Advanced Calculus Midterm Exam

Advanced Calculus Midterm Exam

1. Let {x_k} be a sequence defined recursively by x₁=√

2, and x_k+1=p

2 + x_k, for k = 1, 2, . . . . (a) (10 points) Show by induction that (i) x_k< 2 and (ii) x_k< x_k+1 for all k.

Solution: (i) When k = 1, x1=√ 2 < 2.

Assume that x_k< 2. Then we have x_k+1 =p

2 + x_k<√

2 + 2 = 2.

Thus, the argument of induction implies that x_k< 2 for all k.

(ii) Using (i) twice, we have x_k+1=p

2 + x_k>√

x_k+ x_k=√ 2√

x_k>√ x_k√

x_k= x_kfor all k.

(b) (10 points) Show that lim

k→∞x_kexists and evaluate it.

Solution: The results of (a) implies that {x_k} is an increasing sequence and it is bounded from above by 2. The Monotone convergence theorem implies that lim

k→∞x_k= x exists, and since x =

k→∞limx_k+1= lim

k→∞

p

2 + x_k=√

2 + x, we have 0 = x²− x − 2 = (x − 2)(x + 1) =⇒ x = 2.

2. (16 points) Let S ⊂ Rⁿand x ∈ Rⁿ. Show that x ∈ S if and only if there is a sequence of points {x_k} in S that converges to x. [Hint: You may use the fact that the closure of S is the union of S and all its boundary points, i.e. S = S ∪∂^S.]

Solution: (⇒) If x ∈ S = S ∪∂S, then either

Case (i) : x ∈ S, then the sequence {x_k= x}, for k = 1, 2, . . . , is a sequence in S that converges to x, or

Case (ii) : x ∈∂S, then, there is an x_k∈ B(1

k, x) ∩ S, for each k = 1, 2, . . . satisfying that lim

k→∞x_k= x.

(⇐) For each x /∈ S, there exists an r > 0 such that B(r, x) ⊂ S^c. This implies there does not exist any sequence of points {x_k} in S that converges to x.

Remark: This equivalence says that S ∪∂^{S = S ∪ S0, where S0} denote the set of accumulation points of S.

3. (a) (10 points) Let B(r, 0) = {x ∈ Rⁿ: kxk < r} be the ball of radius r about the origin. Show that B(r, 0) is open in Rⁿ.

Solution: For each p ∈ B(r, 0), we have B(r − kpk, p) ⊂ B(r, 0) since for each y ∈ B(r − kpk, p) we have kyk ≤ ky − pk + kpk < r − kpk + kpk = r, i.e. y ∈ B(r, 0).

Alternative proof: Since 0 ∈ B(r, 0) ⊂ B(r, 0), 0 is an interior point of B(r, 0).

For each p ∈ B(r, 0) \ {0}, by settingρ = min{kpk, r − kpk} > 0, we have B(ρ, p) ⊂ B(r, 0) and conclude that p is an interior point of B(r, 0). Thus, B(r, 0) is open since each of its points is an interior point.

(b) (16 points) Show that if S1and S2are open, so are S1∪ S2and S1∩ S2.

Solution: For each p ∈ S₁∪ S₂, since p is an interior point of S_i, for i = 1 or 2, there exists a B(r, p), for some r > 0, satisfying either B(r, p) ⊂ S_i⊂ S₁∪ S₂. Thus, p is an interior point of S₁∪ S₂, and S₁∪ S₂is open.

For each p ∈ S₁∩ S₂, there exist r₁, r₂> 0 such that B(r₁, p) ⊂ S₁, and B(r_r, p) ⊂ S₂. By taking r = min{r₁, r₂} > 0, we have B(r, p) ⊂ S₁∩ S₂, and conclude that S₁∩ S₂is open.

Advanced Calculus Midterm Exam (Continued) November 24, 2010

(c) (8 points) Show that for any S ⊂ Rⁿ, S^int is open.

Solution: For each p ∈ S^int, there is an r > 0 such that B(r, p) ⊂ S. By (a), every point in B(r, p) is an interior point of B(r, p) ⊂ S, hence, it is also an interior point of S. Thus, B(r, p) ⊂ S^int, and this implies that S^int is open.

Alternative proof:The definition of an interior point says that x ∈ S^int if there exists a ball B(r, x) ⊂ S. Therefore, if x /∈ S^int, then either every ball centered at x intersects both S and S^c, or there exists a ball B(r, x) ⊂ S^c, i.e. x ∈∂^{S ∪ (Sc)int. Since}∂^{S =}∂^Sc,∂^{S ∪ (Sc)int=}∂^{Sc∪ (Sc)int} is closed, S^int is open.

4. (10 points) Let f : S → R^mbe a function satisfying

|f(x) − f(y)| ≤ C|x − y|^λ for all x, y ∈ S,

where C > 0 andλ > 0 are constants. Show that f is uniformly continuous on S.

Solution: For each ε > 0, since |f(x) − f(y)| ≤ C|x − y|^λ for all x, y ∈ S, we choose δ =¡ε C

¢_1/λ

such that if , x, y ∈ S satisfying that |x − y| <δ, then |f(x) − f(y)| ≤ C|x − y|^λ < Cδ ⁼ε. Hence, f is uniformly continuous on S.

