Sequences and Sets
Definitions Let F : A → B be a map of sets.
(a) For any subset E ⊂ A, the image of E under F is F (E) ⊆ B, i.e.
F (E) = {F (x) ∈ B | x ∈ E}
(b) The range of F is F (A).
(c) The domain of F is A.
(d) The codomain of F is B.
(e) F isonto (surjective) if F (A) = B.
(f) For C ⊆ B, the inverse image of C is F−1(C) = {x ∈ A | F (x) ∈ C}.
(g) F isone-to-one (injective) if for each element b ∈ B, F−1(b) has at most one element.
(h) F isbijective if F is injective and surjective.
(i) A and B are in 1 − 1 correspondence, or have the same cardinality, or, also equivalently, the sets are called bijective or equivalent, if ∃ a bijection F : A → B.
Note F is injective ⇐⇒ (F (x1) = F (x2) ⇐⇒ x1 = x2).
Remark Bijective sets are denoted A ↔ B or A ∼ B. The latter notation is justified because this relation is an equivalence relation.
Definition For n ∈ N, let
[n] = {1, 2, 3 . . . , n}.
Let A be a set.
(a) A is finite if ∃ a bijection, A ↔ [n] for some n ∈ N.
(b) A is infinite if it is not finite.
(c) A is countableif A ↔ N.
(d) A is at most countable if A is finite or countable.
(e) A is uncountable if A is not finite and not countable.
Definition Let A be a set. A sequence in A is a function (i.e., map of sets) f : N → A. We write f (n) = an. and {an} (i.e., just the image).
Theorem Every subset of a countable set is at most countable.
Proof Let A be a countable. Let E ⊆ A. If E is finite, then E is at most countable and we’re done. Suppose E is infinite. Since A is countable, A ↔ N. So A may be written in a sequence {an}. Let n1 be the smallest n ∈ N such that an1 ∈ E. In other words,
{n ∈ N | an ∈ E},
because any nonempty subset of N has a least element. For k > 1, let nk be smallest integer greater than nk−1 such that ank ∈ E:
{n ∈ N | n > nk−1, an∈ E}
Now we have inductively labeled all the elements of E. This gives a function N → E. It is bijective, thus showing any subset of a countable set is at most countable.
Proposition Consider the following:
(a) A ∪ (B ∩ C) = (A ∩ B) ∪ (A ∩ C).
(b) (De Morgan’s Law) Let A1, A2, · · · ⊆ X. Then both the following hold:
X\(
∞
[
i=1
Ai) =
∞
\
i=1
(X\Ai),
X\(
∞
\
i=1
Ai) =
∞
[
i=1
(X\Ai).
Note For B ⊆ A, A\B = A − B = Bc.
Theorem A countable union of countable sets is countable.
Proof Let {An} be a sequence of countable sets (i.e., Aj is countable for each j). Let A1 = {a11, a12, a13, . . . }, A2 = {a21, a22, a23, . . . }, and in general, An= {an1, an2, an3, . . . }, then we can write
a11 a12 a13 . . . a21 a22 a21 . . . a31 a32 a33 . . . ... ... ... . ..
.
Note that along the matrix diagonals, the sum of the i, j is constant. Consider aij as an array, we count along the diagonals:
{a11, a21, a12, a31, a22, a13, . . . }.
This is a count of
∞
[
i=1
Ai.
Challenge Find an explicit function f : N →
∞
[
i=1
Ai.
Theorem Let A be a countable set. Then An = A × A × · · · × A (n times) is countable.
Recall An = {(a1, . . . , an) | ai ∈ A}.
Proof (Induction) A1 = A, so A1 is countable. Suppose Ak is countable. Elements of Ak+1 are of the form (x, y), where x ∈ Ak, y ∈ A. For each fixed x0 ∈ Ak, we know
{(x0, y) | y ∈ A} ↔ A and thus it is countable. Thus
{(x, y) | x ∈ Ak, y ∈ A}
is a countable union of countable sets:
{(x, y) | x ∈ Ak, y ∈ A} = [
x0∈Ak
{(x0, y) | y ∈ A}.
Therefore Ak+1 is countable. Hence, the result follows by induction.
Corollary Q is countable.
Proof Z2 is countable by previous theorem. So {ab | a, b 6= 0 ∈ Z} is countable.
