Min Operation Yields Infimum of Parameters/General Case

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {S, \preceq}$ be a totally ordered set.

Let $x_1, x_2, \dots ,x_n \in S$ for some $n \in \N_{>0}$.


Then:

$\min \set {x_1, x_2, \dotsc, x_n} = \inf \set {x_1, x_2, \dotsc, x_n}$

where:

$\min$ denotes the min operation
$\inf$ denotes the infimum.


Proof

We will prove the result by induction on the number of operands $n$.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:

$\min \set {x_1, x_2, \dotsc, x_n} = \inf \set {x_1, x_2, \dotsc, x_n}$


Basis for the Induction

$\map P 1$ is the case:

$\min \set {x_1} = \inf \set {x_1}$


By definition of the min operation:

$\min \set {x_1} = x_1$

From Infimum of Singleton:

$\inf \set {x_1} = x_1$

So:

$\min \set {x_1} = \inf \set {x_1} = x_1$


Thus $\map P 1$ is seen to hold.

This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$\min \set {x_1, x_2, \dots ,x_k} = \inf \set {x_1, x_2, \dots ,x_k}$


from which it is to be shown that:

$\min \set {x_1, x_2, \dots ,x_k, x_{k+1}} = \inf \set {x_1, x_2, \dots ,x_k, x_{k+1}}$


Induction Step

This is the induction step.

Now:

\(\ds \min \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }\) \(=\) \(\ds \min \set {\min \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }\) Definition of Min Operation (General Case)
\(\ds \) \(=\) \(\ds \min \set {\inf \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }\) Induction hypothesis


As $\struct {S, \preceq}$ is a totally ordered set, all elements of $S$ are comparable by $\preceq$.

Therefore there are two cases to consider:


Case 1: $\inf \set {x_1, x_2, \dotsc, x_k} \preceq x_{k + 1}$

Let:

$\inf \set {x_1, x_2, \dotsc, x_k} \preceq x_{k + 1}$.

By definition of the min operation:

$\min \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \inf \set {x_1, x_2, \dotsc, x_k}$


By definition, $\inf \set {x_1, x_2, \dotsc, x_k}$ is a lower bound of $\set {x_1, x_2, \dotsc, x_k}$.

That is:

$\forall 1 \le i \le k : \inf \set {x_1, x_2, \dotsc, x_k} \preceq x_i$

Thus:

$\forall 1 \le i \le k + 1 : \inf \set {x_1, x_2, \dotsc, x_k} \preceq x_i$

So $\inf \set {x_1, x_2, \dots ,x_k}$ is a lower bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.


Let $y$ be any other lower bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.

Then:

$y$ is a lower bound of $\set {x_1, x_2, \dotsc, x_k}$.

By definition of the infimum:

$y \preceq \inf \set {x_1, x_2, \dotsc, x_k}$


It has been shown that the infimum of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$ is:

$\inf \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \inf \set {x_1, x_2, \dotsc, x_k}$


Thus it follows:

$\min \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \inf \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$


Case 2: $x_{k + 1} \preceq \inf \set {x_1, x_2, \dotsc, x_k}$

Let:

$x_{k + 1} \preceq \inf \set {x_1, x_2, \dotsc, x_k}$.

By definition of the min operation:

$\min \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = x_{k + 1}$


By definition, $\inf \set {x_1, x_2, \dotsc, x_k}$ is a lower bound of $\set {x_1, x_2, \dotsc, x_k}$.

That is:

$\forall 1 \le i \le k : \inf \set {x_1, x_2, \dotsc, x_k} \preceq x_i$


By definition of an ordering:

$\preceq$ is transitive.

Thus:

$\forall 1 \le i \le k : x_{k + 1} \preceq x_i $

By definition of an ordering:

$\preceq$ is reflexive.

Thus:

$x_{k + 1} \preceq x_{k + 1}$

It follows that $x_{k + 1}$ is a lower bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.


Let $y$ be any other lower bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.

Then:

$y \preceq x_{k + 1}$.

It has been shown that

$\inf \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = x_{k + 1}$.


Thus it follows:

$\min \set {x_1, x_2, \dotsc, x_k, x_{k + 1}} = \inf \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$


In either case, the result holds.


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\min \set {x_1, x_2, \dotsc, x_n} = \inf \set {x_1, x_2, \dotsc, x_n}$

$\blacksquare$