a N b Qn0 1 c The Cantor set d 1 12 132 1n2 n E N Soluti

(a) N (b) Qn[0, 1] (c) The Cantor set. (d) {1 + 1/2? + 1/32 + + 1/n2, n E N).

Solution

a)The set N is closed, but it is not compact. The sequence (n) in N has no convergent subsequence since every subsequence diverges to infinity

(b)The set Q of rational numbers has no interior or isolated points, and every real number is both a boundary and accumulation point of Q.

which is not compact. Define the sequence (a_n) recursively by setting a₁ = 1 and a_n+1 = 1/( 4a_n) for all n = 1, 2, 3, . . .. Each an is in the set Q [0, 1] but the sequence (a_n) converges to 2 3, which is irrational and hence not in the set Q [0, 1].

(c)We define a nested sequence (F_n) of sets F_n [0, 1] as follows. First, we remove the middle-third from [0, 1] to get F₁ = [0, 1] \\ (1/3, 2/3), or F₁ = I₀ I₁, I₀ = [0, 1 /3] , I₁ = [ 2 /3 , 1] .

Next, we remove middle-thirds from I₀ and I₁, which splits I₀ \\ (1/9, 2/9) into I₀₀ I₀₁ and I₁ \\ (7/9, 8/9) into I₁₀ I₁₁, to get F₂ = I₀₀ I₀₁ I₁₀ I₁₁, I₀₀ = [0, 1/ 9] , I₀₁ = [2 /9 , 1 /3] , I₁₀ = [2/ 3 , 7 /9] , I11 = [8/ 9 , 1].

The Cantor set C is clearly nonempty since the endpoints as, bs of Is are contained in F_n for every finite binary sequence s and every n N. These endpoints form a countably infinite set. What may be initially surprising is that there are uncountably many other points in C that are not endpoints.

The Cantor set is compact

d){1+1/2²⁺1/3²+.....+1/n² } is not compact as the limit does not tends to converge.the limt vale does not belong to the set.

e) {1+1/2+2/3,+3/4 +4/5, . . .} is Compact as the limit tends to converge.