# Supremum + Infimum

Example 1. Prove
$\sup\left\lbrace \dfrac{2n+1}{n+1} \ : \ n\in\mathbb{N}\right\rbrace = 2$
​.
Ex 1: Comment
Hint
Visual
Formal Write-Up
Follow-Up
We cannot use L'Hôpital's rule in this setting as we don't yet have derivatives, let alone use limits! Instead, we must work directly from the definition of supremum.
Show the sequence is bounded above by a number, and then show each number less than that bound is strictly less than a point in the set.
Example for showing
$2-\varepsilon$
is not an upper bound of
$\mathcal{C}$
when
$\varepsilon > 0$
.​
In general, we must assume
$\varepsilon$
​ is an arbitrary number (i.e. we do not get to pick
$\varepsilon$
​ in this problem).
Let
$\varepsilon \in (0,1)$
​ be given and let
$\mathcal{C}$
​ denote the provided set. It suffices to show
$2$
​ is an upper bound of
$\mathcal{C}$
​ and, by the arbitrariness of
$\varepsilon$
​, also show
$2-\varepsilon$
​ is not an upper bound of
$\mathcal{C}$
​. First, observe
 ​​
This shows
$2$
​ is an upper bound for
$\mathcal{C}$
​. By the Archimedean property of
$\mathbb{R}$
​, there is
$k\in\mathbb{N}$
​ such that
 ​$k > \dfrac{1-\varepsilon}{\varepsilon} > 0.$​
This implies
 ​$\varepsilon k > 1-\varepsilon \quad \implies \quad 1 + \varepsilon k > 2-\varepsilon,$​
which can be rearranged to obtain
 ​$1+2k > (2-\varepsilon)(k+1) \quad \implies \quad \dfrac{2k+1}{k+1} > 2-\varepsilon.$​
Since
$k$
is a natural number, it follows that
$(2k+1)/(k+1) \in \mathcal{C}$
, and so the above inequality shows
$2-\varepsilon$
is not an upper bound for
$\mathcal{C}.$
Hence we conclude 2 is the least upper bound of
$\mathcal{C}$
, as desired. ​
 ​$\blacksquare$​
The formal write up likely feels mysterious. It is straightforward to verify the inequalities are correct, but how does this reveal the thought process for coming up with the proof? In this case, we simply start with the inequality we want, and then work backwards. That is, suppose there is
$k$
​ such that
 ​$\dfrac{2k+1}{k+1} > 2-\varepsilon$​
Rearranging this inequality until we can isolate
$k$
​ enables us to put a lower bound on the size of
$k$
​. Moreover, we can use the Archimedean property of
$\mathbb{R}$
​ to verify the existence of such a natural number.
Example 2. Let
$\mathcal{S}$
and
$\mathcal{T}$
be nonempty subsets of
$\mathbb{R}$
such that
$s\leq t$
​ for all
$s\in\mathcal{S}$
​ and
$t\in\mathcal{T}.$
Prove
Ex 2: Comment
Hint
Visual
Formal Write-Up
Since the sets are nonempty and the bounds are provided, we may simply use the fact the supremum of
$\mathcal{S}$
​ and the infimum of
$\mathcal{T}$
​ exist and are finite.
First consider an arbitrary element of
$\mathcal{T}$
​, and relate this to an arbitrary element of
$\mathcal{S}$
​.
Labeled diagram of
$\mathcal{S}$
and
$\mathcal{T}$
with sample choices of
$s$
and
$t$
. First note
$t$
is an upper bound of
$\mathcal{S}$
, and so
$\sup(\mathcal{S})\leq t$
. This implies
$\sup(\mathcal{S})$
is a lower bound for
$\mathcal{T}$
, making
$\sup(\mathcal{S})\leq\inf(\mathcal{T})$
.
Let
$t\in \mathcal{T}$
be given. By hypothesis,
$s \leq t$
for all
$s\in\mathcal{S}$
, and so
$t$
is an upper bound for
$\mathcal{S}.$
This implies
$\sup \mathcal{S}\leq t$
since the supremum of
$\mathcal{S}$
is its least upper bound. Because
$t$
was arbitrarily chosen in
$\mathcal{T}$
, we deduce
$\sup \mathcal{S} \leq t$
for all
$t\in \mathcal{T}$
, i.e.
$\sup \mathcal{S}$
is a lower bound for
$\mathcal{T}$
. As
$\inf \mathcal{T}$
is the greatest lower bound of
$\mathcal{T}$
, we conclude
$\sup \mathcal{S}\leq \inf \mathcal{T}$
, and the proof is complete.
