γ ≥ x for each x ∈ E

1. (a) (4 points) Suppose S is an ordered set, E ⊂ S, and E is bounded above. State the definition of the sup E.

Solution: γ = sup E if γ has the following properties:

(1) γ is an upper bound of E, i.e. γ ≥ x for each x ∈ E;

(2) if s < γ then s is not an upper bound of E, i.e. if s < γ, then there exists an x ∈ E such that s< x.

(b) (8 points) Suppose A ⊂ R is nonempty and bounded below by c ∈ R. Let −A = {−x | x ∈ A}.

Prove that inf A = − sup(−A).

Solution: Since A ⊂ R is nonempty and bounded below, β = inf A exists with the following properties:

(1) β ≤ x for each x ∈ A;

(2) if t > β , then there exists an x ∈ A such that t > x.

Equvalently, we have

(1) −β ≥ −x for each −x ∈ −A;

(2) if s = −t < −β , then there exists an −x ∈ −A such that s = −t < −x.

This implies that β = inf A ⇔ −β = sup(−A), i.e. inf A = − sup(−A).

(c) (4 points) Let S be an ordered set. State the definition when S is said to have the least-upper-bound property.

Solution: S is said to have the least-upper-bound property if every nonempty, bounded above subset E ⊂ S has the least upper bound in S, i.e. If E ⊂ S, E is not empty, and E is bounded above, then sup E exists in S.

2. (a) (8 points) Let X be an infinite set. For p , q ∈ X , define

d(p, q) =

(1 if p 6= q, 0 if p = q.

Prove that this is a metric.

Solution: Since

(1) d(p, q) = 1 > 0 for any p 6= q ∈ X , and d(p, p) = 0, (2) d(p, q) = 1 = d(q, p) for any p 6= q ∈ X ,

(3) d(p, q) = 1 ≤ d(p, r) + d(r, q) for any p 6= q, r ∈ X and d(p, p) = 0 ≤ d(p, r) + d(r, p) for any r ∈ X ,

the function d is a metric on X . (b) (8 points) For x , y ∈ R, define

d(x, y) = |x − y|

1 + |x − y|. Prove that this is a metric.

Solution: Since (1) d(x, y) = |x − y|

1 + |x − y|> 0 for any x 6= y ∈ R, and d(x, x) = 0, (2) d(x, y) = |x − y|

1 + |x − y|= |y − x|

1 + |y − x| = d(y, x) for any x, y ∈ R, (3) d(x, y) − d(x, z) − d(y, z)

= |x − y|

1 + |x − y|− |x − z|

1 + |x − z|− |z − y|

1 + |z − y|

= |x − y| − |x − z| − |z − y|
− 2|x − z| |z − y| − |x − y| |x − z| |z − y|

(1 + |x − y|)(1 + |x − z|)(1 + |z − y|)

≤ 0 for any x, y, z ∈ R,

the function d is a metric on R.

3. Let (X , d) be a metric space and let E ⊂ X .

(a) (4 points) State the definition when a point p ∈ X is called a limit point of E.

Solution: A point p ∈ X is called a limit point of the set E if every neighborhood of p contains a point q 6= p such that q ∈ E, i.e. p is called a limit point of E if N_r(p) ∩ E \ {p} 6= /0 for every r> 0.

(b) (8 points) Let E⁰be the set of all limit points of E. Prove that E⁰is closed.

Solution:

Method 1:If p ∈ E⁰
0

then for each k ∈ N, ∃ pk∈ N_1/k(p) ∩ E⁰\ {p} 6= /0.

Since p_k∈ E⁰, ∃ q_k∈ E ∩ N_1/k(p_k) \ {p_k}

=⇒ d(p, q_k) ≤ d(p, p_k) + d(p_k, q_k) < 2 k,

=⇒ q_k∈ N_2/k(p) ∩ E \ {p}

=⇒ N_2/k(p) ∩ E \ {p} 6= /0 ∀ k ∈ N

=⇒ p ∈ E⁰. Hence, E⁰
0

⊆ E⁰and E⁰is closed.

Method 2: If p /∈ E⁰then ∃ r > 0 such that

N_r(p) ∩ E \ {p} = /0 =⇒ N_r(p) ∩ E⁰\ {p} = /0 =⇒ p /∈ E⁰
0

. Hence, E⁰
0

⊆ E⁰and E⁰is closed.

Method 3: Let p ∈ E⁰
c

. There exists r = r(p) > 0 such that N_r(p) ∩ E \ {p} = /0.

For each q ∈ N_r(p), since 0 < r − d(p, q) ≤ r, we have N_r−d(p,q)(q) ⊂ N_r(p), and

N_r−d(p,q)(q) ∩ E \ {q} ⊂ N_r(p) ∩ E \ {q} ⊂ {p} \ {q} = /0

=⇒ N_r−d(p,q)(q) ∩ E \ {q} = /0

=⇒ q ∈ E⁰
c

.

Since q is an arbitrary point in N_r(p), we have N_r(p) ⊂ E⁰
c

. Hence, p is an interior point of E⁰
c

. Since p is an arbitrary, E⁰
c

is open and E⁰is closed.

(c) (8 points) Prove that E and ¯E= E ∪ E⁰have the same limit points.

Solution:

Method 1: Since E⁰is closed by 3(b) and ¯E = E ∪ E⁰, we have E⁰⊆ ¯E =⇒ E¯
0

= E⁰∪ E⁰
0

= E⁰. Method 2:

(1) Since E ⊆ ¯E, we have E⁰⊆ ¯E
0

.

