Advanced Calculus Midterm Exam
November 18, 20061. (12 points) Let A = {(x, y) ∈ R2| x ∈ Q}.
(a) Find A0, the set of all cluster point of A. Justify your answer.
Solution: Since for any P = (x, y) ∈ R2and for any r > 0, the open ball neighborhood BP(r) of P, satisfies that BP(r) ∩ A \ {P} 6= /0, A0= R2.
(b) Find the boundary Bd(A). Justify your answer.
Solution: Since for any P = (x, y) ∈ R2and for any r > 0, the open ball neighborhood BP(r) of P, satisfies that BP(r) ∩ A 6= /0, and BP(r) ∩¡
R2\ A¢
6= /0, Bd(A) = R2. (c) Find the interior Int(A). Justify your answer.
Solution: Since for any P = (x, y) ∈ R2and for any r > 0, the open ball neighborhood BP(r) of P, satisfies that BP(r) ∩¡
R2\ A¢
6= /0, Int(A) = /0.
2. (10 points) Let {Ua}a∈A be a collection of open subsets in Rp. Show that the set U = ∪a∈AUais open in Rp.
Solution: For any q ∈ U, q ∈ Ua for some a ∈ A . Since Ua is open, there exists a r > 0, such that q ∈ Bq(r) ⊂ Ua⊂ U which implies that U is open in Rp.
3. (10 points) Show that the set∆= {(x, x)|x ∈ R} is closed in R2. Note that the set∆represents the graph of the function f (x) = x in R2.
Solution: For any P = (a, b) /∈∆, letting r = |b − a|
2 , we have r > 0.
Claim:∆∩ BP(r) = /0
The claim obviously implies that R2\∆is open in R2. Hence,∆is closed in R2. Proof of Claim: Observe that
BP(r) ⊂ Q = {(x, y) | 0 ≤ |x − a| < r ; 0 ≤ |x − b| < r}, i.e. Q is a square of length 2rwith center P.
Suppose∆∩ Q 6= /0 and (c, c) is a point in∆∩ Q, then we have 0 ≤ |c − a| < r and 0 ≤ |c − b| < r.
But, by using the triangle inequality, we have
|b − a| = |c − a − (c − b)| ≤ |c − a| + |c − b| < r + r = 2r = |b − a| which is contradiction.
Therefore,∆∩ Q = /0. Since BP(r) ⊂ Q, we have∆∩ BP(r) = /0.
4. (a) (10 points) Let A ⊂ Rp. Show that A = A ∪ A0, where A0is the set of all cluster point of A.
Solution: For any x /∈ A ∪ A0, x /∈ A, and x /∈ A0. From the definition of x /∈ A0, there is an open neighborhood U of x, such that U ∩ A = U ∩ A \ {x} = /0 which also implies that U ∩ A0 = /0.
Therefore, U is an open neighborhood of x satisfying U ∩¡
A ∪ A0¢
= /0, i.e. x ∈ U ⊂¡
A ∪ A0¢c and¡
A ∪ A0¢c
is open or A ∪ A0is a closed. Since A ⊂ A ∪ A0, A ∪ A0is a closed set containing A.
The definition of A implies that A ⊆ A ∪ A0.
On the other hand, for any x /∈ A, x /∈ A and, since A is closed, there is an open neighborhood U of x such that U ⊆¡
Rp\ A¢
which implies that U ∩ A = /0, i.e. x /∈ A0. This implies that A ∪ A0⊆ A.
Hence, we have A = A ∪ A0.
Advanced Calculus Midterm Exam (Continued) November 18, 2005
(b) (4 points) Let F ⊂ Rp be a closed subset. By using (a), prove that F is closed if and only if it contains all its cluster points.
Solution: If F is closed, since F ⊆ F, we have F = F. By using part (a), we get F = F = F ∪ F0 which implies that F0⊆ F.
Conversely, if F0⊆ F, by part (a), we have F = F ∪ F0= F. Since F is closed, F is closed.
