# MATH 320 Chapter Notes - Chapter 3: Limit Point, Nested Intervals, Open Set

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Orest Byskosh Homework 5 Due Monday October 23

Math 320-1, Fall 2017

1. Exercise 3.2.2

(a) Objective is to ﬁnd the limit points of the sets Aand B.

Recall the deﬁnition, a number a is limit point of set S if there exists a neighborhood of a

that contains an inﬁnite number of members of the set S.

xn= (−1)n+2

n

then it follows that :

lim

n→∞

xn= lim{(−1)n+2

n}

this equates to 1 if ais even and -1 if ais oﬀ, this fact stems from using the density of Q.

Therefore, 1 and -1 are the limit points of set A.

For every x∈[0,1], > 0,V(x) = (x−, x +)∩B6=∅, which has inﬁnitely many rational

numbers all of which are limit points of B.

(b) Any point in A has the form −1 + 2

n,1 + 2

n. Take 1∈A, for every > 0,V(1) 6⊂ A. Since −1

is not contained within the set, the set is not closed. Observe there is no open -neighborhood

of 1∈A, therefore Ais not open or closed.

B→[0,1],[0,1] 6⊂ B, (x−, x +)∩B6=∅.V(x) = (x−, x +)6⊂ Band thus we see

that Bis not open or closed.

(c)

{(−1)n+2

n:n= 1,2,3, ...}

From this we see that most of Aare isolated points, except for 1.

B⊂[0,1] →Bhas no isolated points.

(d) A= [A∪ { limit points of A}]=[A∪ {−1}]and B= [0,1].

2. Exercise 3.2.6

(a) False, Q⊆(−∞,−√2) ∪(√2,∞)which are open sets; but, (−∞,√2) ∪(√2,∞)6=R.

(b) The nested interval property is false for closed sets.

In= [n, ∞)which is bounded and compact. Then I1⊇I2⊇... but ∩In=∅.

(c) Let V6=∅be open and x∈V. Then, there exists an -neighborhood (x−, x +)such that

(x−, x +)⊆VSince (x−, x +)⊆Vcontains inﬁnitely many rational numbers, so it

follows that every nonempty open set has rational numbers making the statement True.

(d) Let xn=√3 + 1

n, V ={√3}∪{1

n+√3 : n∈N}.√3is the only limit point of Vand

√3∈V. Therefore, Vis closed, also Vdoes not contain and rational numbers making the

claim False.

(e) Cantor Set C= [0,1] −[(1

3,2

3)∪(1

9,2

9)∪(7

9,8

9)∪...]with [(1

3,2

3)∪(1

9,2

9)∪(7

9,8

9)∪...]being

open. But, if follows than that [(1

3,2

3)∪(1

9,2

9)∪(7

9,8

9)∪...]cis closed. Therefore, C=

[0,1] ∩[(1

3,2

3)∪(1

9,2

9)∪(7

9,8

9)∪...]cis closed, making the statement true.

3. Exercise 3.2.7