where n_{k+1} is the first index such that x_{nk+1} is contained B
(p,δk+1)
.
Then
lim xnk = p. ■
k→ ∞
Another useful result is the following.
Lemma 11.1.19Suppose x_{n}→ x and y_{n}→ y. Then d
(xn,yn)
→ d
(x,y)
.
Proof:Consider the following.
d(x,y) ≤ d(x,xn)+ d(xn,y) ≤ d (x,xn) + d(xn,yn)+ d(yn,y)
so
d(x,y)− d(xn,yn) ≤ d(x,xn)+ d (yn,y)
Similarly
d(xn,yn)− d(x,y) ≤ d(x,xn)+ d (yn,y)
and so
|d(xn,yn)− d(x,y)| ≤ d(x,xn)+ d(yn,y)
and the right side converges to 0 as n →∞. ■
First are some simple lemmas featuring one dimensional considerations. In these, the metric space is ℝ
and the distance is given by
d(x,y) ≡ |x− y|
First recall the nested interval lemma. You should have seen something like it in calculus, but
this is often not the case because there is much more interest in trivialities like integration
techniques.
Lemma 11.1.20Let
[ak,bk]
⊇
[ak+1,bk+1]
for all k = 1,2,3,
⋅⋅⋅
. Then there exists a point p in∩_{k=1}^{∞}
[ak,bk]
.
Proof: We note that for any k,l,a_{k}≤ b_{l}. Here is why. If k ≤ l, then
ak ≤ al ≤ bl
If k > l, then
bl ≥ bk ≥ ak
It follows that for each l,
sup ak ≤ bl
k
Hence sup_{k}a_{k} is a lower bound to the set of all b_{l} and so it is no larger than the greatest lower bound. It
follows that
supak ≤ inf bl
k l
Pick x ∈
[supkak,inflbl]
. Then for every k,a_{k}≤ x ≤ b_{k}. Hence x ∈∩_{k=1}^{∞}
[ak,bk]
.■
Lemma 11.1.21The closed interval
[a,b]
is compact.This means that if there is a collection ofopen intervals of the form
(a,b)
whose union includes all of
[a,b]
, then in fact
[a,b]
is contained inthe union of finitely many of these open intervals.
Proof: Let C be a set of open intervals the union of which includes all of
[a,b]
and suppose
[a,b]
fails
to admit a finite subcover. That is, no finite subset of C has union which contains
[a,b]
. Then this must be
the case for one of the two intervals
[a, a+b2 ]
and
[a+2b,b]
. Let I_{1} be the one for which this is so. Then split
it into two equal pieces like what was just done and let I_{2} be a half for which there is no finite subcover of
sets of C. Continue this way. This yields a nested sequence of closed intervals I_{1}⊇ I_{2}⊇
⋅⋅⋅
and by the
above lemma, there exists a point x in all of these intervals. There exists U ∈C such that x ∈ U.
Thus
x ∈ (a,b) ∈ C
However, for all n large enough, the length of I_{n} is less than min
(|x− a|,|x − b|)
. Hence I_{n} is
actually contained in
(a,b)
∈C contrary to the construction. Hence
[a,b]
is compact after all.
■
As a useful corollary, this shows that ℝ is complete.
Corollary 11.1.22The real line ℝ is complete.
Proof:Suppose
{xk}
is a Cauchy sequence in ℝ. Then there exists M such that
{xk}
_{k=1}^{∞}⊆
[− M, M ]
.
Why? If there is no convergent subsequence, then for each x ∈
[− M,M ]
, there is an open set
(x − δx,x + δx)
which contains x_{k} for only finitely many values of k. Since
[− M, M ]
is compact, there are
finitely many of these open sets whose union includes
[− M,M ]
. This is a contradiction because
[− M, M ]
contains x_{k} for all k ∈ ℕ so at least one of the open sets must contain x_{k} for infinitely many k. Thus there
is a convergent subsequence. By Theorem 11.1.16 the original Cauchy sequence converges to some
x ∈
[− M, M ]
. ■
Example 11.1.23Let n ∈ ℕ. ℂ^{n}with distance given by
d (x,y) ≡ max {|xj − yj|}
j∈{1,⋅⋅⋅,n}
is a complete space. Recall that
|a+ jb|
≡
√ ------
a2 + b2
. Then ℂ^{n}is complete. Similarly ℝ^{n}iscomplete.
To see that this is complete, let
{ k}
x
_{k=1}^{∞} be a Cauchy sequence. Observe that for each j,
{ k}
xj
_{k=1}^{∞}.
That is, each component is a Cauchy sequence in ℂ. Next,
|| k k+p|| || k k+p||
|Re xj − Re xj | ≤ |x j − xj |
Therefore,
{ }
Re xkj
_{k=1}^{∞} is a Cauchy sequence. Similarly
{ }
Im xkj
_{k=1}^{∞} is a Cauchy sequence. It
follows from completeness of ℝ shown above, that these converge. Thus there exists a_{j},b_{j} such
that
kli→m∞Re xkj + iIm xkj = aj + ibj ≡ x
and so x^{k}→ x showing that ℂ^{n} is complete. The same argument shows that ℝ^{n} is complete. It is easier
because you don’t need to fuss with real and imaginary parts.