# Infimum in Ordered Subset

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## Theorem

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

Let $R = \left({T, \preceq'}\right)$ be an ordered subset of $L$.

Let $X \subseteq T$ such that

- $X$ admits an infimum in $L$.

Then $\inf_L X \in T$ if and only if

- $X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$

## Proof

By definition of ordered subset:

- $T \subseteq S$

and

- $\forall x, y \in T: x \preceq' y \iff x \preceq y$

### Sufficient Condition

Let $\inf_L X \in T$.

By definition of infimum:

- $\inf_L X$ is lower bound for $X$ in $L$.

By definition of $\preceq'$:

- $\inf_L X$ is lower bound for $X$ in $R$.

We will prove that

- $\forall x \in T: x$ is lower bound for $X$ in $R \implies x \preceq' \inf_L X$

Let $x \in T$ such that

- $x$ is lower bound for $X$ in $R$.

By definition of $\preceq'$:

- $x$ is lower bound for $X$ in $L$.

By definition of infimum:

- $x \preceq \inf_L X$

Thus by definition of $\preceq'$:

- $x \preceq' \inf_L X$

$\Box$

Hence $X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$

$\Box$

### Necessary Condition

assume that

- $X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$

By definition of infimum:

- $\inf_R X \in T$

Thus by assumption:

- $\inf_L X \in T$

$\blacksquare$

## Sources

- Mizar article YELLOW_0:63