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 infiniteif 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.
Let A = Z = {0, 1, −1, 2, −2, 3, −3, . . .} and N = {1, 2, 3, 4, 5, 6, 7, . . .}.
Define f : N → A by
f (n) =
n
2 if n is even
−n − 1
2 if n is odd.
Then f is an 1 − 1 correspondence and Z ∼ N.
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,
n1 = min{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:
nk = min{n ∈ N | n > nk−1, an ∈ E}
Now we have inductively labeled all the elements of E. This gives a function g : N → E defined by
g(k) = ank.
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 ∩ 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 ̸= 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 ̸= xi∀i.
Let a1 ̸= x11, a1 ∈ {0, 1}. Let a2 ̸= x22. . . Let ai ̸= xii. Then let x = (a1, a2, a3, . . . ) ∈ {0, 1}∞. Observe that for any i, x ̸= xi, since ai ̸= xii, by construction. We get that the set is not countable.
(c) The interval (0, 1) = {x ∈ R | 0 < x < 1} is uncountable.
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) = ∥⃗x − ⃗y∥.
R2, ds(⃗x, ⃗y) =
2
X
i=1
|xi− yi|.
Non-Example On Rn, consider the following metrics that don’t work:
d(⃗x, ⃗y) = |x1− y1|
d(p, q) =
n
X
i=1
pilogpi
qi
d(⃗x, ⃗y) = 0
Definition Let x ∈ X, r ∈ R > 0 and let Br(x) = B(x, r) = {y ∈ X | d(x, y) < r} = Nr(p) denote the (open) ball centered at x of radius r and let B(x, r) = {y ∈ X | d(x, y) ≤ r} denote the closed ball centered at x of radius r.
Definitions Let (X, d) be a metric space.
(a) A point p ∈ X is called an interior point of E if there is a r > 0 such that Br(p) ⊂ E. The set of interior points of E is denoted by Eo.
Thus p ∈ Eo ⇐⇒ ∃ r > 0 such that Br(p) ⊂ E.
Examples Let X = R. Find Eo when E = Z, or 1
n | n ∈ N
, or Q, or (0, 1), or [0, 1), or [0, 1].
(b) E is open(in X) if every point of E is an interior point of E, i.e. E = Eo (c) A neighborhood of p ∈ X is an open set containing p.
(d) A point p ∈ X is called a limit point of E ⊂ X if every neighborhood N of p contains a point q ̸= p such that q ∈ E, i.e. N ∩ E \ {p} ̸= ∅. The set of limit points of E is denoted by E′.
Equivalently, p ∈ E′ ⇐⇒ ∀r > 0, Br(p) ∩ E \ {p} ̸= ∅.
Examples Let X = R. Find E′ when E = Z, or 1
n | n ∈ N
, or Q, or (0, 1), or [0, 1), or [0, 1].
(e) E is closed(in X) if every limit point of E is a point of E, i.e. E′ ⊂ E.
(f) The closure of E is the set ¯E = E ∪ E′.
(g) A point p ∈ X is calleda boundary point of E ⊂ X if every neighborhood N of p intersects both E and Ec = X \ E, i.e. N ∩ E ̸= ∅ and N ∩ Ec̸= ∅. The set of boundary points of E is denoted by ∂E.
Equivalently, p ∈ ∂E ⇐⇒ ∀r > 0, Br(p) ∩ E ̸= ∅ and Br(p) ∩ Ec̸= ∅.
Examples Let X = R. Find ∂E when E = Z, or 1
n | n ∈ N
, or Q, or (0, 1), or [0, 1), or [0, 1].
Remarks
(a) If p ∈ E then p is either an interior point or a boundary point of E, i.e. E ⊆ Eo∪ ∂E.
(b) A set E ⊆ X is open if and only if Ec = X \ E is closed.
Proof
(⇒) Suppose E is open.
For each p ∈ (Ec)′, since p is a limit point of Ec, we have
Br(p) ∩ Ec\ {p} ̸= ∅ for all r > 0 =⇒ Br(p) ∩ Ec̸= ∅ for all r > 0.
This implies that
Br(p) ̸⊂ E for all r > 0 =⇒ p /∈ Eo = E since E is open =⇒ p ∈ Ec. Hence we have proved that (Ec)′ ⊂ Ec and Ec is closed.
(⇐) Suppose Ec is closed.
