Aujero, Gadian, Pamonag

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Aujero, Gadian, Pamonag

Let x0 ∈ R, and let r > 0. The set

B(x0, r) = {x : ||x − x0 || < r}

is called the open ball of radius r centered

at x0. It is also called the r-neighborhood

of x0 which consists the points in Rn whose

distance from x0 is less than r.

We define a number of terms, familiar inour study of R, in terms of this simple

concept.

1. A set E in Rn is open if to each x0 ∈ E there

corresponds an ε > 0 such that B(x0, ε) ⊂E. An open set containing a point x0 is

also called a neighborhood of x0.

2. A set E in Rn is closed if Rn \ E is open.

3. A point x0 ∈ E ⊂ Rn is an accumulation

point, or limit point, of E if every open ball

centered at x0 contains points of E other

than X0. We shall use the terms

“accumulation point” and “limit point”

interchangeably.

4. A point x0 ∈ Rn is a boundary point of a

set E ⊂ Rn provided that for each ε > 0,

B(x0, ε) ∩ E ≠ ∅ and B(x0, ε) \ E ≠ ∅.

5. A set E ⊂ Rn is bounded if there exists

M>0 such that E ⊂ B(0,M).

ExampleThe set

E3 = Q × Q = {(x, y) : x and y are rational}.

• The set is neither open nor closed.

• Every point in R2 is an accumulation point

of E3 and a boundary point of E3.

• E3 is not bounded in R2.

A sequence of points in Rn is a function

f : IN → Rn.

As was the case for sequences in R, we will

frequently use notation such as {xk} to

denote a sequence in Rn. Here each vector

xk can be written as an n-tuple using double

subscript notation:

xk = (xk1 xk2, . . . , xkn).

A sequence {xk} in Rn is bounded if there exists

M ∈ IN such that ||xk|| ≤ M for all k ∈ IN.

We can define convergence of a sequence

{xk} in Rn in the same way as we did for

sequences of real numbers (Definition 2.6).

Note that here the norm plays the same

role that the absolute value did earlier.

Let {xk} be a sequence in Rn. We say {xk}

converges to a point x and write

lim xk = x

or

xk → x as k→∞

provided that for each ε > 0 there exists N ∈ IN

such that || xk – x || < ε whenever k ≥ N.

k→∞

This definition is equivalent to the requirement

that every open ball centered at x contains

all but a finite number of terms of the

sequence: For each ε > 0 there exists N suchthat xk ∈ B(x, ε) for all k ≥ N. It is also

equivalent to the requirements

|| xk − x || → 0

or

d(xk, x) → 0

as k→∞, where d is the euclidean distance.

We can also describe convergence in Rn in

terms of coordinate-wise convergence. T o

see this, observe first that for

x = (x1, x2, . . . , xn) and j = 1, . . . , n,

x j2 ≤ Σ xi

2 = ||x||2 ≤ ( Σ|xi|)2.

Thus

|xj| ≤ ||x|| ≤ Σ |xi|. (7)

n

i=1

n

i=1

n

i=1

Now let

{xk} = {(xk1 , xk2 , . . . , xkn)}

be a sequence in Rn, and let

x = (x1, x2 , . . . , xn) be a point in Rn. By (7)

|xkj − xj| ≤ ||xk − x|| ≤ |xki − xi|. (8)Σn

i=1

If xk → x as k→∞, that is,

||xk − x|| → 0 as k → ∞,

then we see from (8) that for each j = 1, . . . ,n, |xkj − xj| → 0 as k → ∞. Conversely, if foreach j = 1, . . . , n we have |xkj − xj| → 0 ask → 0, then

|xki − xi| → 0

as k→∞, so we see once again from (8) that||xk − x|| → 0 as k→∞. We summarize thisdiscussion as a theorem.

i=1

Σn

Let

{xk} = {(xk1, . . . , xkn)}

be a sequence in Rn and let

x = (x1, x2, . . . , xn) ∈ Rn. Then

lim xk = x

if and only if for each j = 1, . . . , n, limk→∞ xkj = xj .k→∞

Let {xk} and {yk} be sequences in Rn, and let α ∈R. If limk→∞ xk = x and limk→∞ yk = y, then

i. lim k→∞ (αxk) = αx.

ii. lim k→∞ (xk + yk) = x + y.

iii. lim k→∞ (xk ・ yk) = x ・ y.

