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1 Set Theory and Functions

1.1 Basic Denitions and Notation

A set A is a collection of objects of any kind.

We write a 2 A to indicate that a is an element ofA: We express this as a is containedin A.

We write A B if every element of A also is an element in B . We express this as Ais a subset of B

We write A = B ifA and B consist of exactly the same elements. Otherwise we write

A 6= B:

If A B and A 6= B we say that A is a proper subset of B:

We write ? for the empty set, the set that contains no elements at all.

1.2 Operations on Sets

1. The union: Let A and B be two sets. Then, the union, denoted A [ B is the set ofall elements belonging to at least one of the two sets. That is,

A [ B = fxjx 2 A or x 2 Bg :

Given an arbitrary collection of sets Ai; where i is some parameter such that i 2 I(allows for nite or innite numbers) we write [iAi for the union, which is the set ofall elements that belongs to at least one of the sets in the collection,

[iAi = fxj x 2 Ai for some i 2 I g :

2. The Intersection: Given A and B we write A \ B for its intersection- the set of allelements belonging to both A and B:

A \ B = fxjx 2 A and x 2 Bg :

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The generalization for a collection of sets Ai is obviously

\iAi = fxj x 2 Ai for every i 2 I g

3. Disjoint Sets: A and B are said to be disjoint if they have no elements in common.

That is, if

A \ B = ?:

4. The Complement (sometimes called dierence): We (KF uses dierent notation)

write AnB for the set of all elements in A that dont belong to B;

AnB = fxjx 2 A and x =2 Bg

Remark 1 It follows immediately from the denitions that [ and\ satisfy communicativeand associative laws,

A [ B = B [ A

A \ B = B \ A

(A [ B) [ C = A [ (B [ C)

(A \ B) \ C = A \ (B \ C)

Hence, we can simply write A [ B [ C:In addition:

Proposition 1 [ and\ satisfy the distributive law

(A [ B) \ C = (A \ C) [ (B \ C)

(A \ B) [ C = (A [ C) \ (B [ C)Proof. Exercise on Problem Set 1

Sometimes we will also use the Cartesian product:

Denition 1 Given non empty sets A andB; the Cartesian product, denoted A B; is theset of all ordered pairs (a; b) such that a 2 A and b 2 B

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Example 1 If A = fa1; a2; a3; a4g and B = fb1; b2g then

A B = f(a1; b1) ; (a1; b2) ; (a2; b1) ; (a2; b2) ; (a3; b1) ; (a3; b2) ; (a4; b1) ; (a4; b2)g

Example 2 If A = [0; 1] and B = [0; 1] we write A B = [0; 1] [0; 1] or A B = [0; 1]2

;where geometrically the Cartesian product of the two (unit) intervals is the (unit) square.

1.3 Functions

The terminology Ill use is as follows:

Denition 2 A function (or mapping) from a set X to Y (denoted f : X ! Y) is a rule

that assigns a unique element y = f(x) 2 Y to every x 2 X:

Usually, X is referred to as the domain of f:

Remark 2 Sometimes functions are dened in terms of the graph G, which is a collection

of ordered pairs hx; f(x)i 2 X Y:We say that G is the graph of a function if and only iffor every x 2 X there is a unique pair hx; f(x)i 2 G.

We call the element f(x) 2 Y the image of x (under f). This is generalized to subsetsas follows:

Denition 3 If A X we say that

f(A) fy 2 Yjy 2 f(a) for some a 2 Ag

is the (direct) image of A:

Also,

Denition 4 If B Y we say that

f1 (B) = fx 2 Xjf(x) 2 Bg

is the inverse image (or preimage) of B:

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The image and the inverse image can be used to dene some important types of functions:

Denition 5 f : X ! Y is said to be surjective (onto) if f(X) = Y

Denition 6 f : X ! Y is said to be injective (one to one) if f(x1) 6= f(x2) for every pair(x1; x2) 2 X such that x1 6= x2:

Denition 7 f : X ! Y is said to be bijective if it is surjective and injective.

It is more or less immediate that:

Proposition 2 If f is a bijection, then there is a (uniquely dened) function g : Y ! Xdened by letting g (y) 2 X be given by g (y) = f1 (y) for every y 2 Y: This function iscalled the inverse off (since no confusion can arise we will use f1 also to denote the inverse

function in the rest of the course).

In some sense the conclusion is trivial. However, well write down a proof to make sure

we understand what the denitions mean

Proof. Since f is injective there are two possibilities:

1. f1 (y) = ?

2. there exists a unique element x 2 X such that fxg = f1 (y)

To see this, assume that for contradiction that there exists two distinct elements x1 and

x2 such that x1 2 f1 (y) and x2 2 f1 (y) : Then,

fx1; x2g f1 (y) = fx 2 Xjf(x) 2 fygg = fx 2 Xjf(x) = yg

which contradicts the assumption that f is injective. Moreover, since f is surjective

Y = f(X) fy 2 Yjy 2 f(x) for some x 2 Xg ;

which can be rephrased as saying that there exists some x 2 X such that f(x) = y for everyy 2 Y: This rules out f1 (y) = ?; so the result follows

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Denition 8 If f : X ! Y and g : Y ! Z the composition f g is given by the functionh : X ! Y such that h (x) = g (f(x)) for every x 2 X:

Proposition 3 Suppose that f : X ! Y and g : Y ! Z are bijective, then the compositionf g is bijective.

Proof. Fix any z0 2 Z: Since g : Y ! Z is surjective there exists y0 2 Y such that g (y0) = z0:Since f is surjective there exists x0 2 X such that f(x0) = y0: Hence,

g (f(x0)) = g (y0) = z0

Since z0 2 Z was arbitrary g (f(X)) = Z:Suppose that f

g is not injective. Then there exists x1

6= x2 such that g (f(x1)) =

g (f(x2)) : Since f is injective f(x1) = y1 6= y2 = f(x2) : Hence, there exists y1 6= y2 suchthat g (y1) = g (y2) contradicting that g is an injection.

1.4 Sequences I

Just like functions can be dened in terms of ordered pairs (dening the function from its

graph) we may dene an ordered pair as a function

f : f1; 2g ! X:

Similarly, a nite sequence can be dened as a mapping

f : f1;::::;ng ! X:

Now, letting N be the set of natural numbers we can dene: an innite sequence as a mapping

f : N

!X:

It is customary to depart somewhat from standard functional notation when dealing with

sequences and,

1. Write hxiini=1 instead of (f(1) ; f(2) ;:::;f(n)) or hf(i)ini=1

2. Write hxii1i=1 instead ofhf(i)i1i=1

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1.5 Finite, Innite, Countable and Uncountable Sets

1.5.1 Finite vs Innite Sets

A nite set is a set which is either empty or a set for which there exists some natural number

n such that the set has exactly n elements. That is:

Denition 9 The set A is nite if it is empty or if there exists some n 2 N and a bijectionf : f1;:::;ng ! A (this denes a set with exactly n elements too): A set A is innite if it isnot nite.

Remark 3 Since the bijection f : f1;:::;ng ! A denes an inverse f1 : A ! f1;:::;ng

niteness could be dened as a bijection from A onto f1;:::;ng : Both conventions are usedto dene niteness (and similarly for countability).

The remark above also proves part of the following useful claim:

Proposition 4 A set A is nite if and only if there exists a nite set B and a bijection

g : A ! B

Proof. (=)) If A is nite there exists some nite n and a bijection f : f1;:::;ng ! A;implying that g = f1 : A ! f1;:::;ng is well-dened. Letting B = f1;:::;ng proves the rstpart.

((=) Suppose that there exists nite set B and a bijection g : A ! B: Since B is nitethere exists some n 2 N and (using the remark) a bijection h : B ! f1;:::;ng : Since g and hare bijective g h : A ! f1;::::;ng is bijective. Hence, there is a bijection f : f1;:::;ng ! Awhere

f(i) = g h(i) for i = 1;::::;n

We now prove a result sometimes used in combinatorics, which is referred to as the

Pigeonhole principle. If there are more pigeons than pigeonholes at least two pigeons need

to share a hole.

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Proposition 5 Suppose that m; n 2 N: Then:

1. Ifm n there exists an injection (onto map) f : f1;:::;mg ! f1; : : ;ng

2. Ifm > n there is no injection f : f1;:::;mg ! f1;::;ng :

Proof. For part 1, let f : f1;:::;mg ! f1;::;ng be given by

f(j) = j for every j 2 f1;:::;mg f1;::;ng :

Obviously, f is injective

For the part 2, we begin by noting that if n = 1; then the only map from f : f1;:::;mg !

f1;::;ng is f(j) = 1 for every j 2 f1;:::;mg ; which is not an injection for any m > 1.Next, assume that there is no injection f : f1;:::;mg ! f1; : : ;kg for any m > k: We need

to demonstrate that it follows that there is no injection f : f1;:::;mg ! f1;::;k + 1g for anym > k + 1; which will prove the claim by induction.

