Equivalence of Definitions of Minimal Element

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Theorem

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

Let $T \subseteq S$ be a subset of $S$.


The following definitions of the concept of Minimal Element are equivalent:

Definition 1

An element $x \in T$ is a minimal element of $T$ if and only if:

$\forall y \in T: y \preceq x \implies x = y$


That is, the only element of $T$ that $x$ succeeds or is equal to is itself.

Definition 2

An element $x \in T$ is a minimal element of $T$ if and only if:

$\neg \exists y \in T: y \prec x$

where $y \prec x$ denotes that $y \preceq x \land y \ne x$.


That is, if and only if $x$ has no strict predecessor.


Proof

Definition 1 implies Definition 2

Let $x$ be an minimal element by definition 1.

That is:

$(1): \quad \forall y \in T: y \preceq x \implies x = y$

Aiming for a contradiction, suppose that:

$\exists y \in T: y \prec x$

Then by definition:

$y \preceq x \land x \ne y$

which contradicts $(1)$.

Thus by Proof by Contradiction:

$\nexists y \in T: y \prec x$

That is $x$ is a minimal element by definition 2.

$\Box$


Definition 2 implies Definition 1

Let $x$ be a minimal element by definition 2.

That is:

$(2): \quad \nexists y \in T: y \prec x$

Aiming for a contradiction, suppose that:

$\exists y \in T: y \preceq x: x \ne y$

That is:

$\exists y \in T: y \prec x$

which contradicts $(2)$.

Thus:

$\forall y \in T: y \preceq x \implies x = y$

Thus $x$ is a minimal element by definition 1.

$\blacksquare$


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