∀ p ∈ E =⇒ p /∈ Ec
=⇒ p is not a limit point of Ec since Ec is closed
=⇒ ∃ r > 0 such that Br(p) ∩ Ec\ {p} = ∅
=⇒ Br(p) ∩ Ec= ∅
=⇒ Br(p) ⊂ E since Br(p) ⊂ X = E ∪ Ec
=⇒ p ∈ Eo
=⇒ E ⊂ Eo and E is open.
(c) A set E ⊆ X is closed if and only if Ec is open.
(d) If p is a limit point of E ⊆ X = (X, d), then every neighborhood of p contains infinitely many points of E.
Proof Suppose that there exists a neighborhood N of p such that
N ∩ E \ {p} = {q1, . . . , qn} = a set of finite number of points.
Choose r > 0 such that r ≤ min
1≤m≤nd(p, qm) and Br(p) ⊂ N. Then we have Br(p)∩E \{p} = ∅ which contradicts to that p is limit point of E. Hence, every neighborhood of p must contain infinitely many points of E.
(e) A finite set has no limit points.
(f) Example: Let X = R and E = {x ∈ R | a < x ≤ b} = (a, b].
The set of interior points of E is Eo = (a, b), since
for each c ∈ (a, b), by letting r = min{d(c, a), d(c, b)} we have Br(c) ⊆ E.
The set of boundary points of E is ∂E = {a, b}.
The set of limit points of E is E′ = [a, b].
Also note that E is neither open nor closed since b ∈ E \ Eo, and since a ∈ E′\ E.
Theorems
(a) E ⊆ X. ¯E is closed.
Proof For each p ∈ X \ ¯E = X \ (E ∪ E′), since p /∈ E′ and p /∈ E, there exists an open neighborhood N of p such that
N ∩ E = N ∩ E \ {p} = ∅ =⇒ N ⊆ Ec.
Suppose that q ∈ N ∩ E′ ̸= ∅. Then N is a neighborhood of q such that
N ∩ E \ {q} ⊂ N ∩ E = ∅ =⇒ N ∩ E \ {q} = ∅ a contradiction to q ∈ E′. Hence,
N ∩ E′ = ∅ =⇒ N ⊆ E′c
.
Thus we have
N ⊆ Ec∩ E′c
= E ∪ E′c
= X \ ¯E
and p is an interior point of X \ ¯E. Since p is an arbitrary point of X \ ¯E, X \ ¯E is open and ¯E is closed.
(b) E = ¯E ⇐⇒ E is closed.
Proof
(⇒) Done, by proceding theorem (a).
(⇐). Suppose E is closed. Therefore it contains all of its limit points, i.e. E′ ⊆ E. Thus E = E ∪ E¯ ′ = E.
(c) Let E ⊆ F ⊆ X. If F is closed, then ¯E ⊆ F .
Proof Let p ∈ E′. Let U be an open neighborhood of p. There exists q ̸= 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 E′ ⊆ F . Therefore ¯E = E ∪ E′ ⊆ F .
Remark Since E ⊂ ¯E and ¯E is closed,
E =¯ \
F = ¯F , E⊆F ⊆X
F.
(d) ¯E = E ∪ ∂E, i.e. E ∪ ∂E = E ∪ E′
Remarks Let X be a metric space and E be a subset of X.
(a) If E ⊆ F ⊆ ¯E then ¯F = ¯E.
(b) Since Eo ⊆ E, we have Eo ⊆ ¯E.
Compact Sets
Theorems
(a) 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.
For each p ∈ U, there exists an α such that p ∈ Uα. Since Uα is open, p is an interior point of Uα. Thus there exists r > 0 such that B(p, r) ⊆ Uα. Hence B(p, r) ⊆ U .
This shows that p is an interior point of U . (b) Let U1, . . . , Un be open in X. Then
n
\
i=1
Ui is open.
Proof For each p ∈
n
\
i=1
Ui, since p ∈ Ui and Ui is open for all 1 ≤ i ≤ n, p is an interior point of Ui for all 1 ≤ i ≤ n.
Hence, there exists ri > 0 such that B(p, ri) ⊆ Ui for each 1 ≤ i ≤ n.
By letting r = min{ri | i = 1, . . . , n}, we have B(p, r) ⊆
n
\
i=1
Ui.
This showss that
n
\
i=1
Ui is open.
(c) Let {Fα} be closed sets in X. Then
\
α
Fα is closed.
If F1, . . . , Fn are closed, then
n
[
i=1
Fi is closed.