5. (10 points) Let A ⊂ Rp. Show that Bd(A) = A ∩ Ac, where Ac= Rp\ A denotes the complementary set of A in Rp.
Solution: By 4(a), A = A ∪ A0, and Ac= Ac∪ (Ac)0, we have A ∩ Ac
=¡
A ∪ A0¢
∩¡
Ac∪ (Ac)0¢
=¡
A ∩ Ac¢
∪¡
A0∩ Ac¢
∪¡
A ∩ (Ac)0¢
∪¡
A0∩ (Ac)0¢
=¡
A0∩ Ac¢
∪¡
A ∩ (Ac)0¢
∪¡
A0∩ (Ac)0¢ . Hence, a point x ∈ A ∩ Ac
if and only if x ∈¡
A0∩ Ac¢
or x ∈¡
A ∩ (Ac)0¢
or x ∈¡
A0∩ (Ac)0¢
if and only if for any open neighborhood U of x U ∩ A 6= /0 and U ∩ Ac6= /0 if and only if x ∈ Bd(A).
Therefore, we have A ∩ Ac= Bd(A).
6. (10 points) Let X and Y be non-empty sets and let f : X ×Y → R have bounded range in R. Prove that sup
y
infx f (x, y) ≤ inf
x sup
y
f (x, y).
Solution: For each x ∈ X , and y ∈ Y, since inf
x f (x, y) ≤ f (x, y), we have sup
y
infx f (x, y) ≤ sup
y
f (x, y), for all x ∈ X which implies that sup
y
infx f (x, y) ≤ inf
x sup
y
f (x, y).
7. (10 points) Let K ⊂ Rpbe a compact subset. Prove directly (without using Heine-Borel theorem) that K is closed and bounded.
Solution: Proof of closedness: For any x /∈ K, the set {Rp\ Bx(1n) | n ∈ N} is an open covering of K. The compactness of K implies that there is a m ∈ N such that K ⊆ ∪mn=1Rp\ Bx(1n) = Rp\ Bx(m1) which implies that Bx(m1) ∩ K = /0 and Kcis open or K is closed.
Proof of boundedness: Let o denote the origin of Rp. The set {Bo(n) | n ∈ N} is an open covering of Rp, hence it is an open covering of K. The compactness of K implies that there is a m ∈ N such that K ⊂ ∪mn=1Bo(n) = Bo(m) which implies that K is bounded.
8. (10 points) Let C be an open, connected subset of Rp, and W be any set satisfying that C ⊆ W ⊆ C.
Show that W is connected.
Solution: Suppose that W is disconnected and A, B are two open sets separating W. Since A ∩W 6= /0 and W ⊆ C = C ∪C0, we have /0 6= A ∩C =¡
A ∩C¢
∪¡
A ∩C0¢
which implies that A ∩C 6= /0. By using a similar argument, we have B ∩ C 6= /0. This implies that A, B separate C which contradicts to that C being connected. Hence, W is connected.
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Advanced Calculus Midterm Exam (Continued) November 18, 2005
9. Let C1, C2be open, connected subsets of Rp. Assume that C1∩C26= /0.
(a) (4 points) Is it true that C1∩C2is always connected? Why?
Solution: No, consider C1= {(x, y) ∈ R2 | 1 < x2+ y2< 4, y > −1
2} and C2= {(x, y) ∈ R2| 1 < x2+ y2< 4, y < 1
2}, then both C1, C2 are open, connected subsets of R2, while C1∩ C2 is disconnected.
(b) (10 points) Show that C1∪C2is connected.
Solution: Suppose that C1∪ C2 is disconnected and A, B are two open sets separating C1∪ C2. Let x be a point in C1∩ C2, and assume that x ∈ A. Then A ∩ C16= /0 and A ∩ C2 6= /0. Since C1∪C2= A ∪ B, we get either C1∩ B 6= /0 or C2∩ B 6= /0 which would imply A, B separate either C1or C2, respectively. Either case contradicts to that both C1, C2are connected. Hence, C1∪C2 is connected.
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