Subset of Bounded Above Set is Bounded Above

Theorem

Let $A$ and $B$ be sets of real numbers such that $A \subseteq B$.

Let $B$ be bounded above.

Then $A$ is also bounded above.

Proof

Let $B$ be bounded above.

Then by definition $B$ has an upper bound $U$.

Hence:

$\forall x \in B: x \le U$

But by definition of subset:

$\forall x \in A: x \in B$

That is:

$\forall x \in A: x \le U$

Hence, by definition, $A$ is bounded above by $U$.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 1$