Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Thanks in advance.

share|improve this question
    
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

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
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
1  
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
1  
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 $B$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.