Definition An algebraic number is an element x0 ∈ R such that anxn0 + an−1xn−10 + · · · + a1x0+ a0 = 0 for some a0, . . . , an∈ Z.
Theorems
(a) The set of all algebraic numbers is countable.
(b) The set {0, 1}∞ is uncountable.
Note An= {(a1, . . . , an) | ai ∈ A}. Thus A∞= {(a1, a2, . . . , aj, . . . ) | ai ∈ A}.
Proof (Cantor) Suppose {0, 1}∞ is countable. Consider the following naming:
x1 = x11x12x13. . . x1j ∈ {0, 1}
x2 = x21x22x23. . . x2j ∈ {0, 1}
x3 = x31x32x33. . . x3j ∈ {0, 1}
...
In fact, for any i, j ∈ Z+, we have xij ∈ {0, 1}. Our goal is to construct an x ∈ {0, 1} such that x 6= xi∀i.
Let a1 6= x11, a1 ∈ {0, 1}. Let a2 6= x22. . . Let ai 6= xii. Then let x = (a1, a2, a3, . . . ) ∈ {0, 1}∞. Observe that for any i, x 6= xi, since ai 6= xii, by construction. We get that the set is countable.
Remark Same argument works for R. Consider why does this argument not show Q is uncount- able?
Suppose I started counting the reals:
x1 = 0.137294 . . . x2 = 3.123490 . . . x3 = 0.999713 . . .
Going down the diagonal works for the reals but not for the rationals because rationals either terminate or repeat. In other words, you cannot prove the result from changing all the diagonals is rational itself.
Metric Spaces
Definition A set X is a metric space if there exists a map d : X × X 7→ R such that for all points p, q ∈ X:
• d(p, q) ≥ 0; equality ⇐⇒ p = q.
• d(p, q) = d(q, p).
• d(p, q) ≤ d(p, r) + d(r, q) ∀r ∈ X.
Here d is called ametricon X. A metric space is a space that comes with a metric defined on it.
Recall A metric on X must satisfy the following
• d(a, b) ≥ 0, (d(a, b) = 0 ⇐⇒ a = b)
• d(a, b) = d(b, a)
• d(a, b) ≤ d(a, c) + d(c, b)
Example Consider the following spaces and corresponding metrics:
• R, d(x, y) = |x − y|.
• Rn, d(~x, ~y) = k~x − ~yk.
• R2, ds(~x, ~y) =P2
i=1|xi− yi|.
Non-Example On Rn, consider the following metrics that don’t work:
• d(~x, ~y) = |x1− y1|
• d(p, q) =P pilogpi
qi
• d(~x, ~y) = 0
Definition Let x ∈ X, r ∈ R ≥ 0. We define Br(x) = B(x, r) = {y ∈ X | d(x, y) < r}. This is called an open ball centered at x of radius R. Also called a neighborhood. The closed ball is given by B(x, r) = {y ∈ X | d(x, y) ≤ r}.
Definitions Let (X, d) be a metric space.
(a) A neighborhood of p ∈ X is an open ball B(p, r) = Nr(p) = {q ∈ X | d(p, q) < r}.
(b) A point p ∈ X is called a limit point of E ⊂ X if every neighborhood of p contains a point q 6= p such that q ∈ E. The set of limit points of E is denoted by E0.
i.e. p ∈ E0 ⇐⇒ ∀r > 0, Nr(p) ∩ E \ {p} 6= ∅.
Examples Let X = R and E = Z, or 1
n | n ∈ N
, or Q, or (0, 1), or [0, 1), or [0, 1]. Find E0.
(c) E is closed(in X) if every limit point of E is a point of E, i.e. E0 ⊂ E.
(d) A point p ∈ X is calledan interior point of Eif there is a r > 0 such that the neighborhood Nr(p) of p satisfies that Nr(p) ⊂ E. The set of interior points of E is denoted by Eo. i.e. p ∈ Eo ⇐⇒ ∃ r > 0 such that Nr(p) ⊂ E.
Examples Let X = R and E = Z, or 1
n | n ∈ N
, or Q, or (0, 1), or [0, 1), or [0, 1]. Find Eo.
(e) E is open(in X) if every point of E is an interior point of E, i.e. E = Eo (f) The closure of E is the set ¯E = E ∪ E0.
