2) Useful properties of metric spaces Cauchysequencesandcompleteness Compactness Flashcards
DEF 2.1 Cauchy sequence
A sequence (xn) in a metric space (X, d) is a Cauchy sequence if for any ε > 0 there is an N such that d(xn,xm) < ε for all n,m ≥ N
e.g x_n =1/n is it a Cauchy sequence? in R
Cauchy sequence with limit in R
for d(x_n,x_m) = |1/n - 1\m| = |(n-m)/nm| = (n-m)/nm ≤ 1/N^2 ≤ 1/N
If we have n and m at least as big as N
hence if ε>0 and we take N> 1/ε we have d(x_n,x_m). ≤ 1/N < ε whenever n and m are both ≥ N.
thm 2.3 When is a sequence a Cauchy sequence
Suppose that (xn) is a convergent sequence in a metric space (X,d), i.e., there is a limit point x such that d(xn, x) → 0. Then (xn) is a Cauchy sequence.
PROOF: Proof. Take ε > 0. Then there is an N such that d(xn,x) < ε/2 whenever n ≥ N. Now suppose both n≥N and m≥N. Then
d(xn, xm) ≤ d(xn, x) + d(x, xm) = d(xn, x) + d(xm, x) < ε/2 + ε/2 = ε, and we are done WE CAN USE TRIANGLE INEQUALITY
Proposition 2.4. subsequences of Cauchy ones
Every subsequence of a Cauchy sequence is a Cauchy sequence.
Proof. If (xn) is Cauchy and (x_n_k ) is a subsequence, then given ε > 0 there is an N such that d(xn,xm) < ε whenever n,m ≥ N. Now there is a K such that n_k ≥ N whenever k ≥ K. So d(x_n_k,x_n_l) < ε whenever k,l ≥ K.
converge? Cauchy
(X,d) = Q, as a subspace of R with the usual metric. Take x_0 = 2
and define x({n+1} = ((x_n)/2) + 1/x_n.
3/2, 17/12, 577/408, . . . and xn→x where x=(x/2) + (1/x)
,i.e.,x^ =2. But this isn’t in Q.
Thus (xn) is Cauchy in R, since it converges to √2 when we think of it as a sequence in R. So it is Cauchy in Q, but doesn’t converge to a point of Q
Take (X, d) = (0, 1). Then (1/n) is a Cauchy sequence in X
is a Cauchy sequence in X (since it is Cauchy
in R), and has no limit in X. In each case there are “points missing from X”.
def 2.6 Completeness
(Completeness). A metric space (X,d) is complete if every Cauchy sequence in X converges to a limit in X
Thm 2.7 completness of R
The metric space R is complete
In parts of the literature R is simply defined as the completion of Q. In this case one does not have to prove that R is complete, but it is complete by construction. One then has to work a bit to show that it is also a fiel
Example: Consider sequence where distance between consecutive terms halves Cauchy?
d(x_n+1, x_n+2) = 0.5d(x_n, x_(n+1))
by triangle inequality used repeatedly
d(x_m,x_n) ≤ d(x_m, x_m+1) +….+d(x_n-1, x_n)
≤ d(x1,x_2) * [ (1/2)^{m+1} +….+(1/2)^{n}]
geometric series ≤ infinite sum. = (1/2^{m-2}} d(x1,x2). which converges to 0 as n tends to infinity
So we take an N s.t. d(x_1,x_2)/ (2^{m-1}) <
Example 2.9 Open intervals in R are not complete. Closed intervals are.
(a,b) seq (x_n) = b - (b-a)/n converges to b which is not in (a,b)
closed subset complete?[0, ∞
)
Closed subspace of complete space R so yes
(all closed subsets of R^k are complete.)
What about C[a, b] with d1 IS IT COMPLETE?
c[a,b] c[0,2] f_n(x) =
{x^n for x in [0,1]
{1 for x in [1,2]
sketch the graphs!
