Max, min, least upper, and greatest lower bound for sets question

1. Oct 23, 2004

KataKoniK

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...

2. Oct 23, 2004

Hurkyl

Staff Emeritus
0.9 isn't the maximum value; it's smaller than 0.95.

3. Oct 23, 2004

cogito²

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.

4. Oct 23, 2004

master_coda

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)

5. Oct 23, 2004

arildno

Hint: (Arithmetic) averages are useful here..

6. Oct 23, 2004

HallsofIvy

Alright, Arildno, I'll bite. WHY would "arithmetic averages" be useful here?

7. Oct 23, 2004

arildno

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

8. Oct 23, 2004

quasar987

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 ???

Last edited: Oct 23, 2004
9. Oct 23, 2004

Hurkyl

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

10. Oct 23, 2004

arildno

0.9 periodic is no other number than 1.
Hence, it is NOT in (0,1).

11. Oct 23, 2004

quasar987

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.

12. Oct 23, 2004

Hurkyl

Staff Emeritus
Are you sure there are maximum and minimum values?

13. Oct 23, 2004

quasar987

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.

14. Oct 24, 2004

KataKoniK

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.

15. Oct 24, 2004

arildno

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.