For contradiction, suppose such an injection would exist. Then, there must be some

j 2 f1;:::;mg such that f(j) = k + 1 since otherwise h : f1;:::;mg ! f1;::;kg is an injectiondespite m > k + 1 > k; which violates the induction hypothesis. Moreover, it must be the

case that exactly one element j 2 f1;:::;mg is such that f(j) = k + 1 since obviously thereis no injection more than one element is mapped into k + 1 : It follows that the function

eh : f1;:::;mg n fjg ! f1;::;kgdened as

eh (j) = h (j) for every j 2 f1;:::;mg n fjgmust be an injection. At this point we may either just assert that f1;:::;mg n fjg andf1;:::;m 1g are equivalent sets or relabel by constructing the map g : f1;:::;m 1g !f1;::;kg

g (j) =

8 j :

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It is obvious that g is injective if and only ifeh is injective, but g is not injective accordingto the induction hypothesis. The result follows

We can extend this result to

Proposition 6 Suppose that m 2 N: Then:

1. There exists an injection (onto map) f : f1;:::;mg ! N

2. There is no injectionf : N ! f1;::;mg

Proof. For part 1, let f : f1;:::;mg ! N be given by

f(j) = j for every j 2 f1;:::;mg N:

Obviously, f is injective

For part 2, if f : N ! f1; : : ;mg is an injection it follows that f(j) 6= f(i) for everypair i; j 2 N; which implies that f(j) 6= f(i) for every pair i; j 2 f1;:::;m + 1g ; whichcontradicts the previous proposition

I will take the following for granted (see Bartle and Sherbert pages 61-62)

1.5.2 Countable vs. Uncountable Sets

Countably innite sets can be dened in similar fashion to the denition of nite sets in

Denition 9. However, it is more convenient to use the following denition:

Denition 10 A set A is said to be countable if there exists an injective function f : A !N:

Bartle and Sherbert use a dierent denition and establish existence of an injection

f : A ! N as a proposition.

Example 3 Let A = N: Consider the identity function f : N ! N given by

f(i) = i for every i 2 N:

Obviously, this is injective, so N is countable.

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Example 4 Let Z = f::: 1; 0; 1;:::g be the set of positive and negative integers. Let f :Z ! N be given by

f(i) =

8 n). Hence, any nite set is countable

Combining with previous results we conclude that:

Proposition 7 LetA be a countable set. Then, A is (countably) innite if and only if there

exists some bijective f : A ! N:

Proof. ()) For contradiction, suppose f is surjective and that A is nite. If f is bijectivef1 : N

!A exists and is injective (and surjective). If A is nite there exists n

2N and

bijection g : A ! f1;:::;ng : Hence f1 g : N ! f1;:::;ng is injective (and surjective),which contradicts Proposition 6. Hence, A cannot be nite, implying that it is innite

(()Suppose that A is nite and that there exists a bijection f : A ! N: Then

f1 : N ! A

is bijective. Moreover, since A is nite there exists a bijection g : A ! f1;:::mg ; where

m 2 N: Hence,f1 g : N ! f1;:::;ng

is bijective, which is impossible due to Proposition 6.

Theorem 1 If A is countable and B A; then B is countable (if A is nite, then B isnite).

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Proof. First, consider the case with A nite. Without loss, suppose B is non-empty (empty

set if nite and therefore countable by default). If B is innite there exists bijection

f : B ! N

Let b0 2 B A and consider map g : A :! B

g (a) =

81

y x;

implying that1

n< y x

Moreover, there exists v 2 N such that (here we use fact that x 0)

v 1 nx < v

since:

1. Set fm 2 Njm nxg is nonempty (otherwise nx upper bound for N)

2. Set fm 2 Njm nxg is bounded below by nx )exists a least element v 2 fm 2 Njm nxg

x 0: Then, the previous argument establishes that

there exists n; v 2 N such that

x < 0 0: Hence, if n K

jxn

1j

= n 1

n 1 =

1n =

1

n 1

K:

We conclude that xn =n1n

converges to 1:

Example 12 Sometimes limit not so easy to guess, as with

xn =

1 1

n

n

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This can be motivated from probability theory: The chance of of k successes in n Bernoulli

trials is

pkjn =n!

k! (n k)!pk (1 p)nk

So, if the probability of success is 1n ; then the probability of no success out of n draws is

n!