(g) A point p ∈ X is called a boundary point of E ⊂ X if every neighborhood of p intersects both E and Ec= X \ E. The set of boundary points of E is denoted by ∂E.
i.e. p ∈ ∂E ⇐⇒ ∀r > 0, Nr(p) ∩ E 6= ∅ and Nr(p) ∩ Ec6= ∅.
Examples Let X = R and E = Z, or 1
n | n ∈ N
, or Q, or (0, 1), or [0, 1), or [0, 1]. Find
∂E.
Remarks
(a) If p ∈ E then p is either an interior point or a boundary point of E, i.e. E ⊆ Eo∪ ∂E.
(b) Consider E ⊆ X where X is metric space is openif every point of E is an interior point.
(c) An aternative definition for the closedness is: a set E ⊆ X is closed if it contains all of its limit points.
Example Let X = R and E = {x ∈ R | a < x ≤ b}, i.e. the segment (a, b]. The interior points of E are (a, b). To see this, let r = min (d(c, a), d(c, b). Then Nr(c) ⊆ E. The limit points of E are [a, b]. Note E is not open because b is not an interior point. Similarly, it is not closed because a is a limit point but not in E.
Theorem A set E ⊆ A, is closed if and only if A \ E is open.
Proof
(⇒) Suppose E is closed. So E contains all of its limit points. Our goal is to show that Ec = A\E is open, i.e. all points of Ec are interior points. Let p ∈ A \ E. Therefore p is not a limit point of E. Thus there exists a neighborhood U of p such that U ⊆ A \ E. Thus p is an interior point of A \ E. Therefore A \ E is open.
(⇐) Suppose A \ B is open. Let p ∈ A \ E. We must show p is not a limit point of E.
But p is an interior point of A \ E. So there exists an open neighborhood U of p such that U ⊆ A \ E. Thus p is not a limit point of E.
Corollary A set E ⊆ X is open if and only if Ec is closed.
Theorem If p is a limit point of E ⊆ X, then every neighborhood of p contains infinitely many points of E.
Corollary A finite set has no limit points.
Proof (Constructive) Let r1 > 0. Then there exists q1 6= p, q1 ∈ E, such that q1 ∈ B(p, r1). Let r2 = d(p, q1)/2. Then there exists q2 6= p, q2 ∈ E, q2 ∈ B(p, r2). Given ri−1 > 0 there exists qi 6= p, such that qi ∈ E, qi ∈ B(p, ri) where ri = d(p,q2i−1). Note that {qi} ⊆ B(p, ri).
Note The axiom of choice, which is independent of the other axioms of set theory says if you have an infinite collection of sets, it’s possible to assume only element of each if they’re nonempty.
Proof (Alternate to Theorem 2) Let r > 0, B(p, r). Suppose q1, . . . , qn ∈ B(p, r) where qi 6= p.
B(p, r) ∩ E = {q1, . . . , qn}. Let r0 < min{d(p, qi) | i = 1, . . . , n}. Then qi 6∈ B(p, r0). But p is a limit point, so there exists q ∈ B(p, r0) such that q ∈ E, q 6= p. But q ∈ B(p, r). Notice that q 6= qi, since qi 6∈ B(p, r0). Thus B(p, r) ∩ E 6= {q1, . . . , qn}.
Theorem Let {Uα} be a collection of open sets in X. Then it follows that [
α
Uα is open in X.
Proof Let U = [
α
Uα. Let p ∈ U . Then there exists an α such that p ∈ Uα. But Uα is open.
Thus p is an interior point of Uα. Thus there exists B(p, r) ⊆ Uα. Hence B(p, r) ⊆ U . Thus p is an interior point of U .
Theorem Let U1, . . . , Un be open in X. Then
n
\
i=1
Ui
is open.
Remark For intuition, let Un= (−1 n,1
n). Then it follows that
∞
[
i=1
Ui = (−1, 1),
∞
\
i=1
Ui = {0}.
Proof Let p ∈
n
\
i=1
Ui. Then p ∈ Ui for all i. But Ui is open for all i. Thus p is an interior point of Ui. Hence, there exists ri > 0 such that B(p, ri) ⊆ Ui. Let r = min{ri | i = 1, . . . , n}. Then B(p, r) ⊆
n
\
i=1
Ui. Hence
n
\
i=1
Ui is open.
Corollary Let {Fα} be closed sets in X. Then
\
α
Fα is closed. If F1, . . . , Fn are closed, then Sn
i=1Fi is closed.
