Let $X$ be a nonempty set and let $f:X \to \Bbb R$ be such that $f[X]:=\{f(x): x \in X\}$ is a bounded subset of $\Bbb R$.

Question

Let $X$ be a nonempty set and let $f:X \to \Bbb R$ be such that $f[X]:=\{f(x): x \in X \}$ is a bounded subset of $\Bbb R$.

Prove that for any $a \in \Bbb R$:

$$ \sup \{a+f(x):x \in X\} = a+ \sup(f[X]). $$

My attempt so far:

$\sup \{a+f(x):x \in X\} = a + \sup f[X]$

$\sup \{ a+f(x):x \in X\} = a + \sup \{f(x):x \in X\}$

But now I'm confused? What is the sup of $\{f(x):x \in X\}$? How do I complete the proof? My prof doesn't do any examples so this is super difficult so if you could explain how you did it that'd be amazing.

Answer 1

The idea is to work with the definition: the suppremum is the smallest upper bound.

1. $a+\sup f(X)$ is an upper bound for $\sup \{a+f(x):x\in X\}$. This should be obvious, since $a+f(x) \leq a+\sup f(X)$.

2. $a+\sup f(X)$ is the smallest upper bound. Well, from the definition of $\sup f(X)$, for every $\varepsilon>0$ there exists $x_\varepsilon \in X$ such that $f(x_\varepsilon)>\sup f(X)-\varepsilon$. Now you just add $a$ to each one of the sides: $$ a+f(x_\varepsilon)>(a+\sup f(X))-\varepsilon$$ This shows that indeed $a+\sup f(X)$ is the smallest upper bound for $\{a+f(x):x \in X\}$.

Answer 2

The key is to use the definition of the supremum of a (bounded) set as the minimum of all upper bounds for a set.

By assumption $f[X]$ is bounded in the reals so $s:= \sup(f[X])$ is a well-defined real number (by order completeness of $\Bbb R$).

Now if $x \in X$, $f(x) \in f[X]$ so $f(x) \le s$, as $s$ is an upper bound for $f[X]$ by definition. But then $a+f(x) \le a+s$ too, and as $x \in X$ is arbitary (!), $a+s$ is an upperbound for $\{a + f(x): x \in x\}$ and $\sup \{a + f(x): x \in x\}$ is the minimum of the set of that set's upperbounds, so

$$\sup \{a + f(x): x \in x\} \le a+s\tag{1}$$

has been shown.

Now suppose that $L$ is any upperbound for the set $\{a + f(x): x \in X\}$. I'll show that $L-a$ is an upperbound for $f[X]$: let $x \in X$, then $a + f(x) \in \{a + f(x): x \in X\}$ so $a+f(x) \le L$, as $L$ is an upperbound, and so (subtracting $a$ from both sides preserves the order) $f(x) \le L-a$. As $s = \sup\{ f(x): x \in X\}$ is the smallest of all upperbounds for $f[X]$ we again conclude that $s \le L-a$ or equivalently:

$$a+s \le a+(L-a) = L\tag{2}$$

So $a+ s = a + \sup(f[X])$ is smaller than any arbitary upperbound $L$ of $\{a + f(x): x \in X\}$ and so is its supremum (from (1) and (2) together). QED