Question

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

Answer 1

Answer: We have that
`max\{S\} = sup\{S\} = 1` and
`inf\{S\}=0` 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
`\alpha` such that
`\alpha \geq x` for all
`x \in S`. Note that
`1 \leq k` for every
`k \in \mathbb{N}`. Thus, for each
`k \in \mathbb{N}`, we have that
`(1)/(k) \leq 1`. This proves that
`1` is an upper bound for
`S`. Since the set is bounded above, then we have that the supremum must exist. Notice that
`1 \in S` which means 1 is the highest number in
`S`. So we conclude that
`max\{S\} = sup\{S\} = 1`.

Now to find the infimum, we must start by finding a lower bound for
`S`. This will turn out to be quite easy as we just need to notice that if
`k \in \mathbb{N}` then
`(1)/(k) > 0`. Now we have our lower bound! This means S is bounded below and thus its infimum exists. Let us claim that
`inf\{S\}=0`. 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
`L > 0`. Then there is some
`n \in \mathbb{N}` such that
`(1)/(n) < L`. Since
`(1)/(n) \in S`, then
`L` is not a lower bound for
`S`. This shows that
`inf\{S\}=0`. 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.