(2) For the proof of E¯
0

⊆ E⁰, we prove it by contradiction.

Suppose there is a limit point p of ¯Ewhich is not a limit point of E.

Since ¯E = E ∪ E⁰is closed, E¯
0

⊆ ¯E and E¯
0

\ E⁰⊆ ¯E\ E⁰= E ∪ E⁰\ E⁰⊆ E \ E⁰,

=⇒

if p ∈ E¯
0

\ E⁰ =⇒ p ∈ E \ E⁰. Since p /∈ E⁰, there exists r > 0 such that

N_r(p) ∩ E \ {p} = /0.

Claim: N_r(p) ∩ E⁰= /0.

By assuming the claim, we have

N_r(p) ∩ ¯E\ {p} = N_r(p) ∩ E ∪ E⁰
\ {p} ⊆ N_r(p) ∩ E \ {p}
∪ N_r(p) ∩ E⁰
= /0.

This contradicts to the assumption that p ∈ E¯
0

. Hence, E¯
0

\ E⁰= /0 which implies that E¯
0

⊆ E⁰.

Proof of Claim: For each q ∈ N_r(p), since 0 < r − d(p, q) ≤ r, we have N_r−d(p,q)(q) ∩ E \ {q} ⊂ N_r(p) ∩ E \ {q} ⊂ {p} \ {q} = /0

=⇒ N_r−d(p,q)(q) ∩ E \ {q} = /0

=⇒ q /∈ E⁰.

Since q is an arbitrary point in Nr(p), we have N_r(p) ∩ E⁰= /0.

This completes the proof of the claim.

4. Let (X , d) be a metric space and let E ⊂ X .

(a) (8 points) Let p ∈ X and r > 0, prove that N_r(p) = {q ∈ X | d(q, p) < r} is open.

Solution: Let q ∈ N_r(p).

Since 0 < r − d(p, q) ≤ r, we have N_r−d(p,q)(q) ⊂ N_r(p). This implies that q is an interior point of N_r(p).

Since q is an arbitrary point of N_r(p), N_r(p) = {q ∈ X | d(q, p) < r} is open.

(b) (8 points) Let E^obe the set of all interior points of E. If G ⊂ E and G is open, prove that G ⊂ E^o. Solution: For each p ∈ G, since G is open and G ⊂ E, there exists r > 0 such that N_r(p) ⊂ G ⊂ E.

This implies that p is an interior point of E whenever p ∈ G. Hence, we have proved that G ⊂ E^o. (c) (4 points) Let X = R and for x, y ∈ R define d(x, y) = |x − y|. Suppose E = {1

n | n ∈ N} ∪ (1, 2], find E^oand E⁰.

Solution: E^o= (1, 2) and E⁰= {0} ∪ [1, 2].

5. Let (X , d) be a metric space and K ⊂ X .

(a) (4 points) State the definition when K is said to be compact.

Solution: A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover, i.e. A subset K of a metric space X is said to be compact if {G_α} is an open cover of K, then there are {G_α1, . . . , G_αn} ⊂ {G_α} such that K ⊂ ∪ⁿ_i=1G_αi.

(b) (8 points) If K is compact, prove that K is closed.

Solution: For each p /∈ K, since K⊆ X \ {p} =

∞

[

n=1

X\ N_1/n(p) =

∞

[

n=1

{x ∈ X | d(x, p) >1

n}, {X \ N_1/n(p) | n = 1, 2, 3, . . .} forms an open cover of K.

The compactness of K implies that there exists a finite subcover {X \ N_1/n1(p), . . . , X \ N_1/nl(p)}

such that K ⊆

l

[

i=1

X\ N_1/ni(p).

Let m = max{n_i| 1 ≤ i ≤ l}. Since X \ N_1/ni(p) ⊆ X \ N_1/m(p) for each 1 ≤ i ≤ l, we have

K⊆

l

[

i=1

X\ N_1/ni(p) ⊆ X \ N_1/m(p) =⇒ K ⊆ X \ N_1/m(p) =⇒ N_1/m(p) ⊆ K^c.

Hence, K^cis open and K is closed.

(c) (8 points) If K is compact, prove that K is bounded.

Solution: For each p ∈ X , since K ⊆ X =

∞

[

n=1

N_n(p), {N_n(p) | n = 1, 2, 3, . . .} forms an open cover of K.

The compactness of K implies that there exists a finite subcover {N_n1(p), . . . , N_nl(p)} such that K⊆

l

[

i=1

N_ni(p).

Let m = max{ni| 1 ≤ i ≤ l}. Since N_ni(p) ⊆ N_m(p) for each 1 ≤ i ≤ l, we have

K⊆

l

[

i=1

N_ni(p) ⊆ N_m(p) =⇒ K ⊆ N_m(p).

Hence, K is bounded.

(d) (8 points) If K is compact, F ⊂ K and F is closed in X , prove that F is compact.

Solution: Let {G_α} be an open cover of F. Since F^cis open and F⊆^[

α

G_α =⇒ K ⊆ X = F ∪ F^c⊆ ^[

α

G_α
∪ F^c, {G_α} ∪ {F^c} is an open cover of K.

The compactness of K implies that there exists a finite subcover {G_α1, . . . , G_αn, F^c} ⊆ {G_α} ∪ {F^c} such that K ⊆

n

[

i=1

G_αi
∪ {F^c}.

Since F ⊆ K and F ∩ F^c= /0, F ⊆

n

[G_αi
.

This implies that {G_α1, . . . , G_αn} is a finite subcover of F.

Therefore, F is compact .