# Increasing Union of Ideals is Ideal/Chain

## Theorem

Let $R$ be a ring.

Let $\left({P, \subseteq}\right)$ be the ordered set consisting of all ideals of $R$, ordered by inclusion.

Let $\left\{{I_\alpha}\right\}_{\alpha \in A}$ be a non-empty chain of ideals in $P$.

Let $\displaystyle I = \bigcup_{\alpha \in A} I_\alpha$ be their union.

Then $I$ is an ideal of $R$.

## Proof

##### Property 1: $0 \in I$

Since $\left\{{I_\alpha}\right\}_{\alpha \in A}$ is non-empty chain, it must contain some ideal $I_\beta$

Since $I_\beta$ is an ideal, $0 \in I_\beta$.

Thus $0 \in I$.

##### Property 2: $x \in I \implies -x \in I$

If $x \in I$, then $x \in I_\beta$ for some $\beta\in A$.

Since $I_\beta$ is an ideal, $-x \in I_\beta$.

Thus $-x \in I$.

##### Property 3: $x, y \in I \implies x + y \in I$

If $x,y \in I$, then $x \in I_\beta$ for some $\beta\in A$, and $y \in I_\gamma$ for some $\gamma \in A$.

Since $\left\{{I_\alpha}\right\}_{\alpha \in A}$ is totally ordered, $I_\beta \subseteq I_\gamma$ or $I_\gamma\subseteq I_\beta$.

Without loss of generality, we can assume $I_\beta \subseteq I_\gamma$, which gives us $x,y \in I_\gamma$.

Since $I_\gamma$ is an ideal, $x + y \in I_\gamma$.

Thus $x + y \in I$.

##### Property 4: $x \in I \land r \in R \implies rx, xr \in I$

If $x \in I$, then $x \in I_\beta$ for some $\beta\in A$.

If $r \in R$, then since $I_\beta$ is an ideal, $rx, xr \in I_\beta$.

Thus $rx, xr \in I$.

Since $I$ satisfies these 4 properties, it is a ideal of $R$.

$\blacksquare$