Proof Theorem follows by observing that X \ (\
α
Fα) =[
α
(X \ Fα), and
X \ (
n
[
i=n
Fi) =
n
\
i=1
(X \ Fi).
Example For each n ∈ N, let Un= (−1 n,1
n). Note that
∞
[
i=1
Ui = (−1, 1) obtain an open set from an infite union of open sets,
∞
\
i=1
Ui = {0} obtain a closed set from an infite intersecion of open sets.
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 ⊆
n
[
i=1
Gαi.
(d) K is compact if and only if every open cover {Gα} of K contains a finite subcover, i.e.
∃ {Gα1, . . . , Gαk} ⊂ {Gα} such that K ⊆
k
[
i=1
Gαi.
Examples
Let E = X = R and {Un = (n − 1, n + 1) | n ∈ Z}.
Let E = Z, X = R and {Un= (n − 12, n + 12) | n ∈ Z}.
Let E = [0, 1], X = R and {Un = (−2n ,2n) | n ∈ N}.
Let E = (0, 1), X = R and {Un∪ Vn| Un = (0,n+11 ), Vn = (n+11 , 1) for all n ∈ N}.
Let E = {1, 3, 4}, X = R and {Un= (n − 12, n + 12) | n ∈ N}.
Example Let E = X = R. Note that {Un = (n − 1, n + 1) | n ∈ N} is an open cover of R and any finite subcover {Uij | 1 ≤ i1 < i2 < · · · < il} of {Un} cannot cover R since i1− 2 ̸∈
l
[
j=1
Uij. Thus R is not compact.
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[
α
Uα∪{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.
Theorem If {Kα} is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of {Kα} is nonempty, then
\
α
Kα ̸= ∅.
Corollary If {Kn} is a sequence of nonempty compact sets such that Kn ⊂ Kn+1for each n ∈ N,
then ∞
\
n
Kn̸= ∅.
Theorem If E is an infinite subset of a compact set K, then E has a limit point in K.
Theorem (Nested Interval Property) Let {In = [an, bn] | an < bn, n ∈ N} be a sequence of intervals in R such that In⊇ In+1 for all n ∈ N. Then
∞
\
n=1
In ̸= ∅.
Proof Since In ⊇ In+1 for all n ∈ N, we have
an ≤ an+1≤ bn+1 ≤ an ∀ n ∈ N ⇐⇒ a1 ≤ a2 ≤ · · · ≤ an≤ an+1 ≤ · · · ≤ bn+1 ≤ bn ≤ · · · ≤ b2 ≤ b1. This implies that each bn is an upper bound of the set {am | m ∈ N} and x = sup{am | m ∈ N}
exists in R. Hence we have
an ≤ x ≤ bn ∀ n ∈ N ⇐⇒ x ∈ In ∀ n ∈ N and
x ∈
∞
\
n=1
In ̸= ∅.
Theorem (Nested Cells Property) Let {In | n ∈ N} be a sequence of k-cells in Rk such that In⊇ In+1 for all n ∈ N. Then
∞
\
n=1
In ̸= ∅.
Remark For each n ∈ N, let In = [an,1, bn,1] × [an,2, bn,2] × · · · × [an,k, bn,k]. Following the proof of the proceding Theorem to find
an,j ≤ xj ≤ bn,j ∀ n ∈ N and ∀ 1 ≤ j ≤ k ⇐⇒ x = (x1, x2, . . . , , xk) ∈ In ∀ n ∈ N.
Hence
x = (x1, x2, . . . , , xk) ∈
∞
\
n=1
In ̸= ∅.
Theorem Every k-cell I = [a1, b1] × [a2, b2] × · · · × [ak, bk] is compact.
Proof Let
δ =
k
X
j=1
(bj − aj)1/2
= diameter of I = diam (I).
Suppose {Gα} is an open cover of I which contains no finite subcover of I.
Bisecting sides of I into 2k k-cells {Qi}2i=1k with equal sizes such that
I =
2k
[
i=1
Qi.
Let I1 ∈ {Qi}2i=1k such that I1 cannot be covered by any finite subcollection of {Gα}.
Next subdivide I1 and continue the process to obtain a sequence of k-cells such that I ⊃ I1 ⊃ I2 ⊃ · · · ; diam (In) = 1
2diam (In−1) = · · · = 1
2ndiam (I) = δ
2n ∀ n ∈ N (1) In cannot be covered by any finite subcollection of {Gα} (2)
Let
{x} =
∞
\
n=1
In=⇒ x ∈ I ⊂[
α
Gα =⇒ x ∈ Gα1 for some Gα1 ∈ {Gα}.