Proof {Fα} closed. Then X \ Fα is open for all X.
\
α
=?
Inspect the complement
X \ (\
α
Fα) =[
α
(X \ Fα)
is open. To determine if
n
[
i=1
Fi is closed, we inspect its compliment.
X \ (
n
\
i=n
Fi) =
n
\
i=1
(X \ Fi) is open.
Theorem E ⊆ X. ¯E is closed.
Proof We will show X \ ¯E is open. Let x ∈ X \ ¯E. We must show x is in an interior point of X \ ¯E. Since x ∈ X \ ¯E, we know x is not a limit point of E. Thus there exists an open neighborhood U of x such that U is completely contained in X \ E. Also note that U contains no limit points of E. Indeed, let p ∈ E0. Suppose p ∈ U . Thus there exists a point q 6= p, q ∈ U, q ∈ E. However, we know U ⊆ X \ E. Thus there does not exist a point p ∈ E0∩ U . Thus U ⊆ X \ E ∩ X \ E0 = X \ (E ∩ E0) = X \ ¯E. Thus x is an interior point of X \ ¯E This holds for all x ∈ X \ ¯E. Thus X \ ¯E is open. Therefore ¯E is closed.
Theorem E = ¯E ⇐⇒ E is closed.
Proof
(⇒) Done, by previous theorem.
(⇐). Suppose E is closed. Therefore it contains all of its limit points, i.e. E0 ⊆ E. Thus E = E ∪ ¯¯ E = E.
Theorem Let E ⊆ F ⊆ X. If F is closed, then ¯E ⊆ F .
Proof Let p ∈ E0. Let U be an open neighborhood of p. There exists q 6= p such that q ∈ U and q ∈ E. Since E ⊆ F , we know q ∈ F . Therefore p is a limit point of F . Since F is closed, p ∈ F . Thus E0 ⊆ F . Therefore ¯E = E ∪ E0 ⊆ F .
Remark
E =¯ \
F = ¯F , E⊆F ⊆X
F.
Question Is (a, b) open? Consider this interval in R. Consider any c ∈ (a, b). Indeed we can find a neighborhood of c completely contained in (a, b). Now consider the interval in R2. In this case it does not. Thus the question is ambiguous.
Definition Let Y ⊆ X be metric spaces. A subset E ⊆ Y is open relative to Y if and only if there is an open set U ⊆ X such that:
E = U ∩ Y.
Example Let Y = (a, b] and X = R. Then (c, b] is open in Y for any c ∈ (a, b).
Compact Sets
Definition An open cover of E ⊆ X is a collection {Uα} of open sets of X (Uα is open in X) such that E ⊆[
α
Uα.
Example Consider the following examples:
• Let E = X = R. Consider Un = (n − 1, n + 1).
• Let E = Z, X = R. Consider Un= (n − 12, n + 12).
• Let E = [0, 1], X = R. Consider Un = (−2n ,2n).
• Let E = (0, 1), X = R. Consider Un = (0,n1), Vn= (1n, 1). Consider then {Un} ∪ {Vn}.
• Let E = {1, 3, 4}, X = R. Let Un = (n −12, n + 12).
Definition A set K ⊆ X is compact if and only if every open cover {Uα} contains a finite sub cover. (K ⊆ Uα1 ∪ · · · ∪ Uαk)
Theorems Let X be a metric space.
(a) ¯E is closed.
(b) E = ¯E ⇔ E is closed, i.e. Ec= X \ E is open.
(c) ¯E = \
E⊂F, F is closed in X
F, i.e. ¯E is the ”smallest” closed set that contains E.
(d) E is open iff Ec is closed.
(e) ¯E = E ∪ ∂E, i.e. E ∪ ∂E = E ∪ E0
Remark Let X be a metric space and E be a subset of X.
(a) If E 6= ∅ and suppose that E ⊆ F ⊆ ¯E then ¯F = ¯E.
(b) Suppose Eo 6= ∅, is it true that Eo= ¯E?
Definitions Let X be a metric space and K be a subset of X.
(a) For each α, let Gαbe a subset of X. The colloection {Gα} is calleda cover of K if K ⊆ ∪αGα. (b) For each α, let Gα be a subset of X. The colloection {Gα} is called an open cover of K if
each Gα is open and K ⊆ ∪αGα.