1 n!!

1

n

01 1

n

n=

1 1

n

n:

Can show that

1 1n

n ! 1e

:

Proposition 11 If hxni1n=1 converges there is a unique limit.

Proof. Suppose that x and x are limits ofhxn

i1

n=1

and let x < x without loss of

generality. Let " = xx

3 : Because x is a limit ofhxni1n=1 there exists K such that

xn < x + " = x +

x x3

=2

3x +

1

3x

for all n K: Symmetrically, as x is a limit ofhxni1n=1 there exists K such that

xn > x " = x x

x3

=2

3x +

1

3x

Hence,

2

3x +

1

3x < xn x "; which implies that

x xn xK > x "

for every n

K: Hence

jxn xj "

for every n K: Since " was arbitrary, x is the limit ofhxni1n=1 :Decreasing sequences-same argument.

2.3 Subsequences

Denition 17 Let hxni1n=1 be a sequence and fkig1i=1 a strictly monotonic sequence withki 2 N for every i: Then, the sequence

hxknikn = hxk1; xk2;:::::;xkni

(constructed by setting xki = xn for every ki) is called a subsequence of hxni1n=1

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Example 13 Let hxni1n=1 be given by

xn =

8 0 we say that an open ball (" neighborhood) of x 2 R is given

by the setB (x; ") = fy 2 Rj jx yj < "g = (x "; x + ") :

Denition 19 A setA R is open if for every x 2 A there exists an open ball B (x; ") A:

Denition 20 A set A R is closed if the complement B = RnA is open.

Example 14 Let A = R: Fix x and let " = 1: Then any y 2 (x 1; x + 1) 2 R; so R is

open.

Example 15 Consider the open interval A = (a; b) :Pick any x 2 (a; b) and let

" = min fx a; b xg

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then for any y 2 B (x; ")

y > x " = x min fx a; b xg

x + max

fa

x; x

b

g x + a

x = a

y < x + " = x + min fx a; b xg

x + b x = b:

Hence, (a; b) is an open set.

Example 16 Let A = (1; a) and B = (b; 1) : Open, by same argument.

Example 17 Consider [a; b] : Not open since for every " > 0

B (a; ") \ [a; b] = (a "; a]

in nonempty for every " > 0:

Note that

Rn [a; b] = (1; a) [ (b; 1) :

By the previous argument, if x 2 (1; a) we can nd a ball B (x; ") (1; a) and if x is

in (b; 1) we can also nd a ball B (x; ") (b; 1) : Hence, there an open ball B (x; ") (1; a) [ (b; 1) for every x 2 (1; a) [ (b; 1) ; so (1; a) [ (b; 1) : we conclude that [a; b] isa closed set.

Example 18 ? is open (vacuously since there is no point in the set). Hence, R is closed

and open. Since R is also open, it follows that? is open and closed.

Theorem 6 The union of any collection of open sets is open.

Proof. Let I be some indexing set and assume that Ai is open for every i 2 I: If x 2 A =[i2IAi: Then there exists j 2 I and " > 0 such that B (x; ") Aj : Hence,

B (x; ") Aj [i2IAi:

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Theorem 7 The intersection of a nite collection of open sets is open.

Proof. Let x 2 \ki=1Ai: Let "i > 0 be such that B (x; "i) Ai; which exists since x 2

\ki=1Ai

)x

2Ai for every i and each Ai is open. Let

" = min f"1; "2;::;"kg > 0;

and note that

B (x; ") B (x; "i) Ai for every i

implying that

B (x; ") \ki=1Ai:

Example 19 Let

0; 1k

and consider

\ki=1

1k

; 1 +1

k

=

1

k; 1 +

1

k

which is open for every nite k: However

\1i=11k ; 1 + 1k = [0; 1] ;

which is not open.

Theorem 8 The intersection of an arbitrary collection of closed sets is closed.

