This was another proof that I liked.

Let \(A \subseteq \mathbb{R}\) be nonempty and bounded above. Let \(c \in \mathbb{R}\). Define the set \(A + c\) by $$ \begin{align*} c + A = \{ c + a: a \in A\} \end{align*} $$ Then, \(\sup(c+A) = c + \sup A\).

Definitions of bounds: here. How do we prove it? Any time we want to prove something about least upper bounds we’ll need to verify both conditions of the definition of a least upper bound. From the definitions page, we need to verify that \(c + \sup A\) is an upper bound and then prove that it’s the least one.

Proof

Let \(A\) be a bounded non-empty set with a least upper bound \(\sup A\). Let \(c\) be a constant. Let \(c + A\) be the set defined above. Since \(s\) is an upper bound, then we know that for every element \(a \in A\), we must have \(a \leq s\). If we add \(c\) to both sides, then

$$ \begin{align*} a + c \leq \sup A + c. \end{align*} $$

This implies that \(\sup A + c\) is an upper bound for \(A + c\). Next, we want to show \(\sup A + c\) is the least upper bound on \(A + c\). Let \(b\) be any upper bound of \(A + c\). This means that for any element in \(A + c\), we must have

$$ \begin{align*} a + c \leq b. \end{align*} $$

We can re-write the inequality as

$$ \begin{align*} a \leq b - c. \end{align*} $$

But this means that \(b - c\) is an upper bound for \(A\). Therefore, by the definition of a least upper bound, we must have

$$ \begin{align*} \sup A \leq b - c. \end{align*} $$

But we can re-write this as

$$ \begin{align*} \sup A + c \leq b \end{align*} $$

This shows that \(\sup A + c\) is less than or equal to any arbitrary upper bound of \(A + c\). But this just implies that \(\sup A + c\) is the least upper bound of \(A + c\) as we wanted to show. \(\blacksquare\)

References