(c) Let {Gα ⊂ X} be a cover of K. The finite subcollection {Gα1, Gα2, . . . , Gαn} ⊂ {Gα} is called a finite subcover of K if K ⊆ ∪ni=1Gαi.
(d) K is compact if every open cover of K contains a finite subcover.
Examples
(a) Let E = X = R. Consider Un= (n − 1, n + 1). We claim {Un} has no finite subcover. To see this, suppose to the contrary that it does. In other words, R ⊆ Ui1∪ Ui2∪ . . . ∪ Uil, without loss of generality i1 < i2 < · · · < il. But i1 − 2 6∈ ∪lj=1Uij. Thus we have a contradiction and we conclude that R
(b) Let E = [a, b], X = R. Since R ⊆ ∪Un, [a, b] ⊆ Un. Consider {n ∈ Z | n < a}. Let c be the largest element. Similarly, there exist d which is the smallest integer larger than b.
Thus [a, b] ⊆ Uc∪ Uc+1∪ . . . ∪ Ud. Thus {Un} has a finite subcover. We will prove [a, b] is compact.
Theorem Suppose K ⊆ Y ⊆ X. Then K is a compact subset of Y if and only if K is a compact subset of X.
Proof
(⇐) Let K ⊆ X be compact. Let {Vα} be an open cover of K in Y . So Vα ⊆ Y is open, i.e., open Uα ⊆ X such that Vα = Uα ∩ Y . Then K ⊆ ∪αUα. So there exists a finite subcover Uα1, Uα2, . . . , Uαn. But K ⊆ Y . Thus K ⊆ (Uα1 ∩ Y ) ∪ . . . ∪ (Uαn ∩ Y ) = Vα1 ∪ . . . ∪ Vαn. Thus {Vα} has a finite subcover.
(⇒) Similar by taking the intersection of the open cover with Y .
Remark ”K is compact” makes sense. In other words compactness is not relative.
Theorem Let K ⊆ X. If K is compact, then K is closed.
Proof We prove X \ K is open. Let x ∈ X \ K. For all q ∈ K, let rq < d(x, q)
2 . Then B(x, rq) ∩ B(q, rq) = ∅. Note that K ⊆ Uq∈KB(q, rq). Furthermore, K is compact so there exist q1, . . . , qn such that K ⊆ B(q1, rq1) ∪ . . . ∪ B(qn, rqn). Let W ⊆ B(q1, rq1) ∪ . . . ∪ B(qn, rqn). Note that V = B(x, rq1) ∩ . . . ∩ B(x, rqn) does not intersect W . Note x ∈ V . Also note K ⊆ W . Thus V ⊆ X \ K. So x is an interior point of X \ K. So X \ K is open. Thus K is closed.
Theorem Let X be compact, E ⊆ X. If E is closed then E is compact.
Proof Let {Uα} ⊆ X be an open cover of E in X. Note that X \E is open. Then ∪αUα∪{X \E}
covers X. Thus there exists a finite subcover since X is compact of the form Uα1, Uα2, . . . , Uαn, (X\
E). Thus E ⊆ Uα1 ∪ . . . ∪ Uαn. So {Uα} has a finite subcover.
Corollary If F is closed and K is compact then F ∩ K is compact.
Proof F is closed and K is closed and therefore F ∩ K is closed. Therefore F ∩ K is compact by previous theorem.
Theorems Let (X, d) be a metric space and let Y be a subspace of X, i.e. d|Y ×Y is a metric.
(a) A subset E of Y is open relative to Y ⇔ E = Y ∩ G for some open subset G of X.
(b) A subset E of Y is closed relative to Y ⇔ E = Y ∩ F for some closed subset F of X.
(c) A subset K of Y is compact relative to Y ⇔ E is compact relative to X.
Theorems Let X be a metric space and K be a compact subset of X. Then (a) K is closed, i.e. Kc= X \ K is open.
Hint: For each p ∈ Kc, consider
{N1/n(p)c| n ∈ N} = {X \ N1/n(p) | n ∈ N} = {{q ∈ X | d(q, p) > 1/n} | n ∈ N} . (b) K is bounded, i.e. K ⊂ Nr(p) ⊂ X for some r > 0 and p ∈ X.
Hint: For each p ∈ X, consider
{Nn(p) | n ∈ N} = {{q ∈ X | d(q, p) < n} | n ∈ N} .