0 ≤ d_1(f_n,f_m)
= ∫[0,2] |f_n(x)-f_m(x)| = ∫[0,1] |x^n-x^m| = 1/(m+1) - 1/(n+1) ≤ 1/(m+1) ( →0 )
hence (f_n) is Cauchy in (C[0, 2], d1)
Does it converge? If there is an
f ∈C[0, 2] with fn → f as n → ∞, then ∫[0,2]|fn(x) − f(x)| dx → 0, so
∫[0,1] and ∫_[1,2] both tend to zero. So f_n → f in (C[0, 1], d1),
which means that f(x) = 0
on [0, 1] (from an example we did earlier). Likewise, f = 1 on [1, 2], which doesn’t give a continuous limit. So f is not in C[0,1] so this MS is incomplete
True or false: (C[a, b], d1) is incomplete in general
True. Also it is incomplete in the d2 metric, as the same example shows (a similar calculation with squares of functions). We will
see later that it is complete in the d∞ metric
True or false? R^2 is incomplete with any metric d1, d2,d∞;
False: R^2
is also complete with any of the metrics d1, d2 and d∞; since a
Cauchy/ convergent sequence (v_n) = (xn, yn) in R^2
is just one in which both (xn) and (yn) are Cauchy/ convergent sequences in R (cf. Prop. 1.37).
Similar arguments show that R^k
is also complete for k = 1, 2, 3, . . ., and (with the same
proof as for Corollary) all closed subsets of R^k are complete.
Propn 2.11 (Abstract completion)
If a metric space (X, d) is not complete one can always….
Any metric space (X, d) is isometric to a dense subspace of a complete metric space, which is called its abstract completion if
Isometric Embedding:
There exists an isometric embedding f:X→Y, where
Y is a complete metric space. An isometric embedding preserves distances between points.
Dense Subspace:
The image f(X) is dense in Y. This means that the closure of f(X) in Y is the whole space Y. In other words, every point in Y is a limit point of f(X).
Completeness:
The metric space Y is complete. A metric space is complete if every Cauchy sequence in that space converges to a limit within the space.
**
The abstract completion of a metric space provides a way to “complete” the original space by adding points to make it a complete metric space. The process involves embedding the original space into a larger, complete space in a way that preserves distances and ensures that the image is dense.**
Abstract completion proof 2.11
sketch of proof. Based on assigning limits to cauchy sequences
Lemma 2.12 dense sets completeness
Suppose that (X, d) is a metric space and let Y ⊂ X be a dense set
with the property that every Cauchy sequence in Y has a limit in X. Then (X, d) is complete
Proof. Let (x_n) be a Cauchy sequence in X. Now replace x_n with another sequence y_n
in Y such that d(x_n, y_n) < 1/n. Then, by the triangle inequality, y_n is again a Cauchy sequence and converges, by assumption, to some x ∈ X. Then also x_n converges to
x.
Corollary 2.14. abstract completions ex
All abstract completions of a metric space (X, d) are isometric, in other
words, the abstract completions is unique up to isometry.
exercise : unique function f st continuous on x and restriction of function f to X_1 coincides with function f f(x)=f(x) for all x in X_1
for X1 dense subset of X f: X_1 to Y uniformly continuous and f:X to Y
Open cover defn
Of a MS (X, d) is a family of open sets (α∈I) (U_α) such that
[∪_α∈I] U_α = X.
Defn subcover
subcover of a cover is a subset I′ ⊂ I of the index set such that
(U_α)_α∈I′ is still a cover
e.g imagine covering a set by different sets, not necessarily disjoint or a partition. Imagine not needing to use all of these sets to still cover all.
Def 2.15 Compactness
A metric space (X, d) is called compact if every open
cover has a finite subcover
(a space is compact if any infinite open covering is excessive and can be reduced to a finite one)
[0,1] is compact
R, (0,1) is not compact e.g x_n=n shows R is not compact
is [0,1] compact?
(0,1)?
An example of a compact set is [0, 1] and example of
non-compact—all reals or the open interval (0, 1)
On R^m compact IFF closed and bounded
on a MS compact then closed and bounded
Consider the open covering of the interval [0,1] given by the open intervals (0, 1/2) and (1/2, 1). This covering consists of two open sets that together cover the entire interval. Now, by the definition of compactness, we need to show that there exists a finite subcovering. In this case, the original covering itself is already finite, so it serves as its own finite subcovering, demonstrating that [0,1] is compact.
(0,1) is not compact.
Def 2.16 Sequential compactness
A metric space (X, d) is called sequentially compact if every sequence (xn)n∈N in X has a convergent subsequence.
sequential compactness visually:
a space is sequentially compact if there is no room to place infinite number of points sufficiently apart from each other to avoid their condensation to a limit.
Taking x_n = n shows that the set of all reals is not sequentially compact. On the other hand, bounded closed set in R^n every sequence has a convergent subsequence. Therefore, bounded closed sets in R^n are sequentially compact.
