Oh dear, that sounds like something I used to be able to do. Sorry, Emily, I am no help whatsoever.

Okay, wait, finitely many sixes after the decimal, so the least upper bound would probably be 2/3, because for any .66...6 with n-many 6s you can always just add another 6 on the end and get a higher member of the set. And if you mix 0s in there, it'll get lower.

No, that's probably still no help.

I think that the reader was supposed to be surprised by the walling up, it's just that we all know the story now.

Okay, wait, finitely many sixes after the decimal, so the least upper bound would probably be 2/3, because for any .66...6 with n-many 6s you can always just add another 6 on the end and get a higher member of the set. And if you mix 0s in there, it'll get lower.

I suppose that's true. I guess I'm just confused by the use of "least upper bound" here. I mean, there's no number such that all numbers smaller than it in the interval are like that, except 3/5, and that doesn't seem to be the meaning.

I don't see how it could be anything but what -t said. Unless I'm missing something....

Ganked from [link]

The least upper bound, called the supremum, of a set S, is defined as a quantity M such that no member of the set exceeds M, but if ? is any positive quantity, however small, there is a member that exceeds M - ?.

I think it's 2/3. I don't think I can prove it.

(The ?s are epsilons)

I don't think I can prove it.

Can you assume it's not true, and then show that assumption results in a contradiction?

I think all that needs to be shown is that for any positive x, there exists a member of the described set between 2/3-x and 2/3. The proof is left as an exercise for the reader.

Oh, okay. Using that definition, I think it does work. I was looking at

If a property M does not belong to all values of a variable x, but does belong to all values which are less than a certain u, then there is always a quantity U which is the greatest of those of which it can be asserted that all smaller x have property M

and not getting it. Although now I look at it... I guess it does say the same thing. It was that "a variable x" that was confusing me. Thanks, guys! Now I have to attack Dedekind cuts and paradoxes. My brain really isn't up to this stuff.

Let A be the set of numbers in (3/5,2/3) that have decimal expansions containing only finitely many zeros and sixes after the decimal point and no other integer. Find the least upper bound of A. (prove your answer)

My head hurts, and I can't help but wonder what is the practical value of the question. Good luck.

Can't help you there. Dedekind cuts always made my brain hurt. Good luck!

My head hurts, and I can't help but wonder what is the practical value of the question.

Practical value? Did you miss the part where I said History of Math? Practical, shyeah! I do homework on things that were proved impossible 500 years ago!

God I'll be glad when this class is over. The homework just never gets any easier.