Let the set S be given by S = {1/n | n is natural}. Describe maximum, minimum, supremum (least upper bound), and infimum (greatest lower bound) for S (they may not exist).

Answer: We have that [tex] max\{S\} = sup\{S\} = 1 [/tex] and [tex]inf\{S\}=0[/tex] but the minimum of S doesn't exist.Step-by-step explanation: To find the supremum (or least upper bound) of a set, we first need to prove that an upper bound exists. So we must find a number [tex] \alpha [/tex] such that [tex]\alpha \geq x[/tex] for all [tex]x \in S[/tex]. Note that [tex]1 \leq k[/tex] for every [tex]k \in \mathbb{N}[/tex]. Thus, for each [tex]k \in \mathbb{N}[/tex], we have that [tex]\frac{1}{k} \leq 1[/tex]. This proves that [tex] 1 [/tex] is an upper bound for [tex] S [/tex]. Since the set is bounded above, then we have that the supremum must exist. Notice that [tex]1 \in S[/tex] which means 1 is the highest number in [tex]S[/tex]. So we conclude that [tex]max\{S\} = sup\{S\} = 1[/tex]. Now to find the infimum, we must start by finding a lower bound for [tex]S[/tex]. This will turn out to be quite easy as we just need to notice that if [tex]k \in \mathbb{N}[/tex] then [tex]\frac{1}{k} > 0[/tex]. Now we have our lower bound! This means S is bounded below and thus its infimum exists. Let us claim that [tex]inf\{S\}=0[/tex]. As per the definition of infimum, we need to prove that any other lower bound would be smaller than 0. This is not so simple so think of it this way: if we find a number greater than 0 that is also a lower bound for S, then 0 is not the greatest lower bound. Then we will set out to prove that any number greater than 0 is not a lower bound. Let us assume then that [tex]L > 0[/tex]. Then there is some [tex]n \in \mathbb{N}[/tex] such that [tex]\frac{1}{n} < L[/tex]. Since [tex]\frac{1}{n} \in S [/tex], then [tex] L [/tex] is not a lower bound for [tex]S[/tex]. This shows that [tex]inf\{S\}=0[/tex]. Recall that when a set has its infimum as an element, we call it minimum but since we've already proven that every element of S is strictly greater than 0, then the minimum of S doesn't exist.