Greatest lower bound of $\{x+k\mid x\in A\}$

We let $A$ be a nonempty bounded set and define $B=\{ x+k\mid x \in A\}$, where $k$ is a fixed real number.

I’m trying to show that $\operatorname{glb}B = \operatorname{glb}A + k$.

Solutions Collecting From Web of "Greatest lower bound of $\{x+k\mid x\in A\}$"

Since ${\sf glb}(A)$ is a lower bound for $A$, ${\sf glb}(A)+k$ is a lower bound for $A+k$. As ${\sf glb}(B)$ is the greatest of those lower bounds, we deduce

$${\sf glb }(A)+k \leq {\sf glb }(A+k) \tag{1}$$
.
Replacing $(A,k)$ with $(A+k,-k)$, we also have

$${\sf glb }(A+k)-k \leq {\sf glb }(A) \tag{2}$$

Then (1) and (2) give that ${\sf glb }(A+k)={\sf glb }(A)+k$ as wished.

Let $a := \operatorname{inf} A$. For any $y \in A$ then $y \geq a$, and so $y + k \geq a + k : = b$. By definition of $B$, $b$ is thus a lower bound for $B$. Assume there is a larger lower bound for $B$, which means a $c$ such that for all $z \in B, z \geq c > b$. Then $a + k = b < c \leq z = y + k$, $y$ any element of $A$ as the sets $A$ and $B$ are related through a bijection $B = A + k$ (any translation is a bijection), and the inequality holds for any $z \in B$.  But then $y + k \geq c > a + k$ for all $y \in A$ implies $y \geq c -k > a$ for all of $A$, which contradicts the definition of $a$. So $b$ is the greatest lower bound of $B$.