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

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In summary, we discussed how to prove that if $(A,<)$ is a totally ordered set and $a$ is a minimal element of $A$, then $a$ is also the least. We showed that this holds for a set of numbers, where $0$ is the minimal and least element. We then considered how to generalize this to arbitrary totally ordered sets.
  • #1
evinda
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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 $(\forall x \in A)(a \leq x)$? :confused:
 
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  • #2
evinda said:
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.

evinda said:
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.
 
  • #3
Evgeny.Makarov said:
In the second last line, $b$ and $c$ can be equal. Otherwise, yes, it is correct.
Oh yes, right! (Nod)

Evgeny.Makarov said:
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)
 

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

What does it mean to show that a is the least?

Showing that a is the least means proving that a particular value or quantity is the smallest or lowest among a given set of values or quantities.

Why is it important to show that a is the least?

It is important to show that a is the least because it allows us to identify the minimum value or quantity in a given set, which can be useful in various scientific and mathematical applications.

What methods can be used to show that a is the least?

There are several methods that can be used to show that a is the least, such as mathematical proofs, graphs, tables, or experiments. The method used will depend on the specific problem or question being investigated.

Can a be the least in more than one set?

Yes, a can be the least in more than one set. For example, if we have two sets of numbers, a can be the smallest number in both sets. However, it is important to specify which set we are referring to when saying that a is the least.

What are some common misconceptions about showing that a is the least?

One common misconception is that the smallest value in a set is always the least. This is not always true, as the least can also refer to the smallest absolute value or the smallest ratio. Another misconception is that showing a is the least is always a straightforward process, when in reality, it may require complex calculations or experiments.

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