Since Gα1 is open, ∃ r > 0 such that
x ∈ Br(x) ⊂ Gα1. Since x ∈ In and diam (In) = δ
2n =⇒ ∀ y ∈ In we have d(x, y) ≤ diam (In) = δ 2n < r, by choosing n sufficiently large such that δ
2n < r, we have In⊂ Br(x) ⊂ Gα1. This contradicts to the choice of In. Hence,
∃ {Gα1, . . . , Gαk} ⊂ {Gα} such that I ⊆
k
[
i=1
Gαi.
This implies that I is compact.
Definition The set K ⊆ X is boundedif there exists r > 0 and q ∈ X, such that for all p ∈ K, d(p, q) < r.
Theorem Let E be a subset of the Euclidean metric space Rk. Then the following statements are equivalent.
(a) E is closed and bounded.
(b) E is compact.
(c) Every infinite subset of E has a limit point in E.
Proof of (b) ⇐⇒ (c)
(⇒) If E is compact and suppose that S is an infinite subset of E such that S′∩ E = ∅.
This implies that for each q ∈ E, there exists a neighborhood Vq of q such that
Vq∩ S \ {q} = ∅ =⇒ Vq∩ S =
({q} if q ∈ S
∅ if q /∈ S Since
S ⊂ E ⊂ [
q∈E
Vq,
{Vq | q ∈ E} is an open cover of S. Also since |S| = ∞ and each Vq contains at most one point of S, {Vq | q ∈ E} does not contain a finite subcover. This is a contradiction since E is compact and {Vq | q ∈ E} is an open cover of E.
(⇐) Suppose that every infinite subset S of E has a limit point in E.
Claim : E is closed.
Suppose that E is not closed. Then
∃ z ∈ E′\ E.
Thus
∃ xn ∈ E ∩ B1/n(z) \ {z} ̸= ∅ ∀ n ∈ N.
Let S = {xn| n ∈ N}. Since
xn∈ B1/n(z) ∀ n ∈ N ⇐⇒ |xn− z| < 1
n ∀ n ∈ N, S is an infinite subset of E and z is a limit point of S.
Next, note for each y ∈ E, since z /∈ E and |z − y| > 0,
∃ m = m(y) ∈ N such that if n ≥ m then 1
n < |z − y|
2 . Since
|xn− y| ≥ |z − y| − |xn− z| ≥ |z − y| − 1 n ≥ 1
2|z − y|, this imples that
xn∈ B/ |z−y|/2(y) ∀ n ≥ m ⇐⇒ B|z−y|/2(y) ∩ S = ∅ ∀ n ≥ m =⇒ y /∈ S′.
Thus z is the only limit point of S and hence z ∈ E, a contradiction to z ∈ E′ \ E. This proves that E is closed.
Claim : E is bounded.
Suppose that E is not bounded. Then
∀ n ∈ N, ∃ yn ∈ E such that |yn| > n.
Let
S = {yn | n ∈ N}.
Then S is an infinite subset of E that has no limit point in E or in Rk. Hence E must be bounded.
Remarks
(a) The equivalence of (a) ⇐⇒ (b) is known as the
Heine-Borel Theorem A subset of Rk is compact if and only if it is closed and bounded.
(b) (b) =⇒ (a) implies that
Corollary Let K ⊆ R. If K is compact, then sup K exists and sup K ∈ K.
Proof K is bound, hence sup K ∈ R. K is closed, hence K′ ⊆ K. Furthermore, observe that sup K ∈ K′ because it is a limit point.
(c) (b) =⇒ (c) implies that
Bolzano-Weierstrass Theorem Every bounded infinite subset of Rk has a limit point in Rk.
Proof If E ⊆ Rk is infinite and bounded, then E is a subset of a k-cell in Rk. Thus E is an infinite subset of a compact set, and E is bounded. Thus E has a limit point in Rk and specifically in the k-cell.
(d) (b) ⇐⇒ (c) holds in any metric spaces. In fact, we have
Theorem A metric space X is compact if and only if every infinite subset E of X has a limit point in X (note that it is not necessarily in E).
