Upper Bounds are Equivalent implies Suprema are equal

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Theorem

Let $L = \struct {S, \preceq}$ be an ordered set.

Let $X, Y$ be subsets of $S$.

Assume that

$X$ admits a supremum

and

$\forall x \in S: x$ is upper bound for $X \iff x$ is upper bound for $Y$


Then $\sup X = \sup Y$


Proof

We will prove that

$\forall b \in S: b$ is upper bound for $Y \implies \sup X \preceq b$

Let $b \in S$ such that

$b$ is upper bound for $Y$.

By assumption:

$b$ is upper bound for $X$.

Thus by definition of supremum:

$\sup X \preceq b$

$\Box$


By definition of supremum:

$\sup X$ is upper bound for $X$.

By assumption:

$\sup X$ is upper bound for $Y$.

Thus by definition of supremum:

$\sup X = \sup Y$

$\blacksquare$


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