SOLUTION OF HW4
MINGFENG ZHAO
March 09, 2013
1. [10 Points] Let A be a set and x a number. Show that x is a limit point of A if and only if there
exists a sequence x1 , x2 , · · · of distinct points in A that converges to x.
Proof. (=⇒) Assume that x is a limit point of A. By the definition of limit point, for any > 0, there
exists some x ∈ A\{x} such that
|x − x | < .
Take = 1, then there exists some x1 ∈ A\{x} such that
|x − x1 | < 1.
Since x1 6= x, we can take = min
1
2 , |x
− x1 | > 0, then there exists some x2 ∈ A\{x, x1 } such
that
|x − x2 | <
Since x2 6= x, we can take = min
1
3 , |x
1
.
2
− x2 | > 0, then there exists some x3 ∈ A\{x, x1 , x2 }
such that
|x − x3 | <
1
.
3
By induction, there there exists some xn ∈ A\{x, x1 , x2 , · · · , xn−1 } such that
|x − xn | <
1
.
n
Hence we know that xn → x as n → ∞, and xn ∈ A for all n ≥ 1, and xi 6= xj for all i 6= j.
1
2
MINGFENG ZHAO
(⇐=) Assume that there exists a sequence x1 , x2 , · · · of distinct points in A that converges to x. Since
xn → x, as n → ∞, then for any > 0, there exists some N ≥ 1 such that for all n ≥ N , we have
|xn − x| < .
Since xi 6= xj for all i 6= j, then at least for some xK with K ≥ N such that xK 6= x, which implies
that xK ∈ A\{x} and
|xK − x| < .
Therefore, x is a limit point of A.
2. [10 Points] Prove that every infinite set has a countable dense subset. Given an example of a set
A such that the intersection of A with the rational numbers is not dense in A.
Proof. Claim I: Every infinite set has a countable dense subset.
T
Let A an infinite set and An = A [n − 1, n] for all n ≥ 1, it suffices to show that for each An has
a countable dense subset. Now let B be any one of {An : n ≥ 1}, if B is finite set, then B itself is
a finite dense subset of B. Without loss of generality, assume B is infinite. Since B is bounded and
closed, so B is compact. Now for any n ≥ 1, there exists finitely many y1n , · · · , yknn ∈ B such that
B⊂B⊂
kn
[
B n2 (yin ).
i=1
By the definition of B, for each yin , there exists some xni ∈ B such that
|xni − yin | <
1
,
n
which implies that B n2 (yin ) ⊂ B n1 (xni ). Hence we know that
(1)
B⊂B⊂
kn
[
i=1
B n2 (yin ) ⊂
kn
[
i=1
B n1 (xni ).
SOLUTION OF HW4
3
Let C = {xni : 1 ≤ i ≤ kn , n ≥ 1}, then C is a countable subset of B. Now it suffices to show that
C is dense in B. In fact, for any x ∈ B, for any n ≥ 1, by (1), there exists some xnin ∈ C such that
|x − xnin | <
1
.
n
So we know that xnin → x as n → ∞. Hence C is dense in B. Therefore, we know that A has a
countable dense subset.
Example: Let A = R\Q, then A
T
Q = ∅ is not dense in A.
3. [10 Points] Give an example of a set A that is not closed but such that every point of A is a limit
point.
Proof. Example: Let A = Q, then for any x ∈ A, let xn = x +
1
n
∈ A for all n ≥ 1. So we know that
xi 6= xj for all i 6= j, and xn → x as n → ∞. By the result of part a, we know that x is a limit point
of A.
4. [10 Points] Define the derived set of a set A as the set of limit points of A. Prove that the derived
set is always closed. Given an example of a closed set A that is not equal to its derived set. Given an
example of a set A such that the derived set of A is not equal to the derived set of the derived set of
A.
Proof. Let B be the derived set of A, it suffices to show that R\B is open. In fact, for any x ∈ B,
that is, x is not a limit point of A, then there exists some > 0 such that
(x − , x + )
\
A ⊂ {x},
, which implies that for any y ∈ (x−, x+), then y is not a limit point of A, that is, (x−, x+) ⊂ R\B,
which implies that R\B is open, that is, B is closed.
4
MINGFENG ZHAO
S
Example I: Let A = [0, 1] {2}, then A is closed, but the derived set of A is [0, 1] which is not equal
to A.
Example II: Let A = [0, 1]
[
2+
S
1
: n ≥ 1 , then the derived set of A is [0, 1] {2}, and the derived
n
set of the derived set of A is [0, 1].
5. [10 Points] What sets are both open and closed?
Proof. Only ∅ and R are both open and closed sets. First, it is easy to see that ∅ and R are both open
and closed sets. Now let A be a set which is both open and closed. If A = ∅, we are done. Without
loss of generality, assume A 6= ∅. Take x0 ∈ A and fix, since A is open, then there exists some > 0
such that (x0 − , x0 + ) ⊂ A. So we can define
a = inf {x ∈ R : [x, x0 ] ⊂ A},
and b = sup {x ∈ R : [x0 , x] ⊂ A}.
Claim I: a = −∞.
If the Claim I is not true, since (x0 − , x0 + ) ⊂ A, then a ∈ R and a < x0 . Hence there exists
some ak ∈ A such that ak → a as k → ∞. Since A is closed, then a ∈ A. Since A is open, then
there exists some > 0 such that (a − , a + ) ⊂ A, which contradicts with the definition of a. Hence
a = −∞.
Claim II: b = ∞.
If the Claim I is not true, since (x0 − , x0 + ) ⊂ A, then b ∈ R and b > x0 . Hence there exists
some bk ∈ A such that bk → b as k → ∞. Since A is closed, then b ∈ A. Since A is open, then there
exists some > 0 such that (b − , b + ) ⊂ A, which contradicts with the definition of b. Hence b = ∞.
In summary, we know that A = R.
SOLUTION OF HW4
5
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu