(1)Deadline pm (1) Open set Property (b) in Theorem 9.3 (page 63): “If U, V are open subsets of Rn, then, U ∩ V is also open.” Let U1

Deadline: 09/30/5:00pm

(1) Open set Property (b) in Theorem 9.3 (page 63): “If U, V are open subsets of Rⁿ, then, U ∩ V is also open.” Let U₁, · · · , U_m be open sets in Rⁿwhere m is a natural number. Use mathematical induction and the open set property (b) to prove that U₁∩ · · · ∩ U_m is also open.

Proof. When n = 1, U1 is open by assumption. Suppose that the statement is true for n = m. Let U₁, · · · , U_m are open subsets of Rⁿ. By induction hypothesis, U1∩ · · · ∩ U_m is open. Let Um+1 be another open subset of Rⁿ. Since

U₁∩ · · · ∩ U_m∩ U_m+1 = (U₁∩ · · · ∩ U_m) ∩ U_m+1,

by open set property (b) and by the fact that both U₁ ∩ · · · ∩ U_m and U_m+1 are open, (U1∩ · · · ∩ U_m) ∩ Um+1 is open. Thus U1∩ · · · ∩ U_m∩ U_m+1 is open. Thus the statement is true for n = m + 1. By mathematical induction, U₁∩ · · · ∩ U_n is open for any n ∈ N.

 (2) Let S = {(x, y) ∈ R² : xy ≥ 1}. Find int(S). You need to prove that the set int(S)

you found is the right one. Conclude that S is not open by showing S 6= int(S).

Proof. We have proved in class that

int(S) = S \ ∂S.

As long as we show that ∂S = {(x, y) ∈ R² : xy = 1}, int(S) = {(x, y) : xy > 1}. In this case, int(S) 6= S. Notice that the complement of S is S^c= {(x, y) ∈ R²: xy < 1}.

Let A = {(x, y) ∈ R² : xy = 1} For p = (x0, y0) ∈ A, x0y0 = 1. Then either x0 > 0 and y₀ > 0 or x₀< 0 and y₀ < 0. We assume that both x₀ and y₀ are positive. For

 > 0, we choose q= (x0+ /2, y0). Then d(p, q) = 

2 < .

We find q_ ∈ B(p, ). Moreover, (x₀+

2)y₀ = x₀y₀+ 

2y₀ = 1 + 

2y₀ > 1.

Hence q_ ∈ S. We find q_ ∈ B(p, ) ∩ S. This shows that B(p, ) ∩ S is nonempty for any  > 0.

Let us take q_⁰ = (x₀− /2, y₀). Then d(p, q_⁰) = 

2 < .

Hence q_⁰ ∈ B(p, ). Moreover, (x₀−

2)y₀ = x₀y₀− 

2y₀ = 1 − 

2y₀ < 1.

Thus q⁰_∈ S^c. We see that q⁰_∈ B(p, ) ∩ S^c. Hence B(p, ) ∩ S^c is nonempty for any

 > 0.

We conclude that p is a boundary point of S when x0, y0 are both positive. When x₀, y₀ are both negative, we can apply a similar method to show p is a boundary point of S. Thus A ⊂ ∂S.

1

Let p ∈ ∂S. Let us show that p ∈ A. Let U1 = {(x, y) : xy > 1} and U2 = S^c= {(x, y) : xy < 1}. In homework one, we have shown that U₁ is an open subset of R². In fact, we can also show that U2 is open. By definition, R² = U1∪ A ∪ U₂. If p ∈ U₁, by openess of U₁, we can find  > 0 so that B(p, ) ⊂ U1 ⊂ S. (B(p, ) ∩ S^c is empty.) Then p cannot be a boundary point of S. Similarly, if p ∈ U2, then p is never a boundary point of S. Thus the only possibility is that p ∈ A. We conclude that A = ∂S.

 (3) True or false. If the statement is true, prove it; if not, give a counterexample.

(a) If A and B are subsets of Rⁿ, is int(A) ⊂ int(B) when A ⊂ B?

Proof. The statement is true.

Let p be an interior point of A. There exists  > 0 so that B(p, ) ⊂ A. Since A is a subset of B, B(p, ) ⊂ B. Thus p is an interior point of B. We find

int(A) ⊂ int(B).

 (b) Is it true that int(A) ∩ int(B) = int(A ∩ B)?

Proof. The statement is true.

Let p ∈ int(A) ∩ int(B). Then p is an interior point of A and also an interior point of B. There exist ₁, ₂> 0 such that B(p, 1) ⊂ A and B(p, 2) ⊂ B. Take

 = min{1, 2}. Then B(p, ) ⊂ B(p, i) for i = 1, 2. Thus B(p, ) ⊂ B(p, 1) ∩ B(p, 1) ⊂ A ∩ B.

Thus p is an interior point of A ∩ B; p ∈ int(A ∩ B). Thus int(A) ∩ int(B) ⊂ int(A ∩ B)

If p ∈ int(A ∩ B), then we can choose  > 0 so that B(p, ) ⊂ A ∩ B. Thus B(p, ) ⊂ A and B(p, ) ⊂ B. We see that p is an interior point of A and also an interior point of B. We conclude that p ∈ int(A) ∩ int(B). Thus

int(A ∩ B) ⊂ int(A) ∩ int(B).

We conclude that int(A) ∩ int(B) = int(A ∩ B).

 (4) Sketch D1 and D2and find the boundary sets of D1 and D2. You need to prove that

∂D1 and ∂D2 you found is correct.

(a) D₁ = {(x, y) ∈ R² : x²− y²> 1}.

Proof. Let U1 = D1 and U2= {(x, y) : x²− y² < 1} and A = {(x, y) : x²− y²= 1}. Then R² = U₁∪ A ∪ U₂ and D₁^c= U₁∪ A. Let us prove that ∂D₁ = A.

Step I: Show that U₁ and U₂ are open.

We leave it to the readers.

Step II: Show that A ⊂ ∂D1.

Let p = (x0, y0) ∈ A. Then x²₀ − y₀² = 1. Either x0 > 0 or x0 < 0. We assume that x₀ > 0. In this case, x₀ = p1 + y₀² > 1. For any  > 0, we choose q_ = (x₀+ /2, y₀).Then

d(p, q) =  2 < .

Hence q both belong to B(p, ). By  > 0 and x0 > 0, (x₀+ 

2)²− y²₀ = x²₀+ x₀ +²

4 − y²₀ = 1 + x₀ +² 4 > 1.

We find that q_ ∈ D₁. Hence q_ ∈ B(p, ) ∩ D1. We see that B(p, ) ∩ D1 is nonempty for any  > 0.

If 0 <  < 1, we take q_⁰ = (x₀ − /2, y₀). Then d(p, q_⁰) = /2 < . Hence q⁰_∈ B(p, ). By x0> 1,

(x₀− 

2)²− y₀²= 1 − x₀ +²

4 < 1 −  +²

4 = (1 −  2)² < 1.

We find q⁰_∈ D^c₁Hence q_⁰ ∈ B(p, )∩D1^cfor 0 <  < 1. We see that B(p, )∩D^c1is nonempty for 0 <  < 1. If  > 1, we take q⁰_= (x₀−1, y₀). Then d(p, q⁰_) = 1 < .

By x0> 1,

(x₀− 1)²− y₀²= 2 − 2x₀ = 2(1 − x₀) < 0 ≤ 1.

Then q⁰_∈ B(p, ) ∩ D^c₁. Thus B(p, ) ∩ D₁^cis nonempty for  > 1. By the previous result, B(p, ) ∩ D1^c is nonempty for any  > 0. We find that p ∈ ∂D1. Hence

A ⊂ ∂D₁.

Step III: Show that ∂D1⊂ A by openess of U₁ and U2.

Let p ∈ ∂D1. Then p ∈ U1 or p ∈ A or p ∈ U2. If p ∈ U1 or p ∈ U2, by openess of U₁ and U₂, we can show that p is never a boundary point of D₁. (The argument is similar as those in the above exercises.) The only possibility is p ∈ A. As a consequence,

∂D1⊂ A.

We conclude that ∂D₁= A.

 (b) D2 = {(x, y) ∈ R² : x ≤ y}.

Proof. Let U₁ = {x > y} and A = {x = y} and U₂ = {x < y}. Then D₂^c= U₂. By homework one, we know U1 and U2 are open. Claim ∂D2= A. Hence points of U₁ and U₂ are never the boundary points of D₂ due to the openess of U₁ and U2. We see that

∂D2⊂ A.

Let p = (x0, y0) ∈ A. Then x0 = y0. For any  > 0, take q= (x0+ /2, y0) and q⁰_= (x₀− /2, y₀). Since  > 0,

x₀+ 

2 = y₀+ 

2 > y₀, x₀− 

2 = y₀−  2 < y₀. We see that q∈ U₁ while q⁰_∈ U₂. Moreover,

d(p, q) = d(p, q_⁰) =  2 < .

We find that q_ and q_ both belong to B(p, ). Thus q_ ∈ B(p, ) ∩ D2, q⁰_∈ B(p, ) ∩ D2^c.

Thus B(p, )∩D2 and B(p, )∩D^c2are both nonempty for any  > 0. We conclude that p is a boundary point of D₂. Hence

A ⊂ ∂D₂.

We conclude that A = ∂D2. 

(5) Let D = {(x, y) ∈ R² : x²+ 4y² ≤ 4} and define f : D → R by f (x, y) =p

4 − x²− 4y². (a) Find ∂D and int(D).

Proof. Using similar technique as above, one can show that ∂D = {x²+4y²= 4}

and hence int(D) = D \ ∂D = {x²+ 4y²< 4}. The details are left to the reader.

