MHB Can a Minimal Element in a Totally Ordered Set be Considered the Least?

[evinda]

Hello! (Wave)

I want to show that if $(A,<)$ is a totally ordered set, then if $a$ is a minimal element of $A$ then $a$ is also the least.

$a$ is a minimal element of $A$.
That means that $(\forall x \in A)(x \leq a \rightarrow x=a)$

We want to show that $(\forall x \in A)(a \leq x)$.

Since $(A,<)$ is a totally ordered set we have that $( \forall b \in A)(\forall c \in A) (b<c \lor c<b)$.

Is it right so far? (Thinking)

How could we continue in order to show that $(\forall x \in A)(a \leq x)$?

 

[Evgeny.Makarov]

I want to show that if $(A,<)$ is a totally ordered set, then if $a$ is a minimal element of $A$ then $a$ is also the least.

$a$ is a minimal element of $A$.
That means that $(\forall x \in A)(x \leq a \rightarrow x=a)$

We want to show that $(\forall x \in A)(a \leq x)$.

Since $(A,<)$ is a totally ordered set we have that $( \forall b \in A)(\forall c \in A) (b<c \lor c<b)$.

Is it right so far?

In the second last line, $b$ and $c$ can be equal. Otherwise, yes, it is correct.

How could we continue in order to show that $(\forall x \in A)(a \leq x)$?

Prove this when $A$ is a set of numbers, then generalize to arbitrary totally ordered sets.

 

[evinda]

In the second last line, $b$ and $c$ can be equal. Otherwise, yes, it is correct.

Oh yes, right! (Nod)

Prove this when $A$ is a set of numbers, then generalize to arbitrary totally ordered sets.

So we could pick for example $A=n \in \omega$ and consider the relation $(A,\subset)$, right?

$$n=\{0,1,2, \dots,n-1\}$$

We see that the minimal element is $0=\varnothing$ since there is no element $b\in \text{ family of subsets of n}$ such that $b \subsetneq 0$.
This element is also the least since $\forall m \in \text{ family of subsets of n}$ it holds that $\varnothing \subset m$, right? (Thinking)So in this case the proposition holds. But how could we generalize it to arbitrary ordered sets? (Thinking)