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KataKoniK
Oct23-04, 12:51 PM
Hi,

I was wondering if I am doing this correctly. The question asks to state the maximum value, minimum value, least upper bound, and greatest lower bound of a bunch of given sets.

The question I am asking for is this one. {x : x E (0, 1)}

I am a bit confused. Therefore, is the following correct?


Maximum value: 0.9
Minimum value 0.1
Least Upper Bound: I'm not sure
Greatest Lower Bound: Not sure too.

For the least upper bound and greatest lower bound, how do I prove that IT IS the least upper bound and greatest lower bound?

Thanks in advance. Sorry for the stupid question. I'm slowly getting a hang of University math...

Hurkyl
Oct23-04, 12:58 PM
0.9 isn't the maximum value; it's smaller than 0.95.

cogitoČ
Oct23-04, 01:45 PM
Remember this, a set of real numbers does not necessarilly have a maximum or minimum but if it is bounded it always has a greatest lower bound and a least upper bound. When looking for the lub (or the glb) make sure to check if it is a bound for the set and then make sure that no smaller bound will suffice.

master_coda
Oct23-04, 02:26 PM
The maximum of a set is the largest value in the set; however in this case you can show that there is no maximum. Given any value in the set, you can always find a larger value that is also in the set.


The least upper bound is the smallest number that is larger than every number in the set; to show that 'x' is the least upper bound of a set, you thus have to show two things:

1) show that x is larger than every element of the set (in other words, show that x is an upper bound for the set)

2) show that if y < x, then there must be some element of the set which is larger than y (show that no number smaller than x is also an upper bound)

arildno
Oct23-04, 02:29 PM
Hint: (Arithmetic) averages are useful here..

HallsofIvy
Oct23-04, 02:59 PM
Alright, Arildno, I'll bite. WHY would "arithmetic averages" be useful here?

arildno
Oct23-04, 03:30 PM
Let for example a point "x" be in the set; we'll show that there exist a point y in the set so that
x<y<1
(that is, there is no maximum)
just set y=(x+1)/2, this is clearly within the set (0<y<1), in addition, we have x<y

quasar987
Oct23-04, 05:29 PM
I'm also learning this stuff right now Kata. We just had our first exam least week. I'm guessing the mean for the group will be somewhere around 25%.

I also have a question (more like a confusion really) regarding your question.

We know, according to Arildo's argument that the Least Upper Bound is 1, because for anything less than that, there is always an x element of the set as near as we'd like to 1.

But we also have that the number 1 is NOT an element of the set. Therefore we cannot say the 1 is the maximum value. But 0.9 periodic is part of the set, and 0.9 periodic = 1 (!). So 1 IS part of the set after all. The same goes for 0, is it also part of the set. So the writting E = {x : x E [0,1]} is an equivalent writting. Is this right ???

Hurkyl
Oct23-04, 05:32 PM
But 0.9 periodic is part of the set

No it's not, because it is false that 0 < 0.999... < 1. (precisely because 0.999... = 1)

arildno
Oct23-04, 05:32 PM
0.9 periodic is no other number than 1.
Hence, it is NOT in (0,1).

quasar987
Oct23-04, 05:42 PM
Oh ok thanks for clarifying that.

What are the maximum and minimum values then? We never saw that "concept" in class or in the book we use.

Hurkyl
Oct23-04, 06:00 PM
Are you sure there are maximum and minimum values?

quasar987
Oct23-04, 07:02 PM
Usually when the books asks for them, they exist... :)

But I guess not since you can always get closer to 0.9 periodic without quitte getting there.

KataKoniK
Oct24-04, 11:53 AM
Thanks for the help. From the replies, this is what I have compiled, so is the following correct?

Maximum Value: DNE because for every number x < 1, there will always be a number greater than x.

Minimum Value: DNE because for every number x > 0, there will always be a number less than x.

Least Upper Bound: 1 because there will always be an x that will be close to 1, but does not reach it, since the set does not include 1 itself.

Greatest Lower Bound: 0 because there will always be an x that will be close to 0, but does not reach it, since the set does not include 0 itself.

arildno
Oct24-04, 12:10 PM
Max/min almost perfect; you should tweak your sentences into:
"..there will always be a number WITHIN THE SET.."
"Least Upper Bound: 1 because there will always be an x that will be close to 1, but does not reach it, since the set does not include 1 itself.

Greatest Lower Bound: 0 because there will always be an x that will be close to 0, but does not reach it, since the set does not include 0 itself."
These are unfortunate, in particular the "does not reach it" part.
Rather, you should say something like (for LUB):
1 is the least upper bound, because
a) It is an upper bound for the set
AND
b) Any (positive) number strictly less than 1 is within the set, but since such a number is not the maximum value of the set , it certainly cannot be an upper bound for the set.