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 $\operatorname{glb}B = \operatorname{glb}A + k$.

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Will you please elaborate? What has been the result of your trying? Where are you stuck? –  Jonas Meyer Feb 4 '13 at 1:37
sorry about that, didnt know I could flag for technical difficulties, my apologies. –  Peej Gerard Feb 4 '13 at 5:17
@Paul: No worries, the site can be pretty complicated for a newcomer :) –  Zev Chonoles Feb 4 '13 at 5:19
thanks again, trying my best! –  Peej Gerard Feb 4 '13 at 5:21

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.

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Here, does $B=A+k$? –  Peej Gerard Feb 4 '13 at 5:11
yes, $A+k$ is a shorthand for $\lbrace a+k | a\in A \rbrace$ –  Ewan Delanoy Feb 4 '13 at 5:15
thanks, and can you explain the step where you replace $(A,k)$ with $(A+k,-k)$? –  Peej Gerard Feb 4 '13 at 5:16
Which part of “replace” do you not understand ? In other words, if I set $A’=A+k$ and $k'=-k$, then ${\sf glb}(A’)+k’ \leq {\sf glb}(A’)$ by (1), and writing out the values of $A’$ and $k’$ we obtain (2). –  Ewan Delanoy Feb 4 '13 at 5:18
If you don’t like the word “substitution” or “replace”, try to rephrase the argument without using those words. Why are they allowed ? Because the formula we have shown ((1)) is true for ALL and ANY $A$ and $k$. –  Ewan Delanoy Feb 4 '13 at 5:37

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 $B$.

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