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Supremum + Infimum

Example 1. Prove
sup{2n+1n+1 : nN}=2\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
2ε2-\varepsilon
is not an upper bound of
C\mathcal{C}
when
ε>0\varepsilon > 0
.​
In general, we must assume
ε\varepsilon
​ is an arbitrary number (i.e. we do not get to pick
ε\varepsilon
​ in this problem).
Let
ε(0,1)\varepsilon \in (0,1)
​ be given and let
C\mathcal{C}
​ denote the provided set. It suffices to show
22
​ is an upper bound of
C\mathcal{C}
​ and, by the arbitrariness of
ε\varepsilon
​, also show
2ε2-\varepsilon
​ is not an upper bound of
C\mathcal{C}
​. First, observe
This shows
22
​ is an upper bound for
C\mathcal{C}
​. By the Archimedean property of
R\mathbb{R}
​, there is
kNk\in\mathbb{N}
​ such that
k>1εε>0.k > \dfrac{1-\varepsilon}{\varepsilon} > 0.
This implies
εk>1ε    1+εk>2ε,\varepsilon k > 1-\varepsilon \quad \implies \quad 1 + \varepsilon k > 2-\varepsilon,
which can be rearranged to obtain
1+2k>(2ε)(k+1)    2k+1k+1>2ε.1+2k > (2-\varepsilon)(k+1) \quad \implies \quad \dfrac{2k+1}{k+1} > 2-\varepsilon.
Since
kk
is a natural number, it follows that
(2k+1)/(k+1)C(2k+1)/(k+1) \in \mathcal{C}
, and so the above inequality shows
2ε2-\varepsilon
is not an upper bound for
C.\mathcal{C}.
Hence we conclude 2 is the least upper bound of
C\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
kk
​ such that
2k+1k+1>2ε\dfrac{2k+1}{k+1} > 2-\varepsilon
Rearranging this inequality until we can isolate
kk
​ enables us to put a lower bound on the size of
kk
​. Moreover, we can use the Archimedean property of
R\mathbb{R}
​ to verify the existence of such a natural number.
Example 2. Let
S\mathcal{S}
and
T\mathcal{T}
be nonempty subsets of
R\mathbb{R}
such that
sts\leq t
​ for all
sSs\in\mathcal{S}
​ and
tT.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
S\mathcal{S}
​ and the infimum of
T\mathcal{T}
​ exist and are finite.
First consider an arbitrary element of
T\mathcal{T}
​, and relate this to an arbitrary element of
S\mathcal{S}
​.
Labeled diagram of
S\mathcal{S}
and
T\mathcal{T}
with sample choices of
ss
and
tt
. First note
tt
is an upper bound of
S\mathcal{S}
, and so
sup(S)t\sup(\mathcal{S})\leq t
. This implies
sup(S)\sup(\mathcal{S})
is a lower bound for
T\mathcal{T}
, making
sup(S)inf(T)\sup(\mathcal{S})\leq\inf(\mathcal{T})
.
Let
tTt\in \mathcal{T}
be given. By hypothesis,
sts \leq t
for all
sSs\in\mathcal{S}
, and so
tt
is an upper bound for
S.\mathcal{S}.
This implies
supSt\sup \mathcal{S}\leq t
since the supremum of
S\mathcal{S}
is its least upper bound. Because
tt
was arbitrarily chosen in
T\mathcal{T}
, we deduce
supSt\sup \mathcal{S} \leq t
for all
tTt\in \mathcal{T}
, i.e.
supS\sup \mathcal{S}
is a lower bound for
T\mathcal{T}
. As
infT\inf \mathcal{T}
is the greatest lower bound of
T\mathcal{T}
, we conclude
supSinfT\sup \mathcal{S}\leq \inf \mathcal{T}
, and the proof is complete.
\blacksquare
Example 3. Show
00
is the infimum of
S={3/k:kN}.\mathcal{S} = \{ 3/k : k\in\mathbb{N}\}.
Hint
Visual
Formal Write-Up
The proof can be broken into two parts, showing
infS0\inf\mathcal{S}\geq 0
and showing
infS0\inf \mathcal{S} \leq 0
.
