- #1
evinda
Gold Member
MHB
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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 $(\forall x \in A)(a \leq x)$?
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 $(\forall x \in A)(a \leq x)$?