(e) (b) =⇒ (a) holds in any metric spaces. In fact, we have
Theorems Let X be a metric space and K be a compact subset of X. Then (i) K is closed, i.e. Kc= X \ K is open.
Hint: For each p ∈ Kc, note that
{B1/n(p)c| n ∈ N} = {X \ B1/n(p) | n ∈ N} = {{q ∈ X | d(q, p) > 1/n} | n ∈ N}
is an open cover of K.
(ii) K is bounded, i.e. K ⊂ Br(p) ⊂ X for some r > 0 and p ∈ X.
Hint: For each p ∈ X, note that
{Bn(p) | n ∈ N} = {{q ∈ X | d(q, p) < n} | n ∈ N} is an open cover of Rk or K.
Definition Let Y ⊆ X be metric spaces. A subset E ⊆ Y is open relative to Y if and only if
∃ an open set U ⊆ X such that E = U ∩Y ⇐⇒ ∀ p ∈ E, ∃ r = r(p) > 0 such that Br(p)∩Y ⊂ E.
Example Let Y = (a, b] and X = R. Then (c, b] is open in Y for any c ∈ (a, b).
Example Note that (a, b) is open in R and (a, b) is closed in R2.
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 ⇔ K is compact relative to 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.
Example Let E = {p ∈ Q | 2 < p2 < 3}. Is E closed in R? Is it bounded?
Example Let X = (Q, d) be a metric space with d(p, q) = |p − q|, and let K = {p ∈ Q | 2 < p2 < 3}.
(a) Show that X is not complete.
(b) Show that K is closed and bounded in Q, but K is not compact.
(c) Show that K is open.
Example Let A be any set. Suppose A is infinite. Define
d(p, q) =
(0 if p = q, 1 otherwise.
Is A closed? Is A bounded. Yes on both accounts. Notice that B(p,12) = {p}. Thus A = [
p∈A
B(p,12).
Notice that this cannot be reduced to a finite subcover. Hence, A is not compact.
Definition Let X be a metric space. A subset E is said to beperfect if E is closed and if every point of E is a limit point of E, i.e. E is perfect if E = E′.
Theorem Let P be a nonempty perfect set in Rk. Then P is uncountable.
Proof Since P has limit points, P must be infinite.
Suppose P is countable and let
P = {xn| n ∈ N}.
We shall construct a sequence {Vn} of neighborhoods as follows.
Let V1 = Br(x1) ⊂ Rk be any neighborhood of x1 ∈ P = P′. Since x1 ∈ P′, we have V1∩ P′ = V1∩ P ̸= ∅ and note that V1∩ P′ contains infinite points of P.
For each n ∈ N, suppose Vn has been constructed such that Vn∩ P′ = Vn∩ P ̸= ∅. Since every point of P is a limit point of P, there exists a neighborhood Vn+1 such that
(1) Vn+1⊂ Vn (2) xn∈ V/ n+1
(3) Vn+1∩ P ̸= ∅
By (3), Vn+1 satisfies our induction hypothesis, and the construction can proceed.
Put
Kn= Vn+1∩ P.
Since Vn is closed and bounded, Vn is compact. Since xn ∈ V/ n+1, no point of P lies in
∞
\
n=1
Kn. Since Kn⊂ P, this implies that
∞
\
n=1
Kn= ∅.
But each Kn is nonempty, by (3), and Kn ⊃ Kn+1, by (1); this contradicts to Theorem 2.36.
Example (The Cantor Set) Let K0 be [0, 1]. Let K1 be K0 minus the middle third. Let K2 be K1 minus the middle third of the two interval of K1, etc. The cantor set C is
C =
∞
\
i=1
Ki.
Properties of the Cantor Set:
C is nonempty since it is an intersection of nested closed intervals which we showed to be nonempty.
C is closed because arbitrary intersections of closed set is closed.
C is perfect: closed and every point of C is a limit point of C.
C is bijective with {0, 1}∞ and therefore C is uncountable.
C has no interior.
C is totally disconnected.
Connected Sets
Definitions Let X be a metric space.
(a) Let A, B ⊆ X, then A and B are separated if A ∩ ¯B = ∅ and ¯A ∩ B = ∅.
(b) E ⊆ X is connected if it is not the union of two nonempty separated sets. If E is not connected, it is called disconnected and A and B are a separation of E, i.e. E is disconnected if and only if
∃ A, B ⊆ E such that A ∩ ¯B = ∅, ¯A ∩ B = ∅ and A ∪ B = E.
