1. Bolzano-Weierstrass Theorem and Sequential Compactness A sequence (an) in Rp is bounded if there exists M > 0 such that
kank ≤ M for all n ∈ N.
Let us write an = (a1n, · · · , apn) for n ≥ 1. Then
Proposition 1.1. Then (an) is bounded in Rp if and only if (a1n), · · · , (apn) are bounded sequences in R.
Proof. This is left to the reader.
A sequence (bk) in Rp is a subsequence of (an) if there exists an increasing function f : N → N such that
bk = af (k) for k ≥ 1.
We denote f (k) by nk for any k ≥ 1. Hence a subsequence of (an) is also denoted by (ank).
Lemma 1.1. Let f : N → N be an increasing function. Then f (k) ≥ k for any k ≥ 1.
Proof. One can prove this lemma by Mathematical induction.
Proposition 1.2. Let (an) be a sequence in a metric space (M, d) convergent to a point a ∈ M.
Then all subsequences (ank) are also convergent to a.
Proof. Since (an) is convergent to a, for > 0, there exists N∈ N such that d(an, a) < whenever n ≥ N. Let (ank) be a subsequence of (an). If k ≥ N, nk ≥ k ≥ N. Then d(ank, a) < . Therefore (ank) is convergent to a.
Theorem 1.1. (Bolzano-Weierstrass Theorem I) A bounded sequence in Rp has a convergent sub- sequence.
Definition 1.1. Let (an) be a sequence in a metric space (M, d). A point a of M is said to be a cluster point of (an) if there exists a subsequence (ank) of (an) so that limk→∞ank = a. In other word, a is a cluster point of (an) if it is the limit of a subsequence of (an).
Using Definition 1.1, Bolzano-Weiertrass theorem is also equivalent to:
Theorem 1.2. (Bolzano-Weierstrass Theorem II) A bounded sequence in Rp has a cluster point.
A subset S of Rp is said to be bounded if there exists M > 0 such that kxk ≤ M for all x ∈ S.
Lemma 1.2. Let K be a closed and bounded subset of Rp. Then any sequence (an) in K has a cluster point in K.
Proof. Let (an) be a sequence in K. Since K is bounded, there exists M > 0 such that kxk ≤ M for all x ∈ K. Since (an) is a sequence in K, an ∈ K and hence kank ≤ M for all n ≥ 1. We see that (an) is a bounded sequence. By Bolzano-Weierstrass Theorem, we can choose a point a ∈ Rp and a subsequence (ank) of (an) such that limk→∞ank= a. Then a is an adherent point of K. Since K is closed, a ∈ K. By definition, a is a cluster point of (an).
This lemma leads to the notion of sequential compactness.
Definition 1.2. Let (M, d) be a metric space. We say that M is sequentially compact if and only if any sequence in M has a cluster point.
Let (M, d) be a metric space and N be a subset of M. For each x, y ∈ N, we set dN(x, y) = d(x, y).
Since d is a metric on M, dN is a metric on N. The pair (N, dN) is again a metric space and called a metric subspace of (M, d).
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Definition 1.3. A subset N of a metric space (M, d) is called sequentially compact if the metric subspace (N, dN) is sequentially compact.
We obtain another equivalence of Bolzano-Weierstrass Theorem.
Theorem 1.3. (Bolzano-Weierstrass Theorem III) A subset K of Rp is sequentially compact if and only if it is closed and bounded.
Proof. We have proved that if K is closed and bounded, then K is sequentially compact in Lemma 1.2. Now let us prove the converse. Assume that K is sequentially compact.
Suppose K is unbounded. We can choose a sequence (an) in K such that
(1.1) kank ≥ n for all n ≥ 1.
Since K is sequentially compact, choose a subsequence (ank) of (an) convergent to a cluster point a ∈ K by Bolzano-Weierstrass Theorem II. (1.1) implies that
kankk ≥ nk≥ k
for any k ≥ 1. Since (ank) is convergent, it is bounded. There exists M > 0 so that kankk ≤ M for all k ≥ 1. This implies that k ≤ M for all k ≥ 1 which is impossible.
Let us prove that K is closed. Let a be an adherent point of K. Choose a sequence (an) in K such that limk→∞an= a. Since K is sequentially compact, there exists a subsequence (ank) of (an) such that limk→∞ank = b for some b ∈ K. By Proposition 1.2, b = a since (an) is convergent.
Thus a = b ∈ K. This shows that K is closed.
2. Proof of Completeness of Rp.
Definition 2.1. A sequence (an) in a metric space (M, d) is a Cauchy sequence if for any > 0 there exists N∈ N such that
d(an, am) < whenever n, m ≥ N.
Proposition 2.1. A convergent sequence (an) in a metric space (M, d) is always a Cauchy sequence.
Proof. Let (an) be a sequence in (M, d) convergent to a point a. For > 0, we choose N∈ N such that d(an, a) < /2. For n, m ≥ N, we have
d(an, am) ≤ d(an, a) + d(am, a) < 2+
2 = .
Hence (an) is a Cauchy sequence.
Definition 2.2. A metric space (M, d) is said to be a complete metric space if all of its Cauchy sequences are convergent.
The following example shows that not every metric space is complete.
Example 2.1. Let (Q, dQ) be the metric subspace of (R, d) where d(x, y) = |x − y| for any x, y ∈ R.
Define (xn) by
xn= 1 + 1 1!+ 1
2!+ · · · + 1
n!, n ≥ 1.
Show that (xn) is a Cauchy sequence in (Q, dQ) and (xn) is divergent in (Q, dQ).
Proposition 2.2. Let (M, d) be a metric space and (an) be a Cauchy sequence in (M, d). If (ank) is a subsequence of (an) such that (ank) is convergent to a point a ∈ M, then (an) is also convergent to a.
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Proof. Since (an) is a Cauchy sequence, for > 0, there exists N0 ∈ N such that d(an, am) < /2 whenever n, m ≥ N0. Since (ank) converges to a, for the mentioned above, choose N00 ∈ N such that d(ank, a) < /2 whenever k ≥ N00. Let N = max{N0, N00}. When n, k ≥ N, nk ≥ N0, N00. Hence
d(an, a) ≤ d(an, ank) + d(ank, a) < 2 +
2 = .
We find that d(an, a) < whenever n ≥ N. Therefore (an) is convergent to a. Lemma 2.1. Any Cauchy sequence (an) in Rp is bounded.
Proof. We will prove the case when p = 1. In fact the proof works for any p ∈ N. Choose = 1 and find N ∈ N such that
|an− am| < 1 whenever n, m ≥ N.
We choose m = N. Then |an− aN| < 1 for any n ≥ N. Thus |an| < 1 + |aN| for any n ≥ N. Let M = max{|a1|, · · · , |aN −1|, 1 + |aN|}.
Then |an| ≤ M for all n ≥ 1.
By Lemma 2.1, if (an) is a Cauchy sequence in Rp, then it is bounded. By Bolzano-Weierstrass Theorem, (an) has a subsequence (ank) which is convergent to some point a in Rp. By Proposition 2.2, (an) is convergent to a. Thus we proved that
Theorem 2.1. (Completeness of Rp) Rp with the Euclidean metric is a complete metric space.