Proof. Let I be some indexing set and assume that Ai is closed for every i 2 I: By deMorgans laws

Rn (\i2IAi) = [i2IRnAi:Each set RnAi is open, so (by previous theorem) [i2IRnAi is open, proving that Rn (\i2IAi)is open, which by denition means that \i2IAi is closed.

Theorem 9 The union of a nite collection of closed sets is closed.

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Proof. By de Morgans laws

Rn [ki=1Ai = \ki=1RnAi:Each set RnAi is open and k is nite, so (by previous theorem) \

k

i=1RnAi is open, provingthat Rn [ki=1Ai is open, which by denition means that [ki=1Ai is closed.Denition 21 (ALT 1) Given set A R; x 2 R is called a cluster point of A if for every" > 0 there exists y 2 B (x; ") \ An fxg.

Proposition 13 A set A is closed if and only if it contains all its cluster points.

Proof. ()) Suppose not. Then, there exists an cluster point x 2 RnA; which is an openset. Hence, there exists " > 0 such that B (x; ") RnA, implying that B (x; ") \ A = ?;contradicting that x is a cluster point of A:

(() Suppose that A contains all cluster points. Pick y 2 RnA: Then, y is not a clusterpoint, so there exists open ball B (y; ") RnA:True for all y 2 RnA; so RnA is open. Hence,A is closed.

Almost the same result stated in terms of sequences.

Proposition 14 A is closed if and only if the limit of every converging sequence hxni withxn 2 A for all n belongs to A:

Proof. We will prove this from the previous result by showing that

CLAIM: A contains all its cluster points,the limit of every converging sequence hxniwith xn 2 A for all n belongs to A:

()

) Suppose thathxn

iconverges to x:Then, for every " > 0 there exists K such that

xn 2 B (x; ") for n K; and, by hypothesis, xn 2 A: Hence, x is a cluster point of A: Byassumption A contains its cluster points, so limn!1 xn = x

2 A:Suppose that x =2 A is a cluster point. Let "1 > 0 and pick x1 2 B (x; "1) n fxg

arbitrarily. As x1 6= x there exists "2 > 0 such that x1 =2 B (x; "1) ; implying that thereexists x2 2 B (x; "2) B (x; "1) because x is a cluster point: Recursively, if xk 6= x

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and xk 2 B (x; "k) there exists "k+1 > 0 such that xk =2 B (x; "1) : Hence, there existsxk+1 2 B (x; "k+1) B (x; "1) : This denes an innite sequence which obviously convergesto x; so x 2 A if the limit of every converging sequence is in A:

An alternative denition is.

Remark 7 The proof shows that given any set A R; x 2 R is a cluster point of A if forevery " > 0 there exists an innite set X B (x; ") \ A. This is an alternative denitionoften used for a cluster point.

2.5 Compact Sets

Denition 22 An open cover of a set A in R is a collection C = fCig of sets such thatA [i2ICi

Denition 23 C0 is a subcover of C if each set Cj inC 0 also is inC andC 0 is a cover of A:

Denition 24 A inR is said to be compact ifeveryopen coverC has a nite open subcov-eringC 0:

Theorem 10 (Heine Borel) A

R is compact if and only if A is closed and bounded.

Remark 8 Every analysis book says that result is "closely related to Bolzano Weierstrass",

yet proofs tend to not use that result. Here is an argument that makes explicit use of Bolzano

Weierstrass.

Proof. ()) Suppose that A is compact. We note that if Ci = (i; i) then

[1i=1Ci

is an open cover of R: Hence,

A R [11 Ci

so f(i; i) ji 2 Ngis an open cover of A as well. Since it is compact, there exists some niteK such that

A [Ki=1Ci = (K; K) ;

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so A is bounded. Let y =2 A and let

Ci = Rn

y 1i

; y +1

i

Obviously, Gi is open (y 1i; y + 1

i is closed) and we have that [1i=1Ci = R which covers

R and therefore also A: By compactness there is some nite K such that

A [Ki=1Ci = Rn

y 1K

; y +1

K

)

A \

y 1K

; y +1

K

= ?)

A \

y 1K

; y +1

K

= ?

