Problem 2. Show that the “closed” ball Br(x0) = {x Rn ; ||x x0|| r} in Rn center

Question

Problem 2. Show that the “closed” ball Br(x0) = {x Rn ; ||x x0|| r} in Rn centered at x0 Rn of radius r is closed (in the sense that its complement is open).

Problem 3. Verify the following:

(i) The union of any number of open sets is open;

(ii) The intersection of finitely many open sets is open (can you give an example

in R for which the intersection of infinitely many open sets is closed?)

Write down the corresponding statements for closed sets and proof them (recall that a set A is closed, if its complement Rn A is open; what is the complement of a union of sets? Of the intersection of sets?)

Problem 4. Let {xk} and {yk} be sequences in Rn which converge to L1 and L2 respectively. Verify the limit rules:

(i) The sum/difference sequence {xk ± yk} converges to L1 ± L2. (ii) The product sequence {xkyk} converges to L1L2.

Problem 5. Determine the limits of the following sequences as k :

(i) xk = (cos(1/k),ek) in R2
(ii) xk =(1/k,log(1+1/k),1/k+7)inR3

(iii)xk=(k , k ,..., k )inRn k+1 k+2 k+n

Problem 6. Verify that the n-dimensional sphere
Srn(x0) = {x Rn+1 ; ||x x0|| = r}

of radius r > 0 and center x0 Rn+1 is a closed (in fact compact) subset of Rn+1. Problem 7. Calculate the closures of the sets (0, 1]{2}, R2{0}, and {(x1, x2, x3)

R3 ; x23 > 1} and provide a proof of your answer.
Problem 8. Verify that the closed interval [a, b] R is compact.

Explanation / Answer

complement of closed ball is open

what we need to prove is if a point y is such that its distance from x_0 > r

then there exists a neighbour hood of y (open ball with center y ) such that all points in the neighbourhood are at distance > r from x_0

take any point y outside the closed ball (in the complement)

since y is not in the ball

d(y,x_0) > r

let d(y,x_0) = s

s>r

there exists p such that s>p>r

now s-p >0

now consider the open ball with center y and radius (s-p)

for all the points z in the ball

d(z,y) < (s-p)

so d(z,x_0) > r

because if not then d(z,y) <=  d(z,x_0) +d(z,y) <= r + s - p < s (but d(z,y) = s)

So that means for all points z in the open ball around y d(z,x_0) > r

hence they all lie in the complement

this is true for any y in the complement

hence the complement is open.

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