 ​$\blacksquare$​
Example 3. Show
$0$
is the infimum of
$\mathcal{S} = \{ 3/k : k\in\mathbb{N}\}.$
Hint
Visual
Formal Write-Up
The proof can be broken into two parts, showing
$\inf\mathcal{S}\geq 0$
and showing
$\inf \mathcal{S} \leq 0$
.
Example animation for showing
$\varepsilon=0.06$
​ is not a lower bound for
$\mathcal{S}$
​. The core idea is, no matter the value of
$\varepsilon$
​, we can by the Archimedean property of
$\mathbb{R}$
always find a natural number
$k$
​ big enough to ensure
$3/k<\varepsilon$
​.
We claim
$\inf \mathcal{S} = 0$
and verify this by first showing
$\inf \mathcal{S} \geq 0$
and then
$\inf \mathcal{S}\leq 0.$
Consider any element
$\alpha \in \mathcal{S}$
. By the definition of
$\mathcal{S}$
, there is some
$n\in \mathbb{N}$
such that
$\alpha = 3/n$
. Since
$n \in \mathbb{N}$
, we have
$n>0$
, which implies
$\alpha = 3/n > 0$
. Because
$\alpha$
was arbitrarily chosen in
$\mathcal{S}$
, it follows that
$0$
is a lower bound for
$\mathcal{S}$
. Thus,
$\inf \mathcal{S} \geq 0$
since the infimum of
$\mathcal{S}$
is, by definition, the greatest lower bound.
Conversely, we now show
$\inf \mathcal{S} \leq 0.$
$\inf \mathcal{S} > 0$
. This implies there is
$\varepsilon > 0$
such that
$\inf \mathcal{S}= \varepsilon$
. By the Archimedean Property of
$\mathbb{R}$
, there is
$k\in \mathbb{N}$
such that
$k\varepsilon > 3$
, and so
$\varepsilon > 3/k$
. However,
$3/k\in \mathcal{S}$
$\varepsilon$
is a lower bound of
$\mathcal{S}$
. This contradiction proves the initial assumption was false. Hence
$\inf \mathcal{S} \leq 0$
, and, together with the fact
$\inf \mathcal{S} \geq 0$
, we conclude
$\inf \mathcal{S} = 0.$
 ​$\blacksquare$​
Example 4. Let
$\mathcal{C}$
be a nonempty subset of
$\mathbb{R}$
that is bounded from below. Prove
$\inf \mathcal{C} = - \sup(-\mathcal{C})$
.
Formal Write-Up
Follow-Up
We prove the relation by using two inequalities. Let
$\beta \in -\mathcal{C}$
be given. Then there exists
$\alpha \in \mathcal{C}$
such that
$\beta = -\alpha.$
Because
$\inf \mathcal{C}$
is a lower bound for
$\mathcal{C}$
,
$\alpha \geq \inf \mathcal{C}$
. Flipping signs reveals
$\beta =-\alpha \leq -\inf \mathcal{C}$
. Since
$\beta \in -\mathcal{C}$
was chosen arbitrarily, it follows that
$-\inf \mathcal{C}$
is an upper bound for the set
$-\mathcal{C}$
. The fact
$\sup(-\mathcal{C})$
is the least upper bound for
$-\mathcal{C}$
implies
$\sup(-\mathcal{C}) \leq -\inf \mathcal{C}$
, and so
$-\sup(-\mathcal{C}) \geq \inf \mathcal{C}$
.
Now let
$\gamma \in \mathcal{C}$
be given. Observe
$-\gamma \in -\mathcal{C}$
. Thus,
$-\gamma \leq \sup (-\mathcal{C})$
by definition of the supremum. Thus,
$\gamma \geq -\sup(-\mathcal{C})$
. Because
$\gamma \in \mathcal{C}$
was chosen arbitrarily, it follows that
$-\sup(-\mathcal{C})$
is a lower bound for
$\mathcal{C}$
. The fact
$\inf \mathcal{C}$
is the greatest lower bound for
$\mathcal{C}$
implies
$-\sup(-\mathcal{C}) \leq \inf \mathcal{C}$
. Combining our results reveals
$-\sup(-\mathcal{C}) \leq \inf \mathcal{C} \leq -\sup(-\mathcal{C})$
, from which the result follows.
 ​$\blacksquare$​
It is a good habit to let the reader know how you plan to tackle the problem with a short sentence or phrase. For example, contradiction proofs often begin with the phrase By way of contradiction, suppose...''