2) Useful properties of metric spaces Cauchysequencesandcompleteness Compactness

Question 1

DEF 2.1 Cauchy sequence

Answer 1

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

Question 2

e.g x_n =1/n is it a Cauchy sequence? in R

Answer 2

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.

Question 3

thm 2.3 When is a sequence a Cauchy sequence

Answer 3

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

Question 4

Proposition 2.4. subsequences of Cauchy ones

Answer 4

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.

Question 5

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.

Answer 5

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

Question 6

Take (X, d) = (0, 1). Then (1/n) is a Cauchy sequence in X

Answer 6

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”.

Question 7

def 2.6 Completeness

Answer 7

(Completeness). A metric space (X,d) is complete if every Cauchy sequence in X converges to a limit in X

Question 8

Thm 2.7 completness of R

Answer 8

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

Question 9

Example: Consider sequence where distance between consecutive terms halves Cauchy?

Answer 9

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}) <

Question 10

Example 2.9 Open intervals in R are not complete. Closed intervals are.

Answer 10

(a,b) seq (x_n) = b - (b-a)/n converges to b which is not in (a,b)

Question 11

closed subset complete?[0, ∞
)

Answer 11

Closed subspace of complete space R so yes

(all closed subsets of R^k are complete.)

Question 12

What about C[a, b] with d1 IS IT COMPLETE?

Answer 12

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

Question 13

True or false: (C[a, b], d1) is incomplete in general

Answer 13

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

Question 14

True or false? R^2 is incomplete with any metric d1, d2,d∞;

Answer 14

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.

Question 15

Propn 2.11 (Abstract completion)
If a metric space (X, d) is not complete one can always….

Answer 15

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.**

Question 16

Abstract completion proof 2.11

Answer 16

sketch of proof. Based on assigning limits to cauchy sequences

Question 17

Lemma 2.12 dense sets completeness

Answer 17

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.

Question 18

Corollary 2.14. abstract completions ex

Answer 18

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

Question 19

Open cover defn

Answer 19

Of a MS (X, d) is a family of open sets (α∈I) (U_α) such that
[∪_α∈I] U_α = X.

Question 20

Defn subcover

Answer 20

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.

Question 21

Def 2.15 Compactness

Answer 21

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

Question 22

is [0,1] compact?
(0,1)?

Answer 22

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.

Question 23

Def 2.16 Sequential compactness

Answer 23

A metric space (X, d) is called sequentially compact if every sequence (xn)n∈N in X has a convergent subsequence.

Question 24

sequential compactness visually:

Answer 24

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.

Question 25

A useful example of a covering for a set

Answer 25

consider {Bε(x) : x is in X} ε>0 if all balls union cover X then X is complete

Question 26

T or F: bounded subsets on R^n are complete

Answer 26

Bounded closed sets every sequence has a convergent subsequence so true

(In a bounded closed set in R^n, every sequence has a convergent subsequence. This is a consequence of the Bolzano-Weierstrass theorem, which states that every bounded sequence in R^n has a convergent subsequence. The closedness of the set ensures that the limit of the convergent subsequence is also within the set

Question 27

Compact sets in discrete metric space?

Answer 27

In a discrete metric space, every subset is both open and closed.
(In a discrete metric space, the distance between distinct points is always 1. every subset is open, as every singleton set is an open ball.

Compactness:
In a discrete metric space, every subset is open, as every singleton set is an open ball.
Compactness in a metric space means that every open cover has a finite subcover.
Since every subset is open, any open cover trivially has a finite subcover, making every subset compact. Cover of centred at the finite points centred with distances 1 between them

Sequential Compactness:
In the discrete metric space, any sequence is eventually constant, and a constant sequence trivially converges.
Therefore, every sequence in a discrete metric space has a convergent subsequence, making it sequentially compact.

In summary, in a discrete metric space, all subsets are compact and sequentially compact due to the unique properties of the discrete metric.

Question 28

lemma 2.18 sequentially compact and balls

Answer 28

Let (X, d) be a sequentially compact metric space. Then for every ε > 0
there exist finitely many points x1, . . . , xn such that {Bε(xi) | i = 1, . . . , n} is a cover

Proof: assume not, then there is an ε > 0 s.t. for any finite points x_1,..x_n collection of Bε(xi) doesn’t cover. Ie Union from i=1 to n not equal to X.

From n=1 inductively adding points in the complements we end up with an infinite sequence of points x_k s.t d(x_i,x_k) ≥ ε. The sequence cannot have a Cauchy subsequence (required for convergence) in contradiction with the sequential compactness of X.

Question 29

THM 2.19 Sequential compactness vs compactness

Answer 29

A metric space (X, d) is compact if and only if it is sequentially compact

e.g can show MS not compact by not having a convergent subsequence (e.g. not complete then definitely not sequentially compact)

Question 30

Thm 2.19 proof

Answer 30

proof
diagrams in notes

Question 31

DEf 2.20 Boundedness

Answer 31

A subset A ⊂ X of a metric space is called bounded
if there exists x_0 ∈ X and C > 0 such that for all x ∈ A we have d(x_0, x) ≤ C

By triangle inequality ref point x_0 can be any point in X. So if A ⊂ X is bounded and x0 ∈ X, then there
exist a C > 0 such that d(x0, x) ≤ C for any x ∈ A.

ALT: X is contained in B_C(x_0)

Question 32

where does the bounded defn come from?