Hence, for every y =2 A , y 2 RnA there exists " > 0 such that

y 1K

; y + 1K

RnA; so

RnA is open. Hence, A is closed.(() For contradiction, suppose A is closed and bounded, but that that there is an open

cover C = fCg for which there is no nite subcover.For any n, consider the collection of open sets

Cn = fC;ng where C;n =

x 2 Cj infy2RnC

jx yj > 1n

CASE 1: Suppose that there exists K such that CK is a subcover ofC. Since A is boundedby there exists m such that A [m; m] so let

x1 = m and B

x1;1

K

=

m 1

K; m + 1

K

x2 = m + 12K

and B

x2;

1

K

=

m 1

2K; m + 3

2K

::::

xn = m + (n 1)2K

and B

xn;

1

K

=

m + (n 3)

2K; m + (n + 1)

2K

::::

x4mK+1 = m + (4mK+ 1 1)2K

= m + 2m = m

Hence,

B

xn;1K

4mK+1n=1

is a nite cover ofA: Moreover, each xn 2 Cn;K by the hypothesisthat CK is a subcover of C. Pick any y 2 B

xn;

1K

: Then,

infx2Rnxn2Cn;K

jx yj infxn2Cn;K

jx xnj jy xnj > 0

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since jy xnj < 1K as y is in B

xn;1K

and infxn2Cn;K jx xnj 1K by construction of

C;K:

We conclude that B

xn;1K

Cn;K: As B xn; 1K4mK+1n=1 is an open cover it followsthat fCn;Kg

4mK+1

n=1 is an open cover of A: Obviously, fCn;Kg4mK+1

n=1 is a nite subcover of

Cn = fC;ng ; and by assumption Cn = fC;ng is a subcover ofC: Hence, A is compact.CASE 2: Suppose A is not covered by Cn = fC;ng for any n: Then there is a

sequence fxng with xn 2 A and xn 2 Rn[ Cn for every n 2 N: Let fxkng be a convergentsubsequence with limit x: Then we note that:

1. Since A is closed x 2 A

2. Since C is a covering of A there exists such that x 2 C

3. C is open so there exists " > 0 such that B (x; ") C

4. Since there is no n such that x 2 [C;n there is no n such that

x 2 Cn =

x 2 C j infy2RnC

jx yj > 1n

, there is no n such that B

x;1

n

C

[if B x; 1n C then jx yj 1n > 1n+1 for every y =2 C implying that x 2Cn+1]

5. Statements there exists " > 0 such that B (x; ") C and there exists no n suchthat B

x; 1

n

C contradict each other.We conclude that there cannot be any convergent subsequence to fxng ; which (since

fxng is bounded) contradicts the Bolzano Weierstrass theorem. It follows that A must becovered by Cn = fC;ng for some n; which by analysis in CASE 1 means that A is compact.

2.6 Cauchy Convergence

An alternative way of thinking about convergent sequences is as a sequence with terms that

get closer and closer together further out in the sequence.

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Denition 25 hxni1n=1 is said to be a Cauchy sequence if given any " > 0 there exists Ksuch that

jxn xmj < "

for every m; n K:

The set of real numbers is a simple example of a complete metric space. Such spaces

have many important properties, one being that the two convergence criteria are equivalent

(which in the end gives a nice characterization of compactness).

The easy direction is:

Lemma 1 Every convergent sequence of real numbershxn

i1

n=1is a Cauchy sequence.

Proof. Fix " > 0 and let hxni1n=1 have limit x: As hxni1n=1 converges to x there existsK"3

2 N such thatjxn xj < "

2

jxm xj < "2

for every n; m K "2 : Thus (triangle inequality "direct path shorter than any other path")jxn xmj = j(xn x) + (xm x)j

j(xn x)j + j(xm x)j

0 and let K be such that jxn xmj < " for every n; m K: Then

jxn xKj < "

for every n; implying that

xn xK < " and xK xn < "

xK " < xn < xK + "

for every n K: The setfx1; x2;:::;xK "; xK + "g

is obviously bounded above and below and any bounds a; b to this set also bounds xK+ ":

Theorem 11 hxni1n=1 converges if and only if it is a Cauchy sequence.

Proof. Let hxni1n=1 be a Cauchy sequence. By Lemma above it is then bounded above andbelow. Applying the Bolzano-Weierstrass Theorem we know that there exists a convergent

subsequence hxknikn . Let the limit of the convergent subsequence be denoted by x: Fix" > 0: Since hxkni converges to x there exists K1 2 fkig1i=1 such that

jxn xj