# Union of Bounded Above Real Subsets is Bounded Above

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## Theorem

Let $A$ and $B$ be sets of real numbers.

Let $A$ and $B$ be bounded above.

Then $A \cup B$ is also bounded above.

## Proof

Let $A$ and $B$ both be bounded above.

Then by definition $A$ and $B$ both have an upper bound $U_A$ and $U_B$ respectively.

Suppose $U_A \le U_B$.

Then:

- $\forall a \in A: a \le U_B$

and also, by definition:

- $\forall b \in B: b \le U_B$

and so $U_B$ is an upper bound for $A$.

Otherwise, suppose $U_A > U_B$.

Then:

- $\forall b \in B: b \le U_A$

and also, by definition:

- $\forall a \in A: a \le U_A$

Let $x \in A \cup B$.

Then from the above, either $x \le U_A$ or $x \le U_B$.

So either $U_A$ or $U_B$ is an upper bound for $A \cup B$.

Hence, by definition, $A \cup B$ is bounded above by either $U_A$ or $U_B$.

$\blacksquare$

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 2$