Math 320 Introduction to Real Analysis II Assignment 2 by bbu90505

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									                               Math 320
                    Introduction to Real Analysis II
                        Assignment 2 Solutions

Q1. Let {Vα }α∈A be an open cover of K. We have 0 ∈ K so there should
exist one α0 in A, such that 0 ∈ Vα0 . Vα0 is an open set so this implies
that for some positive , we should have (− , ) ⊂ Vα0 . There are just finite
number of natural numbers n, such that n ≤ 1 . For each such n, there is
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an αn such that n ∈ Vαn . We take Vα0 , and these finite collection and we
get a finite subcover of at most 1 + 1 open sets. Therefore K is compact.

Q2. We use the previous problem, K was compact and also it had just one
limit point which was 0.

Q3. We take the set of all open segments (1/n, 1) for n ∈ N. This is a
countable open cover of (0, 1) without any finite subcouver, because among
any finite number of natural number there is a biggest number so if (1/nk , 1)
for 1 ≤ k ≤ t is a finite subcover then by taking the maximum among nk ’s,
we have:
                          M := M ax{nk |1 ≤ k ≤ t}
,
                               (1/nk , 1) = (1/M, 1)
                          1≤k≤t

As we see this won’t cover the whole (0, 1) interval.
               √ √          √     √
Q4. Let S = ( 2, 3) ∪ (− 3, − 2), then E = {p ∈ Q|p ∈ S}. Therefore
E is bounded in Q, and since Q is dense in R every point of Q is a limit
point of Q (here Q is a metric space). Thus E is closed. To prove that E is
not compact, we form an open cover of E as follows:

                 {Gα } = {Nr (p)|p ∈ E&(p − r, p + r) ⊂ S}

If E is compact then there are finitely many indices α0 , . . . , αn such that

                                  E⊂           G αi
                                       1≤i≤n
                                                        √
we take p = max1≤i≤n pi , then p is the closest point to 3. But Nr (p) lies
                √
in E, so [p + r, 3) cannot be covered since Q is dense in R. Therefore E is
not compact. It’s easy to see also E is open in Q.


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Q5. Let X be a separable metric space, and S be a countable dense subset
of X. Let for each s ∈ S, {Nr (s)}r∈Q be a collection of open balls with center
at s and radius rational numbers. Therefore this is a countable collection
of open sets around s. It’s clear that any open set which contains s should
contain one of these open balls. Now we take all this collection of open balls
for each s in S. To prove this is a base. suppose x is point in X and U is
an open set containing x. U is open so There exist an positive such that
x ∈ N (x) ⊂ U . S is dense in X so we should have one s0 in S such that
s0 ∈ N 4 (x). Now it is clear that x ∈ N 2 (s0 ) ⊂ N (x) ⊂ U (why?). This
complete our proof, as we built a base.

Q6. Fix δ > 0, and pick x1 ∈ X. Having chosen x1 , . . . , xj ∈ X, choose
xj+1 , if possible, so that d(xi , xj+1 ) ≥ δ for i = 1, . . . , j. If this process
cannot stop, then consider the set A = {x1 , x2 , . . . , xk }. If p is a limit point
of A, then a neighborhood Nδ/3 (p) of p contains a point q = p such that
q ∈ A. q = xk for only one k ∈ N . If not, d(xi , xj ) ≤ d(xi , p) + d(xj , p) ≤
δ/3 + δ/3 < δ, and it contradicts the fact that d(xi , xj ) ≥ δ for i = j. Hence,
this process must stop after finite number of steps.
Suppose this process stop after k steps, and X is covered by

                           Nδ (x1 ), Nδ (x2 ), . . . , Nδ (xk )

that is, X can therefore be covered by finite many neighborhoods of radius
δ. Take δ = 1/n(n = 1, 2, 3, . . .), and consider the set A of the centers of
the corresponding neighborhoods. Fix p ∈ X. Suppose that p is not in
A, if Nr (p) be a neighborhood of p . Note that Nr/2 (p) can be covered
by finite many neighborhoods Ns (x1 ), . . . , Ns (xk ) of radius s = 1/n where
n = [2/r] + 1 and xi ∈ A for i = 1, . . . , k. Hence, d(x1 , p) ≤ d(x1 , q) +
d(q, p) ≤ r/2 + s < r where q ∈ Nr/2 (p) ∩ Ns (x1 ). Therefore, x1 ∈ Nr (p)
and x1 = p since p is not in A. Hence, p is a limit point of A if p is not in
A, that is, A is a countable dense subset, that is, X is separable.

Q7. By Exercise 23, 24 on pg. 45 in Rudin, X is separable. Hence X has
a countable base. It follows that every open cover of X has a countable
subcover {Gn }, n = 1, 2, . . . . This is because if {Bn } denote the countable
basis of X, and A be an open covering of X. For each positive integer n
for which it is possible to choose an element An of A containing the basis
element Bn . The collection A of the sets An is countable, since it is indexed
with a subset J of positive integer numbers. Furthermore it covers X: given
point x in X we can choose an element A of A containing x. Since A is open
there is a basis element Bn such that x ∈ Bn ⊂ A. Because Bn lies in an

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element of A, the index n belongs to the set J, so An is defined; since An
contains Bn , it contains x. Thus A is a countable subcollection of A that
covers x.
   Suppose X is not compact. Therefore, no finite subcollection of {Gn }
covers X.
   For each n, define

                         Fn := X\(G1 ∪ · · · ∪ Gn ).
   Therefore, Fn = ∅, otherwise G1 ∪ · · · ∪ Gn ⊇ X.
   Also, we have that F1 ⊇ F2 ⊇ · · · , but since ∞ Gn = X, thus
                                                          n=1
  ∞
  n=1 Fn = ∅.
   For each n, pick an element in Fn and form a set E. Then E is infinite,
and so E has a limit point, say x ∈ E .
   Then x ∈ E ⊆ X would imply that there exists N such that x ∈ GN .
But since x is a limit point, then there exists y such that y ∈ GN ∩ E.
   Since y ∈ GN , then y ∈ FN , FN +1 , . . . But y ∈ E would mean that
                               /
y ∈ F1 ∪ F2 ∪ · · · ∪ FN −1 . But |GN ∩ E| ≤ N + 1.
   Since there are only finitely many points in GN ∩ E, so x can’t be a limit
point of E. Contradiction.




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