Set Notation/Inf Problem

[daon]

What does it mean if I'm given a set A, and am asked to prove that inf(A+y) = inf(A) + y.

y was not quantified, but I am assuming it is a real number as previous questions say so. But what does it mean to "add" a real number to a set? Does it distribute to all elements of the set? If so, how would I go about showing this?

Thank you.
-Daon

 

[JakeD]

What does it mean if I'm given a set A, and am asked to prove that inf(A+y) = inf(A) + y.

y was not quantified, but I am assuming it is a real number as previous questions say so. But what does it mean to "add" a real number to a set? Does it distribute to all elements of the set? If so, how would I go about showing this?

Thank you.
-Daon

You are correct that A+y is the set of all sums x+y where x is from A.

To prove the result, let a = Inf(A) and z = Inf(A+y). Show that a+y > z implies that z cannot be the greatest lower bound for A+y, a contradiction. Likewise a+y < z implies a cannot the GLB for A. So a+y = z.

 

[pka]

Here is a second way, keeping with the foregoing notation.
If a=glb(A) then \(\displaystyle \left( {\forall x \in A} \right)\left[ {a \le x} \right] \Rightarrow \left( {\forall x \in A} \right)\left( {\forall y} \right)\left[ {a + y \le x + y} \right].\)
This means that a+y is a lower bound for A+y.
This means that there is a glb for A+y, call it b.
If a+y<b then a<b-y which implies \(\displaystyle \left( {\exists t \in A} \right)\left[ {a \le t < b - y} \right] \Rightarrow \left[ {t + y < b} \right] \\\)
But that is a contradiction.

 

[daon]

Pka, I am having trouble figuring out why a<b-y implies there is a an element of A (t) between a and b-y.

 

[pka]

Because a is the greatest lower bound of A.
The number b-y is greater than a, so b-y is not a lower bound for A.
Therefore, there must be a term of A less than b-y.