Finite Subsets form Ideal

Theorem
Let $X$ be a set.

Let $\mathit{Fin}\left({X}\right)$ be the set of all finite subsets of $X$.

Then $\mathit{Fin}\left({X}\right)$ is ideal in $\left({\mathcal P\left({X}\right), \subseteq}\right)$

where $\mathcal P\left({X}\right)$ denotes the power set of $X$.

Non-Empty
By Empty Set is Subset of All Sets:
 * $\varnothing \subseteq X$ abd $\varnothing$ is finite.

By definition of $\mathit{Fin}$:
 * $\varnothing \in \mathit{Fin}\left({X}\right)$

Thus by definition:
 * $\mathit{Fin}\left({X}\right)$ is non-empty.

Directed
Yhis follows from Finite Subsets form Directed Set.

Lower
Let $x \in \mathit{Fin}\left({X}\right)$, $y \in \mathcal P\left({X}\right)$ such that
 * $y \subseteq x$

By definition of $\mathit{Fin}$:
 * $x$ is a finite set.

By Subset of Finite Set is Finite:
 * $y$ is a finite set.

By definition of power set:
 * $y \subseteq X$

Thus by definition of $\mathit{Fin}$:
 * $y \in \mathit{Fin}\left({X}\right)$

Hence $\mathit{Fin}\left({X}\right)$ is ideal in $\left({\mathcal P\left({X}\right), \subseteq}\right)$