Equivalence of Definitions of Principal Ideal

From ProofWiki
Jump to navigation Jump to search



Theorem

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

Let $I$ be an ideal in $S$.


Then

the definitions of principal ideal are equivalent,

That means that

$\exists x \in I: x$ is upper bound for $I$

if and only if

$\exists x \in S: I = x^\preceq$

where $x^\preceq$ denotes the lower closure of $x$.


Proof

Sufficient Condition

Assume that

$\exists x \in I: x$ is upper bound for $I$

We will prove that

$I \subseteq x^\preceq$

Let $y \in I$.

By definition of upper bound:

$y \preceq x$

Thus by definition of lower closure of element:

$y \in x^\preceq$

$\Box$

We will prove that

$x^\preceq \subseteq I$

Let $y \in x^\preceq$.

By definition of lower closure of element:

$y \preceq x$

Thus by definition of lower set:

$y \in I$

$\Box$

Thus by definition of set equality:

$\exists x \in S: I = x^\preceq$

$\Box$

Necessary Condition

Assume that

$\exists x \in S: I = x^\preceq$

By definition of reflexivity:

$x \preceq x$

Thus by definition of lower closure of element:

$x \in I$

Let $y \in I$.

Thus by definition of lower closure of element:

$y \preceq x$

$\blacksquare$


Sources