Let E ⊂ Rn and let x ∈ Rn. Then x is a limit point of

E if and only if there exists a sequence {xk} of

distinct points of E such that limk→∞ xk = x.

Limit Points As was true in R, we can

characterize limit points in Rn in the language

of sequences.

Suppose x is a limit point of E.

Then for each ε > 0 there exists xk ≠ x in E suchthat ||xk − x|| < ε.

Choose x1 ∈ E such that 0 < ||x1 − x || < 1.

Inductively, having chosen xk, choose xk+1 in Esuch that

0 < || xk+1 − x || < ½ || xk − x ||.

Then limk→∞xk = x, xk ≠ x, and if k ≠ j, xk ≠ xj. Theconverse is obvious.

A set E ⊂ Rn is closed if and only if it contains all

its limit points.

(⇒): Let E be a closed set, and let {xn} be a

sequence in E that converges to x0 ∈ X.

We must show that x0 ∈ E.

Suppose that x0 ∈ E is not true, we assume that

x ∈ Ec.

Since Ec is open, there is an ε > 0 for which

B(x0, ε) is a subset of Ec. Since {xn} → x0, let ňbe such that n > ň ⇒ d(xn, x0) < ε, ε > 0. Then

for n > ň we have both xn ∈ E and xn ∈ B(x0,

ε) ) is a subset of Ec, a contradiction.

(⇐): Suppose E is not closed.

We must show that E does not contain all its

limit points.

Since E is not closed, Ec is not open.

Therefore there is at least one element x of Ec

such that every ball B(x0, ε) contains at least

one element of (Ec)c = E. For every n ∈ IN, let

xn ∈ B(x, 1/n) ∩ E. Then we have a sequence

{xn} in E which converges to x ɆE | i.e., x is a

limit point of E but is not in E, so E does not

contain all its limit points.

Every bounded sequence {xk} in Rn contains a

convergent subsequence.

Let {xk} be a bounded sequence in Rn,

say ||xk || ≤ M for all k ∈ IN.

For each k, let xk = (xk1, . . . , xkn). Then for eachk ∈ IN and j = 1, . . . , n, |xkj| ≤ M. Thus, for all

j, the sequence of real numbers {xkj} is

bounded.

Let j = 1.By the Bolzano-Weierstrass theorem

applied to the bounded sequence

{xkj} (k = 1, 2, 3, . . . ) there exists a sequence

of integers

1 ≤ k(1, 1) < k(1, 2) < . . .

and a number x1 such that

xk(1,i),1→ x1 as i→∞.

Observe that the sequence {xk(1,i),1} is just a

subsequence of the sequence {xk1}.

Next let j = 2.

The sequence {xk(1,i),2} is also bounded by M, so

again, by the Bolzano-Weierstrass theorem

there is a subsequence {k(2, i)} of {k(1, i)}

such that xk(2,i),2 converges.

L et

x2 = lim i→∞ xk(2,i),2.

Now, the sequence {xk(1,i),1} converges to x1, so

the same is true of any subsequence of this

sequence. In particular {xk(2,i),1} → x1 as i→∞.

We continue this process. Having obtained a

sequence

{xk(j,i),j} with {xk(m,i),m} → xm for all m ≤ j,

we use the Bolzano-Weierstrass theorem yet

again to obtain a further subsequence {xk(j+1,i),j+1} that converges to a point xj+1 ∈ R.

The process stops when j + 1 = n. Letting ki =

k(n, i) and x = (x1, . . . , xn), we have

limi→∞ xki = x by Theorem 11.15.

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