Supremum of Set of Real Numbers is at least Supremum of Subset

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Theorem

Let $S$ be a set of real numbers.

Let $S$ have a supremum.

Let $T$ be a non-empty subset of $S$.


Then $\sup T$ exists and:

$\sup T \le \sup S$


Proof 1

The number $\sup S$ is an upper bound for $S$.

Therefore, $\sup S$ is an upper bound for $T$ as $T$ is a non-empty subset of $S$.

Accordingly, $T$ has a supremum by the Continuum Property.


The number $\sup S$ is an upper bound for $T$.

Therefore, $\sup S$ is greater than or equal to $\sup T$ as $\sup T$ is the least upper bound of $T$.

$\blacksquare$


Proof 2

By the Continuum Property, $T$ admits a supremum.

It follows from Supremum of Subset that $\sup T \le \sup S$.

$\blacksquare$


Proof 3

$S$ is bounded above as $S$ has a supremum.

Therefore, $T$ is bounded above as $T$ is a subset of $S$.

Accordingly, $T$ admits a supremum by the Continuum Property as $T$ is non-empty.


We know that $\sup T$ and $\sup S$ exist.

Therefore by Suprema of two Real Sets:

$\forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon \iff \sup T \le \sup S$


We have:

\(\ds \forall \epsilon \in \R_{>0}: \, \) \(\ds 0\) \(<\) \(\ds \epsilon\)
\(\ds \leadsto \ \ \) \(\ds \forall \epsilon \in \R_{>0}: \forall t \in T: \, \) \(\ds t\) \(<\) \(\ds t + \epsilon\)
\(\ds \leadsto \ \ \) \(\ds \forall \epsilon \in \R_{>0}: \forall t \in T: \, \) \(\ds t\) \(<\) \(\ds s + \epsilon\)
\(\, \ds \land \, \) \(\ds s\) \(=\) \(\ds t\)
\(\ds \leadsto \ \ \) \(\ds \forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: \, \) \(\ds t\) \(<\) \(\ds s + \epsilon\) as $T \subseteq S$
\(\ds \leadsto \ \ \) \(\ds \sup T\) \(\le\) \(\ds \sup S\)

$\blacksquare$


Proof 4

By definition $\sup S$ is an upper bound for $S$.

Thus:

$\forall x \in S: x \le \sup S$

As $T \subseteq S$ we have by definition of subset that:

$\forall x \in T: x \in S$

Hence:

$\forall x \in T: x \le \sup S$

So by definition $\sup S$ is an upper bound for $T$.

So $\sup S$ is at least as big as the smallest upper bound for $T$

Thus by definition of supremum:

$\sup T \le \sup S$

$\blacksquare$


Sources

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