 (b) The point p(0, 1) belongs to both of D and ∂D. Is it make sense to you to

define/compute fx(0, 1) and fy(0, 1)? What happen when we compute f (0, 1 + h) − f (0, 1)

h

for h > 0? (We have discussed several examples in class. Please do one by yourself).

Remark: The function f (x, y) is undefined at (0, 1+h) for h > 0. It is nonsense to talk about f (0, 1 + h) and about f (0, 1 + h) − f (0, 1)

h .

(c) Let U be the subset of D consisting of points (x, y) so that fx(x, y) and fy(x, y) both exist. In other words,

U = {(x, y) ∈ D : fx(x, y) and fy(x, y) both exist.}.

Find U. Is U open? If not, find int(U ).

Proof. U = int(D) = {x²+ 4y² < 4} and U is open. Moreover,

f_x= −x

p4 − x²− 4y², f_y = −4y p4 − x²− 4y².

The proof is given as follows. Let p = (x0, y0). Then p is an interior point of D.

We can find  > 0 so that B(p, ) ⊂ D. For |t| < , we set F (t) = f (x₀+ t, y₀), G(t) = f (x₀, y₀+ t).

For |t| < , F (t) =

q

4 − 4y₀²− (x₀+ t)², G(t) = q

4 − x²₀− (y₀+ t)².

To show that f_x(p) and f_y(p) exist, it is equivalent to show that F and G are differentiable at 0. In this case, fx(p) = F⁰(0) and fy(p) = G⁰(0). We will only prove that F is differentiable at 0 here. The rest is left to the reader. For |t| < ,

F (t) − F (0)

t = −2x₀− t

p4 − 4y₀²− (x₀+ t)²+p4 − 4y²₀− x²₀

Since lim_t→0(p4 − 4y₀²− (x₀+ t)²+p4 − 4y₀²− x²₀) = 2p4 − 4y₀²− x²₀ > 0, we find

limt→0

F (t) − F (0)

t = −x₀

p4 − 4y₀²− x²₀. This shows that f_x(p) exists and equals −x₀

p4 − 4y²₀− x²₀.

 (6) Define a function f : R²→ R by

f (x, y) =







xy(x²− y²)

x²+ y² if (x, y) ∈ R²\ (0, 0) 0 if (x, y) = (0, 0).

(a) Show that f_x(0, 0) = lim

h→0

f (h, 0) − f (0, 0)

h , f_y(0, 0) = lim

h→0

f (0, h) − f (0, 0) h

both exist.

Proof. We leave to the readers to check fx(0, 0) = 0 and fy(0, 0) = 0.  (b) Let U be the set of all points (x, y) in R² such that both f_x(x, y) and f_y(x, y)

exist; find U and evaluate fx(x, y) and fy(x, y) for all (x, y) ∈ U. Is U open?

Proof. We have shown that f_x(0, 0) and f_y(0, 0) exist. We can also show that for any p 6= (0, 0), fx(p) and fy(p) both exist. Let p = (x0, y0) with x0y0 6= 0.

Then

F (t) = f (x0+ t, y0) = (x0+ t)y0((x0+ t)²− y₀)²

(x₀+ t)²+ y²₀ , |t| < .

Here  =px²₀+ y²₀(Why? Please think about it). Since (x0+t)y0((x0+t)²−y₀)² and (x0+ t)²+ y₀² are polynomial functions on R, they are differentiable at 0.

By the quotient rule, F is differentiable at 0 with F⁰(0) = y₀(x⁴₀+ 4x²₀y₀²− y₀⁴)

(x²₀+ y²₀)²

Similarly, we can show that fy(p) exists and fy(p) = x0(x⁴₀− 4x²₀y₀²− y⁴₀)

(x²₀+ y²₀)² . We conclude that

fx(x, y) =







0 if (x, y) = (0, 0)

y(x⁴+ 4x²y²− y⁴)

(x²+ y²)² if (x, y) 6= (0, 0).

and

fy(x, y) =







0 if (x, y) = (0, 0)

x(x⁴− 4x²y²− y⁴)

(x²+ y²)² if (x, y) 6= (0, 0).. Hence U = R² is open.

 (c) Show that

f_xy(0, 0) = lim

h→0

fx(0, h) − fx(0, 0)

h , f_yx(0, 0) = lim

h→0

fy(h, 0) − fy(0.0) h

both exist. Is it true that f_xy(0, 0) = f_yx(0, 0)?

Using the previous results, we find f_x(0, h) = −h and f_y(h, 0) = h. Thus fx(0, h) − fx(0, 0)

h = −h − 0

h = −1, f_y(h, 0) − f_y(0.0)

h = h − 0

h = 1.

We see that f_xy(0, 0) = −1 while f_yx(0, 0) = 1. Thus f_xy(0, 0) 6= f_yx(0, 0).