1 Compute with proofs the suprema and infima of the followin

1) Compute, with proofs, the suprema and infima of the following
sets. Keep in mind that 0 is not in N.

a) {(n/m+n):m,n in N}

b) {(n/2n+1): n in N}

2) Compute, without proofs, the suprema and infima of the following
sets. Keep in mind that 0 is not in N

a) {((4+x)/x) : x>= 1}

b) {(sqrt(x+1)/x) : x>=2}

a) A = {n/(m + n) : m, n N}.

N = {1, 2, 3, . . .}, it’s clear that n/(m+n) n/(1+n) < 1.

Of course n 1+n gets arbitrarily close to 1, so sup A = 1.

Likewise, since the numerator is always positive, n/(m+n) > 0, but, when n = 1, the fraction 1/(m+1) gets arbitrarily close to 0, so inf A = 0.

Note that n/(2n+1) < 1/2 ; the left hand side gets arbitrarily close to 1/2 , so sup A = 1/2

Also (again using the convention N = {1, 2, 3, . . .}), we know that n/(2n+1) 1/3

since 1/3 A, clearly inf A = 1/3 .

a) A = {((4+x)/x) : x>= 1}

when x = 1; (4+x)/x = 5 which is the highest value for set A

Hence sup A = 5

As x increases, the value of (4+x)/x gets closer to 1 but always remain above it as x>=1

Therefore, inf A = 1

b) A = {(sqrt(x+1)/x) : x>=2}

For x = 2, sqrt(x+1)/x = sqrt(3)/2 which is the highest value of set A

Hence sup A = sqrt(3)/2

As x increases, sqrt(x+1)/x gets closer to 1 but always remain above 1 as sqrt(x+1) > x

Therefore, inf A = 1