bounded set

The below definitions hold for any partial order e.g. also $(M,\subset )$.

Bounded sets

Let $A\subset \mathbb{R}$
$A$ is bounded from above iff

$$\exists C\in \mathbb{R},\forall a\in A:a\le C$$

$C$ is an upper bound for $A$.
$A$ is bounded from below iff

$$\exists c\in \mathbb{R},\forall a\in A:a\ge c$$

$c$ is a lower bound for $A$.

We say that $A$ is bounded iff it is bounded from above and from below.

Infimum

The infimum (greatest lower bound) of a set $A\subset \mathbb{R}$ that is bounded from below is the largest real number $m$ such that $m$ is a lower bound for $A$:

$$\begin{array}{rlr}\text{(i)}& \forall a\in A:m\le a& \text{(m is a lower bound for A)}\\ \text{(ii)}& \forall c\in \mathbb{R}\text{ with }\forall a\in A:c\le a:c\le m& \text{(m is the greatest of all lower bounds) }\end{array}$$

Denoted by $\inf A$.
If a set is not bounded from below, we define $\inf A=-\infty$. For the empty set, we define $\inf \varnothing =+\infty$.

Supremum

The supremum (least upper bound) of a set $A\subset \mathbb{R}$ that is bounded from above is the smallest real number $M$ such that $M$ is an upper bound for $A$:

$$\begin{array}{rlr}\text{(i)}& \forall a\in A:a\le M& \text{(M is an upper bound for A)}\\ \text{(ii)}& \forall C\in \mathbb{R}\text{ with }\forall a\in A:a\le C:M\le C& \text{(M is the least of all upper bounds) }\end{array}$$

Denoted by $\sup A$.
If a set is not bounded from above, we define $\sup A=+\infty$. For the empty set, we define $\sup \varnothing =-\infty$.

What's the difference between minimum / maximum and infimum / supremum?

$\min [a,b]=\inf [a,b]=\inf [a,b)=\inf (a,b]=\inf (a,b)=a$
and
$\max [a,b]=\sup [a,b]=\sup [a,b)=\sup (a,b]=\sup (a,b)=b$
However,
$\min (a,b]$, $\min (a,b)$, $\max [a,b)$, $\max (a,b)$ do not exist, because there is no second smallest / largest element
For any candidate $x$ for $\min (a,b]$, you can always find something smaller like $\frac{a+x}{2}$, which is between $a$ and $x$ and still in the interval.
That’s exactly why the concept of infimum / supremum are useful. It’s what the elements approach but not never reach.

Does $M:=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}$ have an inf/sup/min/max?

Supremum/maximuim: $1$
Infimum: $0$
Minimum: does not exist, because zero is never reached.

Does $A:=\left\{\frac{8}{n^2}-\frac{4}{n}+1:n\in \mathbb{N}\right\}$ have an inf/sup/min/max?

Completeness axiom of the real numbers $\mathbb{R}$

Every non-empty subset $A\subset \mathbb{R}$ that is bounded from above has a supremum in $\mathbb{R}$, i.e. $\exists T\in \mathbb{R}$ such that $T=\sup A$ (similarily for infimum).

Not every subset of $\mathbb{R}$ has a supremum in $\mathbb{R}$

For example: The set $A=\{x\in \mathbb{Q}:x^2\le 2\}$ is bounded, i.e. $-\sqrt{2}\le x\le \sqrt{2}$.
So $\sup A=\sqrt{2}\notin \mathbb{Q}$, and similarily $\inf A=-\sqrt{2}\notin \mathbb{Q}$, so they do not exist in $\mathbb{Q}$.
→ Infimum and Supremum are always defined in a certain space.
We can approach $\sqrt{2}$ abritrarly closely with rational numbers, but never reach it.
But, we can define $\sup A=\sqrt{2}$ in $\mathbb{R}$, because $\sqrt{2}\in \mathbb{R}$.

Note

In $\mathbb{Z}$: bounded → has minimum and maximum (because of discreteness)
In $\mathbb{Q}$: bounded → may not have $\sup$/$\inf$
In $\mathbb{R}$: bounded → always has $\sup$/$\inf$ (completeness axiom)

A field with a partial order is complete if every $A\subset \Omega$ that is bounded from above has $\sup A\in \Omega$

The field of real numbers $(\mathbb{R},+,\cdot )$ is a complete field with linear order.
In fact, it is the unique complete field with linear order (up to isomorphism).
$\mathbb{Q}$ is not a complete field, as e.g. $\{x^2\le 2\}$ has no supremum in $\mathbb{Q}$.

Link to original

Rational numbers are dense in the reals.

a)

Transclude of archimedian-property#^814f46

b)

$$\forall ϵ>0,\exists n\in \mathbb{N}:\frac{1}{n}<ϵ$$

c)

$$\forall x,y\in \mathbb{R},x<y,\exists \frac{m}{n}\in \mathbb{Q}:x\le \frac{m}{n}\le y$$

→ If I pick two real numbers, there is always a rational number in between them.