Example animation for showing
ε=0.06\varepsilon=0.06
​ is not a lower bound for
S\mathcal{S}
​. The core idea is, no matter the value of
ε\varepsilon
​, we can by the Archimedean property of
R\mathbb{R}
always find a natural number
kk
​ big enough to ensure
3/k<ε3/k<\varepsilon
​.
We claim
infS=0\inf \mathcal{S} = 0
and verify this by first showing
infS0\inf \mathcal{S} \geq 0
and then
infS0.\inf \mathcal{S}\leq 0.
Consider any element
αS\alpha \in \mathcal{S}
. By the definition of
S\mathcal{S}
, there is some
nNn\in \mathbb{N}
such that
α=3/n\alpha = 3/n
. Since
nNn \in \mathbb{N}
, we have
n>0n>0
, which implies
α=3/n>0\alpha = 3/n > 0
. Because
α\alpha
was arbitrarily chosen in
S\mathcal{S}
, it follows that
00
is a lower bound for
S\mathcal{S}
. Thus,
infS0\inf \mathcal{S} \geq 0
since the infimum of
S\mathcal{S}
is, by definition, the greatest lower bound.
Conversely, we now show
infS0.\inf \mathcal{S} \leq 0.
By way of contradiction, suppose
infS>0\inf \mathcal{S} > 0
. This implies there is
ε>0\varepsilon > 0
such that
infS=ε\inf \mathcal{S}= \varepsilon
. By the Archimedean Property of
R\mathbb{R}
, there is
kNk\in \mathbb{N}
such that
kε>3k\varepsilon > 3
, and so
ε>3/k\varepsilon > 3/k
. However,
3/kS3/k\in \mathcal{S}
, which contradicts the fact
ε\varepsilon
is a lower bound of
S\mathcal{S}
. This contradiction proves the initial assumption was false. Hence
infS0\inf \mathcal{S} \leq 0
, and, together with the fact
infS0\inf \mathcal{S} \geq 0
, we conclude
infS=0.\inf \mathcal{S} = 0.
\blacksquare
Example 4. Let
C\mathcal{C}
be a nonempty subset of
R\mathbb{R}
that is bounded from below. Prove
infC=sup(C)\inf \mathcal{C} = - \sup(-\mathcal{C})
.
Formal Write-Up
Follow-Up
We prove the relation by using two inequalities. Let
βC\beta \in -\mathcal{C}
be given. Then there exists
αC\alpha \in \mathcal{C}
such that
β=α.\beta = -\alpha.
Because
infC\inf \mathcal{C}
is a lower bound for
C\mathcal{C}
,
αinfC\alpha \geq \inf \mathcal{C}
. Flipping signs reveals
β=αinfC\beta =-\alpha \leq -\inf \mathcal{C}
. Since
βC\beta \in -\mathcal{C}
was chosen arbitrarily, it follows that
infC-\inf \mathcal{C}
is an upper bound for the set
C-\mathcal{C}
. The fact
sup(C)\sup(-\mathcal{C})
is the least upper bound for
C-\mathcal{C}
implies
sup(C)infC\sup(-\mathcal{C}) \leq -\inf \mathcal{C}
, and so
sup(C)infC-\sup(-\mathcal{C}) \geq \inf \mathcal{C}
.
Now let
γC\gamma \in \mathcal{C}
be given. Observe
γC-\gamma \in -\mathcal{C}
. Thus,
γsup(C)-\gamma \leq \sup (-\mathcal{C})
by definition of the supremum. Thus,
γsup(C)\gamma \geq -\sup(-\mathcal{C})
. Because
γC\gamma \in \mathcal{C}
was chosen arbitrarily, it follows that
sup(C)-\sup(-\mathcal{C})
is a lower bound for
C\mathcal{C}
. The fact
infC\inf \mathcal{C}
is the greatest lower bound for
C\mathcal{C}
implies
sup(C)infC-\sup(-\mathcal{C}) \leq \inf \mathcal{C}
. Combining our results reveals
sup(C)infCsup(C)-\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...''