Answer 32

Imagine bounded above: least upper bound and bounded below: largest lower bound
ie if bounded above and below we ensure that the distance between any two elements or functions is never infinite so we need to show always have d(x,y) not infinite. If we can choose any x,y that are giving ∞ then its not bounded! disproval of bounded

Question 33

To show set unbounded:

Answer 33

strategies seen:
(bounded
if there exists x_0 ∈ X and C > 0 such that for all x ∈ A we have d(x_0, x) ≤ C)

so show that there is some x that maybe converges to infinity

e.g d(1,n) converges to infiniy so [0,∞) unbounded

Question 34

THM 2.22 compact, closed and bounded

Answer 34

Suppose that A ⊂ X is a compact subset of a metric space. Then A is closed and bounded

(growing balls eventually cover)
Proof: last years
First we show A is bounded. Choose any x0 ∈ X and note that the set B_n(x0) indexed by n ∈ N is an open cover of A. Hence, there exists a finite sub-cover:
B_{n_1}(x_0), . . . , B_{n_N}(x0). Hence, A ⊂ B_C(x_0), where C = max{n_1, . . . , c_N }. Hence, A is
bounded.

Next assume that (x_k) is a sequence in A that converges in X. Since A is compact there
exists a subsequence that converges in A. Hence, the limit of xk must also be in A.
Therefore, A is closed.

Question 35

Thm 2.23 (Heine–Borel).

Answer 35

A subset K ⊂ R^m is compact if and only if it is closed
and bounded.

compactness implies closedness and boundedness. If K is closed and bounded we know that it is sequentially compact .

Question 36

heine borel used when?

Answer 36

A subset **K ⊂ R^m only

Otherwise we might consider coverings for compactness or convergent subsequences

Question 37

Q compact?

Answer 37

The set of rational numbers Q is not compact because it is not closed and bounded in the real number line. It is also not complete, as there exist Cauchy sequences of rational numbers that do not converge to a rational number. The closure of Q in the real number line is the set of real numbers, denoted by R. The intersection of Q with any closed interval in R will result in a set that is not compact, as Q itself is not compact.

Question 38

continuous functions and bounded

Answer 38

exercises?2023

Any continuous function on a compact set is bounded.

Any continuous function f : K → X from a compact space K to a metric space X is uniformly continuous

Question 39

thm proved elsewhere might be useful

Answer 39

A is a compact set in MS then A is complete

Question 40

Example of a compact set
An example of a non compact set

Answer 40

er the compactness criteria for Euclidean space as stated in the Heine–Borel theorem, the interval A = (−∞, −2] is not compact because it is not bounded. The interval C = (2, 4) is not compact because it is not closed (but bounded). The interval B = [0, 1] is compact because it is both closed and bounded.

Question 41

Sense of compactness
Why is [0, 1] compact
Why is R not?

Answer 41

We know we it is closed and bounded it’s compact

But idea: sequential compactness Thus, if one chooses an infinite number of points in the closed unit interval [0, 1], some of those points will get arbitrarily close to some real number in that space. For instance, some of the numbers in the sequence
1/2, 4/5 1/3, 5/6, 1/4, 6/7, … accumulate to 0 (while others accumulate to 1). Since neither 0 nor 1 are members of the open unit interval (0, 1), those same sets of points would not accumulate to any point of it, so the open unit interval is not compact. Although subsets (subspaces) of Euclidean space can be compact, the entire space itself is not compact, since it is not bounded. For example, considering R^1 (the real number line), the sequence of points 0,  1,  2,  3, … has no subsequence that converges to any real number.

Question 42

Compactness of Q and R

Answer 42

the space of rational numbers
Q not compact, because it has infinitely many “punctures” corresponding to the irrational numbers, and the space of real numbers R is not compact either, because it excludes the two limiting values +−∞

Question 43

closed vs complete?

Answer 43

A metric space is complete if every Cauchy sequence converges (to a point already in the space). A subset F of a metric space X
is closed if F contains all of its limit points; this can be characterized by saying that if a sequence in F
converges to a point x in X
, then x must be in F
. It also makes sense to ask whether a subset of X is complete, because every subset of a metric space is a metric space with the restricted metric.

It turns out that a complete subspace must be closed, which essentially results from the fact that convergent sequences are Cauchy sequences. However, closed subspaces need not be complete.

If X is a complete metric space, then a subset of X
is closed if and only if it is complete

Question 44

complete vs closed examples

Answer 44

closed subspaces need not be complete. For a trivial example, start with any incomplete metric space, like the rational numbers Q
with the usual absolute value distance. Like every metric space, Q
is closed in itself, so there you have a subset that is closed but not complete. If taking the whole space seems like cheating, just take the rationals in [0,1], which will be closed in Q but not complete.

Question 45

when will funct attain max/min

Answer 45

A function will be bounded and attain a maximum/minimum on a closed and bounded interval. This is known as the Extreme Value Theorem. If a function is continuous on a closed interval [a, b], then it is guaranteed to attain both a maximum and a minimum value on that interval. Additionally, if the function is bounded, it means that its range is limited within a certain range of values.

Question 47

property of discrete, Metric space
Open closed

Answer 47

In a discrete metric space, every subset is both open and closed due to the nature of the discrete metric. Let me explain:

Open Sets:
In the discrete metric space, the open ball of radius r>0 around any point x consists only of x itself if r<1, because the distance between distinct points is always 1.
This implies that every singleton set (containing only one point) is an open set, as it can be covered by the open ball of radius
0.5 centered at that point.
Closed Sets:
Closed sets in a metric space are often defined as complements of open sets. If a set is open, its complement is closed.In the discrete metric space, every set (including singletons) is open. Therefore, the complement of any set is closed.

In summary, the discrete metric space has the property that every subset, including singletons, is open. Consequently, since closed sets are complements of open sets, every subset is also closed in the discrete metric space. This is a distinctive characteristic of the discrete metric spac