5. (16 points) Show that f : Rⁿ→ R^k is continuous (in the sense ofε^-δ definition) if and only if for each open subset U in R^k the set f⁻¹(U) = {x ∈ Rⁿ: f(x) ∈ U } is open.

Solution: (⇒) Let U be an open subset in R^k and p be a point in f⁻¹(U). Given ε > 0, since B(ε, f(p)) ∩U is open, there exists a ball B(η, f(p)) ⊂ B(ε, f(p)) ∩U, for someη> 0. The continuity of f at p implies that there is a ball B(δ, p), for some δ > 0, such that f(B(δ^{, p)) ⊂ B(}η^{, f(p)) ⊂} B(ε, f(p)) ∩U ⊂ U. This implies that B(δ^{, p) ⊂ f−1(U), and f−1}(U) is open .

(⇐) For each p ∈ Rⁿ, and eachε> 0, since the set B(ε, f(p)) is open in R^k, f⁻¹(B(ε, f(p))) is an open set containing p. There exists aδ> 0 such that B(δ^{, p) ⊂ f−1(B(}ε, f(p))), i.e. f(B(δ^{, p)) ⊂ B(}ε^{, f(p)),} and f is continuous at p.

6. (10 points) Suppose that f : Rⁿ→ R^m is continuous on U ⊂ Rⁿ and g : R^m→ R^k is continuous on f(U ) ⊂ R^m. Show that the composite function g(f) : Rⁿ→ R^kis continuous on U.

Solution: For each p ∈ U, and anyε> 0, since g is continuous at f(p), there existsδ1> 0, such that if y ∈ f(U ) satisfying that |y − f(p)| <δ1, then |g(y) − g(f(p))| <ε. Also, since f is continuous at p, there exists aδ > 0, such that if x ∈ U satisfying that |x − p| <δ, then |f(x)− f(p)| <δ1which implies that |g(f(x)) − g(f(p))| <ε, i.e. the composite function g(f) is continuous at each p ∈ U.

Page 2

Advanced Calculus Midterm Exam (Continued) November 24, 2010

7. (10 points) Let S be a compact subset of Rⁿand let f : S → R^mbe continuous at every point of S. Show that the image set

f(S) = {f(x) : x ∈ S}

is also compact.

Solution: Suppose {y_k} is a sequence in f(S). This implies that, for each k, there is an x_k ∈ S such that y_k= f(x_k). Since S is compact, by the Bolzano-Weierstrass theorem, {x_k} has a convergent subse- quence {x_kj} that converges to a point a ∈ S. Since f is continuous at a, lim

j→∞y_kj = lim

j→∞f(x_kj) = f(a) ∈ f (S). Thus, every sequence in f(S) has a subsequence whose limit lies in f(S). This implies that f(S) is compact.

8. (10 points) Let S be a connected subset in Rⁿ. Show that the closure of S is also connected.

Solution: Suppose that S is disconnected and (U,V ) is a disconnection of S. Suppose that U ∩ S 6= /0 and V ∩ S 6= /0, then it is easy to see that (U ∩ S,V ∩ S) is a disconnection of S. This contradicts to that S is connected. Therefore, either U ∩ S = /0, or V ∩ S = /0. Assume that V ∩ S = /0, since S ∪∂S = S = U ∪ V, this implies that V ⊂∂S, and U = S. But, this implies that U ∩ V = S ∩ V 6= /0 which contradicts to that (U,V ) is a disconnection of S. It is easy to see that U ∩ S = /0 will also lead to a contradiction. Therefore, S is connected.

Note: (1)In general, the converse is not true. e.g. Let S = [0, 1) ∪ (1, 2). Then S = [0, 2] is connected while S is not.

(2)A subset A ⊂ S is said to beopen relative to the set Sif there exists an open set U ⊂ Rⁿsuch that A = U ∩ S.

The definition of connectedness of S ⇐⇒ is equivalent to that S cannot be a disjoint union of two nonempty open subsets relative to S, i.e. S cannot be expressed as S = A ∪ B, where /0 6= A = U ∩ S, and /0 6= B = V ∩ S, A ∩ B = /0, and U,V are open subsets of Rⁿ.

proof of (⇒) Suppose that S = A ∪ B, where /0 6= A = U ∩ S, and /0 6= B = V ∩ S, A ∩ B = /0, and U,V are open subsets of Rⁿ.

⇒ A = S \ B = S \ (V ∩ S) = S \V = S ∩V^c, and B = S \ A = S \ (U ∩ S) = S \U = S ∩U^c.

⇒ A ∩ B =¡

S ∩V^c¢

∩¡ S ∩V¢

= /0, and A ∩ B =¡

S ∩V^c¢

∩¡ S ∩V¢

= /0.

Hence, S is disconnected.

proof of (⇐) Suppose that S is disconnected, and S = S₁∪ S₂, where /0 6= S_i, i = 1, 2, and S₁∩ S₂= /0, S₁∩ S₂= /0.

⇒ S1= S2

c∩ S, and S2= S1

c∩ S are disjoint nonempty open subsets relative to S, and S = S1∪ S2.

Page 3