⇐⇒ ∃ A, B ⊂ E relative open subsets to E such that A ∩ B = ∅ and A ∪ B = E.
⇐⇒ ∃ A ⊂ E such that A ̸= ∅, A ̸= E and A is both relative open and closed to E.
Examples
(a) The set N ⊂ R is disconnected, since we can take A = {x ∈ R|x < 3/2} and B = {x ∈ R|x > 3/2}.
(b) The set H = {1/n | n ∈ N} is disconnected.
(c) The set E = {(x, y) ∈ R2|x, y ∈ Q} ⊂ R2 is disconnected, since we can take A = {(x, y) ∈ E|x <√
2} and B = {(x, y) ∈ E|x >√ 2}.
Theorem A subset E ⊂ R is connected iff it has the following property: If x, y ∈ E and x < z < y, then z ∈ E, i.e. E ⊂ R is connected iff E is an interval.
Proof
(=⇒) If there exist x ∈ E, y ∈ E, and some z ∈ (x, y) such that z /∈ E, then E = Az∪ Bz where Az = E ∩ (−∞, z) ⊂ E, Bz = E ∩ (z, ∞) ⊂ E.
Since
x ∈ Az and y ∈ Bz =⇒ Az ̸= ∅ and Bz ̸= ∅.
Also since Az ⊂ (−∞, z) and Bz ⊂ (z, ∞), we have
Az∩ ¯Bz ⊂ (−∞, z) ∩ [z, ∞) = ∅ and ¯Az∩ Bz ⊂ (−∞, z] ∩ (z, ∞) = ∅.
Hence Az and Bz are a separation of E and E is not connected.
(⇐=) Suppose E is not connected. Then there are nonempty separated sets A and B such that
A ∪ B = E.
Pick x ∈ A, y ∈ B and assume (without loss of genrality) that x < y. Define z = sup A ∩ [x, y] =⇒x ≤ z ≤ y.
Since
z ∈ ¯A and ¯A ∩ B = ∅ =⇒z /∈ B =⇒ z < y since y ∈ B.
In particular, we have
x ≤ z < y and either z /∈ A or z ∈ A.
Now
if z /∈ A =⇒ x < z < y and z /∈ A ∪ B = E, if z ∈ A =⇒ z /∈ ¯B, since ¯Bc is open and y ∈ B
=⇒ ∃ z1 ∈ ¯/ B such that z < z1 < y
=⇒ z1 ∈ B, z/ 1 ∈ A since z = sup A ∩ [x, y]/
=⇒ x < z1 < y and z1 ∈ A ∪ B = E,/ E is not an interval in either case.
Remarks
(a) Rudin proves that E ⊆ R is connected if and only if it has the “interval property”.
E ⊆ R has the interval property if ∀ x < y ∈ E and ∀ z satisfying x < z < y we have z ∈ E.
(b) Theorem [a, b] ⊆ R is connected.
Proof If not, there exists a separation A, B of [a, b] with a ∈ A.
Since A ⊂ R is nonempty bounded above by b,
s = sup A exists, s ≤ b and s ∈ A′ ⊂ ¯A =⇒ s ̸∈ B since ¯A ∩ B = ∅.
Thus we have
s ∈ A since A ⊂ A ∪ B = [a, b] =⇒ ¯A ⊂ [a, b] = [a, b].
Furthermore, since A ∩ ¯B = ∅, we have s /∈ ¯B which implies that s is an interior point in [a, b] \ ¯B. Hence there exists ϵ > 0 such that B(s, ϵ) ⊆ [a, b] \ B = A, i.e. (s − ϵ, s + ϵ) ⊆ A which contradicts to that s = sup A since s < s + ϵ/2 ∈ A.
(c) Corollary (a, b), [a, b), (a, b] ⊆ R are connected.
(d) Theorem Rk is connected.
Proof If not, there exists a separation A, B of Rk. Let x ∈ A and y ∈ B and consider the line segment S joining a and y, i.e.
S = {x + t(y − x) | t ∈ [0, 1]}.
Let A1 = {t ∈ R | x + t(y − x) ∈ A} and let B1 = {t ∈ R | x + t(y − x) ∈ B}. It is easily seen that A1 and B1 are disjoint nonempty open subsets of R and provide a disconnection for [0, 1], contradiction to the preceding theorems.
(e) Corollary The only subsets of Rk which are both open and closed are ∅ and Rk.
