# Find the supremum and infimum of S, where S is the set S = {√n − [√n]}

• cooljosh2k2
In summary: Since 1 is not a part of the set, a cannot be an upper bound. This contradiction proves that 1 is the supremum.
cooljosh2k2

## Homework Statement

Find the supremum and infimum of S, where S is the set

S = {√n − [√n] : n belongs to N} .

Justify your claims. (Recall that if x belongs to R, then [x] := n where n is the largest integer less than or equal to x. For example, [7.6] = 7 and [8] = 8)

## The Attempt at a Solution

I found my infimum to be 0 and my supremum to be 1, but how do i go about proving them? Help please.

Well, first show 0 is a lower bound of the set. This is not difficult based on the definition of the floor function. After that, show that it is the greatest lower bound by using an epsilon argument.

Then you do the same for the upper bound which is slightly more complicated.

Thanks, i got it!

Is it sufficient enough to say for the supremum:

If 1 is an upper bound of S that satisfies the stated condition and if v < 1, then ε = 1-v. Since ε > 0, there exists Sε ∈ S, such that v = 1- ε < Sε, therefore v is not an upper bound of S, and we conclude that 1 = sup S

and for the infimum:

If 0 is a lower bound of S and if t > 0, then ε = t-0. Since ε > 0, there exists Sε ∈ S, such that t = ε - 0 > Sε. Therefore, t is not a lower bound of S, and we conclude that 0 = inf S

Is this right?

No what you have there is not a proof. You just stated there exist without proof. With your " proof" I could show that 2 is the supremum:-).

Here is how your proof should look like...
By definition of [x], x>=[x] so the terms of the sequence are always positive thus 0 is indeed a lower bound.

Suppose a was a lower bound and a>0. Taking n=4 we see that a cannot be a lower bound since a>0 . Thus our assumption that a is a lower bound is false.

Proofs require a bit more work. Your thinking is good but your steps are not sufficient; try to use what I gave you.

Last edited:
So would i be able to use the same reasoning for the supremum:

By definition of {x}, {x} < 1 since x - [x] = {x}, therefore {x} never reaches 1. So 1 is an upper bound.

Let b be an upper bound where b < 1. If you take ?, (what value would i be able to use here to prove this, since 1 isn't part of the set)

Can i use a limit to prove sup S = 1, since 1 isn't contained in the set?

cooljosh2k2 said:
So would i be able to use the same reasoning for the supremum:

By definition of {x}, {x} < 1 since x - [x] = {x}, therefore {x} never reaches 1. So 1 is an upper bound.

Let b be an upper bound where b < 1. If you take ?, (what value would i be able to use here to prove this, since 1 isn't part of the set)

First of all, you can't just copy my proof and expect everything to work out. I told you that proving the supremum was a bit more complicated.

I gave you the above proof to show you the general structure of your arguments. I expected you to gain some more incite into the problem.

Secondly,I don't understand your notation what is {x} and how exactly is it equal to x-[x] ? Usually when you put curly brackets it denotes a set.

I will give you an outline of what do you.

1) Show 1 is an upperbound of the set
2) Show that it is the least upper bound that is if a<1 then a is not an upper bound. Use the same sort of epsilon arguments you had originally.

In my proof the reason I picked a number is because it provided me with a straightforward contradiction. In the supremum case you can't use this method directly but you can use a modified version since as you already noted 1 is not in the set.

This is where assuming that a<1 and a is an upperbound will be helpful. As you previously did, a=1- $$\alpha$$. Now try deriving a contradiction.

## 1. What is the definition of supremum and infimum in mathematics?

The supremum of a set is the smallest upper bound, while the infimum is the largest lower bound. In other words, the supremum is the highest value that is still less than or equal to all the elements in the set, and the infimum is the lowest value that is still greater than or equal to all the elements in the set.

## 2. How do you find the supremum and infimum of a set?

To find the supremum and infimum of a set, you first need to determine the upper and lower bounds of the set. Then, you can compare these bounds to the elements in the set to find the smallest upper bound and the largest lower bound.

## 3. What is the difference between supremum and maximum?

The supremum of a set is the smallest upper bound, while the maximum is the largest element in the set. In other words, the supremum may or may not be an actual element in the set, whereas the maximum always is.

## 4. Can the supremum and infimum of a set be equal?

Yes, the supremum and infimum of a set can be equal. This can happen when the set only has one element, or when the set has a finite number of elements and the supremum and infimum are both equal to one of the elements.

## 5. How do you find the supremum and infimum of the set S = {√n − [√n]}?

To find the supremum and infimum of this set, we first need to determine the upper and lower bounds. In this case, the lower bound is 0 and the upper bound is 1. Then, we can compare these bounds to the elements in the set. We can see that the supremum is 1, as it is the smallest upper bound that is still less than or equal to all the elements in the set. The infimum is 0, as it is the largest lower bound that is still greater than or equal to all the elements in the set.

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