Let $X$ be a nonempty subset of $\mathbb{R}$. Assume that $X$ has a supremum. What is the relation between $\sup(X+r)$ and $\sup(X)$? The full question: Let $X$ be a nonempty subset of $\mathbb{R}$.
Assume that $X$ has a supremum. Prove that the set $X+r:=\{x+r\ :\ x\in X\}$ has a supremum. What is the relation between $\sup(X+r)$ and $\sup(X)$? You should prove your claim.
If $X$ has a supremum, then it is bounded by above: $X\leq M \Rightarrow X+r \leq M+r$. Since $X+r$ is bounded by above, then $X+r$ has a supremum.
Let's consider two cases: $r \geq 0$ and $r<0$. First, let $r \geq 0$. Then,
$$x+r \geq x\ \forall x\in X$$
$$\sup\{x+r:x \in X\} \geq \sup\{x:x \in X\}$$
$$\sup(X+r) \geq \sup X$$
Similarly for $r<0$,
$$x+r < x\ \forall x\in X$$
$$\sup\{x+r:x \in X\} < \sup\{x:x \in X\}$$
$$\sup(X+r) < \sup X$$
This seems trivial to me so I'm not sure how to prove my claim. Is my proof correct and sufficient?
 A: Since $X$ is non-empty, $X+r$ is non-empty as well, and since $X$ is bounded above, $X+r$ is also bounded above (if $b$ is an upper bound of $X$, then $b+r$ is an upper bound of $X+r$), therefore, $\sup(X+r)$ exists. 
Now, we claim that $\sup(X+r) = \sup(X)+r$. To prove this, note that
$$\forall x \in X \ \big(x\leq \sup(X)\big)
\ \Rightarrow \
\forall x \in X \ \big( x+r \leq \sup(X) + r\big)$$
that is, $\sup(X) + r$ is an upper bound of the set $X+r$ and hence
$$\sup(X+r) \leq \sup(X) + r \tag{1}$$
by definition of $\sup(X+r)$.
Similarly, from
$$\forall x \in X \ \big(x + r \leq \sup(X+r)\big)
\ \Rightarrow \
\forall x \in X \ \big( x \leq \sup(X+r) - r\big)$$
we see that $\sup(X+r) - r$ is an upper bound for $X$, which implies that $\sup(X) \leq \sup(X+r) - r$, meaning,
$$\sup(X) + r \leq \sup(X+r). \tag{2}$$
Finally, combine $(1)$ and $(2)$ to get the desired result.
A: This is correct, but there’s something stronger: $\sup(X+r) = \sup X + r$. 
How? Well, $\sup X \geq y \iff \sup X + r \geq y+r$; write every element of $X+r$ as $y+r$ and show that $\sup X + r$ is an upper bound. Using the $\iff$ and the definition of $\sup X$), show that it’s the least upper bound, i.e., the supremum.
