Minimal Element in Toset is Unique and Smallest

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Theorem

Let $\left({S, \preceq}\right)$ be a totally ordered set.

Let $m$ be a minimal element of $\left({S, \preceq}\right)$.


Then:

$(1): \quad m$ is the smallest element of $\left({S, \preceq}\right)$.
$(2): \quad m$ is the only minimal element of $\left({S, \preceq}\right)$.


Proof

By definition of minimal element:

$\forall y \in S: y \preceq m \implies m = y$

As $\left({S, \preceq}\right)$ is a totally ordered set, by definition $\preceq$ is connected.

That is:

$\forall x, y \in S: y \preceq x \lor x \preceq y$

It follows that:

$\forall y \in S: m = y \lor m \preceq y$

But as $m = y \implies m \preceq y$ by definition of $\preceq$, it follows that:

$\forall y \in S: m \preceq y$

which is precisely the definition of smallest element.

Hence $(1)$ holds.

$\Box$


Suppose $m_1$ and $m_2$ are both minimal elements of $S$.

By $(1)$ it follows that both are smallest elements.

It follows from Smallest Element is Unique that $m_1 = m_2$.

That is, $(2)$ holds.

$\blacksquare$


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