi ¹ x j for i ¹ j and {x j} converges to x
as j tends to infinity. A consequence of this is that x is in [c, d].
Take x1 in B(x, 1) such that x1 ¹ x and so d (x1 , x) > 0.
This point x1 exists by definition of accumulation point. As we shrink
the Ball B(x, 1/ n), we shall exclude the point x1 .
For instance there exists an integer n 2 such that 1/ n 2
< d (x1 , x) , then by virtue of x being an
accumulation point of A, there exists x2 in B(x, 1/
n 2) such that x2 ¹ x and
so d(x2 , x) > 0. Obviously x2 ¹ x1 for otherwise if x2 = x1 then d(x2 , x1)
= 0 and we have d (x1 , x)£ d (x1 , x2 )
+ d(x2 , x) < 0 + 1/ n 2 = 1/ n
2 contradicting 1/ n 2 < d (x1 , x).
In this way, there exists n 3 such that 1/ n 3 < d
(x2 , x) , x2 , x1
Ï B(x, 1/ n
3) and there exists x2 in B(x, 1/ n 3)
such that x3 ¹ x. So inductively,
we find integers 1 < n 2 < n 3 ¼ and points x1
, x2 , x3 , ¼ such that xj Î
B(x, 1/ n j), x j ¹ x j for i ¹ j. Then obviously {x j} converges
to x as j tends to infinity since for any open set U containing x
there exists an integer J such that x Î B(x, 1/ nJ) Í U. Therefore, for all j
> J, xj Î B(x, 1/
n j) Í B(x, 1/ nJ)
Í U. Now based on this
sequence we are going to construct a subsequence of {a n}
converging to x. Start with x1 , consider a -1 (x1). Choose i1 in a -1 (x1). Then a(i1)
= x1. Next observe that since not all a
-1 (xj) for j > 1 can be bounded above by i1
because otherwise a -1({xj :
j > 1}) would be finite which implies that {xj : j > 1}
is finite contradicting that {xj : j > 1} is infinite since
the sequence {x j} is a sequence of distinct terms. Thus there is a j2
> 1 such that 
is not bounded by i1. There exists i2
in 
such that i2 > i1 and a(i2)
= 
.
Next not all a -1 (xj) for j
> j 2 can be bounded above by i2 . So there exists j3
> j 2 such that 
is not bounded by i2. So there
exists i3 in 
such that i3 > i2
and a(i3) = 
. In this way we obtain a subsequence 
of {x
j} and this subsequence is equal to the subsequence 
of {a n}.
That means 
for n = 1, 2, ¼ . Since {x j} converges to x any subsequence
of it also converges to x. Hence
converges to x. Therefore, 
also
converges to x. This completes the proof.
Remark. The Bolzano-Weierstrass Theorem for bounded sequence in R n follows the same proof above by replacing R by R n, [c, d] by a large closed disk or ball and using the Heine-Borel Theorem for R n.
2. We can use the Bolzano-Weierstrass Theorem to prove the Extreme Value Theorem.
A consequence of the compactness of the domain on continuity.
Uniform Continuity
We shall stick to the one variable case. Let D be a subset of R.
Definition 9. A function f : D ® R is said to be uniformly continuous if given e > 0, there exists a d > 0 such that for any x, y in D, |x - y| < d Þ | f (x) - f (y) | < e.
The next result is a consequence of the closed and bounded interval being a compact set of R.
Notice that uniform continuity implies continuity.
Theorem 9. If the function f : [a, b] ® R is continuous, then it is also uniformly continuous.
Proof. The most important result we use here is the compactness of [a, b]. That means we are going to produce a family of open cover of [a, b]. Since f is continuous at each x in [a, b], given e > 0, there exists a d(x) > 0 (d here may depend on x) such that for any y in [a, b], | y - x| < d(x) Þ | f ( y) - f (x) | < e/2. This means whenever y is in the open set B(x, d(x)) = {z: |z - x| < d(x)}Ç[a, b] then | f ( y) - f (x) | < e/2. Therefore the collection C = {B(x, d(x)/2): x Î [a, b] } is an open cover for [a, b] . Since [a, b] is compact by the Heine-Borel Theorem (Theorem 5), C has a finite subcover say B = {B(x1, d(x1)/2), B(x2, d(x2)/2), ¼, B(xn, d(xn)/2),} where n is some positive integer. Now let d = min { d(x1)/2, d(x2)/2), ¼, d(xn)/2)}. Take any x, y in [a, b] such that | y - x| < d. Since B covers [a, b], x Î B(xk, d(xk)/2) for some 1 £ k £ n.
Therefore, | f ( xk ) - f (x) | < e/2 -------------- (1)
Now, let us see how far away from xk is y.
| y - xk| = | y -x + x - xk| £ | y -x | + | x - xk| < d + d( xk)/2 £ d( xk)/2 + d( xk)/2 = d( xk).
Hence y Î B(xk, d(xk)) and we have
| f ( y ) - f (xk) | < e/2 . -------------- (2)
Therefore,
| f ( y ) - f (x)| = | f ( y ) - f (xk) + f ( xk ) - f (x) |
£ | f ( y ) - f (xk) | + | f ( xk ) - f (x) | by the triangle inequality
< e/2 + e/2 = e by (1) and (2) above.
Hence, f is uniformly continuous.
This notion of uniform continuity proves useful to tell us that any continuous function on a closed and bounded interval is Riemann integrable.
Theorem 10. If the function f : [a, b] ® R is continuous, then it is Riemann integrable on [a, b].
Proof. If f : [a, b] ® R is continuous, then it is also uniformly continuous. Therefore given any e > 0, there exists d > 0 such that for all x, y in [a, b],
|x - y| < d Þ | f (x) - f (y) | < e/(b-a). ------------ (3)
Let P : a = x0 < x1 < x2 < ¼ < xn = b be a partition with norm ||P|| < d that is,
||P|| = max{ |xi - xi-1 | : i = 1, ¼ , n }< d. For i = 1, ¼ , n, let M i = sup{ f (x) : x Î [x i-1 , xi ]}. Then since f is continuous on [x i-1 , xi ], for each i, by the Extreme Value Theorem, M i = f (ci) for some ci in [x i-1 , xi ]. Similarly for each i = 1, ¼ , n, let we let m i = sup{ f (x) : x Î [x i-1 , xi ]}. Then again by the Extreme Value Theorem, for each i = 1, ¼ , n, there exists di in [x i-1 , xi ] such that mi = f (di). Then the upper Riemann sum with respect to P is
and the lower Riemann sum with respect to P is
.
Then the difference
by (3) since |ci - di| £ ||P|| < d, 1£ i £ n
Therefore, U(P) - L(P) < 
.
Hence, Riemann's condition holds and so by Theorem 1 in Riemann Integral and Bounded function, f is Riemann integrable. This completes the proof.