How to Prove That sup(A ∪ B) = max{sup A, sup B}

For nonempty sets $A,B\subseteq\mathbb{R}$ that are bounded above, the identity follows directly from the definition of a supremum as a least upper bound. The larger of the two individual suprema bounds every element of the union, and no smaller number can bound the union.

The assumptions matter: if a set is empty or unbounded above, its supremum may not exist as a real number. Those cases are addressed below.

Proof using the least-upper-bound definition

Assume $A$ and $B$ are nonempty and bounded above. Write

$$\alpha=\sup A,\qquad \beta=\sup B,$$

and set

$$M=\max\{\alpha,\beta\}.$$

We prove that $M$ is the supremum of $A\cup B$ by checking two properties: it is an upper bound, and it is no larger than any other upper bound.

1. $M$ is an upper bound for $A\cup B$.

Since $\alpha\le M$ and $\alpha$ is an upper bound for $A$, every $a\in A$ satisfies $a\le\alpha\le M$. Likewise, every $b\in B$ satisfies $b\le\beta\le M$. Every element of $A\cup B$ belongs to at least one of those sets, so every element of the union is at most $M$.

2. No upper bound for $A\cup B$ is less than $M$.

Let $U$ be any upper bound for $A\cup B$. Then $U$ is an upper bound for $A$ and for $B$, because each set is contained in the union. By the defining property of the supremum,

$$\alpha\le U\quad\text{and}\quad\beta\le U.$$

Therefore $M=\max\{\alpha,\beta\}\le U$.

The number $M$ is an upper bound for the union and is less than or equal to every upper bound for it. Hence it is the least upper bound:

$$\boxed{\sup(A\cup B)=\max\{\sup A,\sup B\}.}$$

Worked example

Take $A=(0,1)$ and $B=[-2,0]$. Their suprema are

$$\sup A=1,\qquad \sup B=0.$$

Thus the formula gives $\sup(A\cup B)=\max\{1,0\}=1$. Indeed, $A\cup B=[-2,1)$, whose least upper bound is $1$. Notice that $1$ is not an element of the union. A supremum need not be attained; it is the least upper bound, not necessarily a maximum.

A useful alternative viewpoint

The union has exactly the upper bounds that are upper bounds for both sets. In symbols, a real number $U$ bounds $A\cup B$ if and only if it bounds $A$ and bounds $B$. The least such number must therefore be the larger of the two least upper bounds. This viewpoint is useful for unions of more than two sets when the relevant supremum exists.

For a finite collection of nonempty sets $A_1,\ldots,A_n$, each bounded above, the same reasoning gives

$$\sup\left(\bigcup_{i=1}^n A_i\right)=\max_{1\le i\le n}\sup A_i.$$

For an infinite collection, the corresponding expression is generally a supremum rather than a maximum:

$$\sup\left(\bigcup_{i\in I}A_i\right)=\sup_{i\in I}\sup A_i,$$

when the sets and suprema are interpreted so that both sides exist. There need not be an index whose individual supremum equals the supremum over the whole collection.

Common mistakes and edge cases

• Confusing supremum with maximum. The value $\sup A$ can be a limit point not contained in $A$, as in $A=(0,1)$. The proof does not require either individual supremum to be attained.
• Forgetting boundedness. If either set is unbounded above, so is the union, and there is no real-valued supremum. For example, if $A=\mathbb{R}$, then $A\cup B$ has no real supremum.
• Ignoring the empty set. Under the usual real-number definition, $\sup\varnothing$ is not a real number: every real number is an upper bound of the empty set, but there is no least real upper bound. So the displayed identity, stated with real-valued suprema, requires both sets to be nonempty.

In the extended real numbers, one often defines $\sup\varnothing=-\infty$ and allows the value $+\infty$ for sets unbounded above. With those conventions the identity remains valid, including when one or both sets are empty or unbounded above.