Question

For a real number a and non-empty subset of reals B, define: a + B = { a + b : b is in B }. Show that if B is bounded above, then sup( a + B ) = a + sup B

What I have so far:
since B is bounded, ∀b∈B, b ≤ m, for some real number m

∀b∈B, a + b ≤ a + m.

Since a + b is an arbitrary element in a + B, a + B is bounded above by a + m.

Since R has the least upper bound properties, both a + B and B has a supremum.
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The part I'm currently having trouble with is sup(a + B) =a + sup(B)

My attempt so far is,
∀b∈B, b ≤ sup(B) by property of supremum.

∀b∈B, a + b ≤ a + sup(B)

hence, sup(a+B) ≤ a + sup(B)

Answer 1

@ganeshie8

Answer 2

For every \( \epsilon > 0\) ther is a \( b_\epsilon \) suich that
\[
b_\epsilon > \sup(B) -\epsilon\\
\sup(a+B) \ge a + b_\epsilon > a +\sup(B) -\epsilon
\]
Since this is true for every \(\epsilon >0\), then
we are done

Answer 3

\(b_\epsilon \in B\)

Answer 4

how is sup(a + B) >/ a + b_e ?

Answer 5

What I understand so far is, sup(B) - e is not an upper bound of B. This means there is an element b_e in B such that
sup(B) - e < b_e <= sup(B)

Answer 6

adding a to the inequality gives
a + sup(B) - e <a + b_e <= a + sup(B)

Answer 7

oh wait, so a + b_e is some element of a + B, so by definition of sup, a + b_e <= sup(a + B). so we have:
a + sup(B) - e < a + b_e <= sup(a + B)

Answer 8

ok, so now the question is how does that imply that
a + sup(B) <= sup(a + B) ??

Answer 9

@AngusV Openstudy is not showing text symbols properly. And yeah, i hate proving the obvious too :/

Answer 10

I hate questions like these. It's stupidly obvious that it is so. The question translates to:

"If you take a number from a set of numbers and add it to another number, then you can't have a result greater than the number you've added + the largest number from the set."

I'm trying to think of a way not to complicate things beyond that level. Your reasoning is perfectly correct because "a" is a constant real number outside B / the set of numbers. If a belongs to B and b belongs to B - then yeah, sup (a+b) =/= a+b.

Don't overthink it, just write that
∀b∈B,b≤sup(B)
because of sup(B) property of whatever - and then add "a" to both sides so that
a+b≤a+sup(B)
, where
a∈R
Since b is defined as any one element from B, it's like saying
a+B≤a+sup(B)
If your teacher has problems understanding why 2+3=3+2=5, I'm sorry for him/her and his/her degree.

Answer 11

Gods fluttering damn this piece of pellet

Answer 12

@AngusV I think you can just refresh the page, and all symbols will show up properly

Answer 13

Oh !

Answer 14

So yeah, in essence you can say it's due to "the property of inequations" or whatever.

Answer 15

That if a < b then a+x < b+x

Answer 16

thank you both @AngusV and @eliassaab . I appreciate your time :)