Show sup(A+B) = s+t

[blissful jolt]

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[indigo scroll]

I was thinking $t \leq u-a ; s \leq u -b $; so $t+s \leq 2u-(a+b)$

[sand lance]

ƒ(Why am. I here)=I don't Know

[indigo scroll]

I ultimately want to get $sup(A) + sup(B)= u$ somehow

[sand lance]

ƒ(Why am. I here)=I don't Know

[lament folio]

u upper-bounds a+b

[indigo scroll]

what ?

[lament folio]

Actually nvm

I wanted to replace a+b in the inequality

With u

But that didn't work

[indigo scroll]

Alternatively, I also have $a+b \leq u \leq sup(A) + sup(B)$

[sand lance]

ƒ(Why am. I here)=I don't Know

[lament folio]

Well since it is true for any a and b

Use the sups instead

[indigo scroll]

hmm, how

[lament folio]

You replace a with sup A

[indigo scroll]

I can start by subtracting sup(A). from all sides

[lament folio]

And b with sup B

And normally everything simplifies

[indigo scroll]

The thing is

what I want it so essentially squeeze sup(A)+sup(B) between something and u

[lament folio]

yes

but you don't really need to

what you need to show is that sup(A) + sup(B) <= u

that is, sup(A) + sup(B) is the lowest upper-bound

since the first question shows it is an upper-bound already

[indigo scroll]

yeah

[lament folio]

so with the method I suggested, you end up with : s + t <= 2u - s - t

which basically yields the conclusion

the reason for that being valid is that s + t <= 2u - a - b would have to hold for ANY a and b

included the case where they get as large as they can get

[indigo scroll]

hmm

how did you get that though

[lament folio]

.

[indigo scroll]

ye, but I need $a+ b \leq s+t \leq 2u -(a+b)$ , right

[sand lance]

ƒ(Why am. I here)=I don't Know

[lament folio]

Question 1 gives you the first one already

s + t is an upper bound of A + B

[indigo scroll]

yes

ooh

[lament folio]

Question 1: s + t is an upper-bound of A+B

Question 2 and 3: it is also the lowest upper-bound

[indigo scroll]

hmm

okay

so $s+t \leq u$

[sand lance]

ƒ(Why am. I here)=I don't Know

[indigo scroll]

so s+t=u

got it

[lament folio]

No not equal u

[indigo scroll]

as $u$ is the least upper bound

[sand lance]

ƒ(Why am. I here)=I don't Know

[lament folio]

It means: if u is an upper-bound, then s + t <= u

But given that s + t is an upper-bound, since it applies to any u, s + t is the lowest upper-bound

[indigo scroll]

got it

thanks so much!

[lament folio]

Good

Np buddy

[indigo scroll]

Can I close the channel now?

[lament folio]

Yes

[indigo scroll]

.close

[tender shadow]

Unable to parse the channel name

[lament folio]

With a plus sign

[indigo scroll]

+close

[runic elbow]

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[lament folio]

Not a dot

[runic elbow]

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[lament folio]

Red button too