Supremum/infinum/max/min concerns x axis or y axis?

[kahwawashay1]

When we talk about supremum/infinum/etc, does it mean the largest/smallest number on the x-axis or y axis??

Ok so first my professor was explaining how when x^2 < 2 , the supremum is root of 2 (so he was talking about how the domain has a least upper bound).

But then he said that when we look at 1/n, where n is 1,2,3,... , the infinum is 0...so in this case i guess he meant that the range is bounded from below (since 1/n approaches 0 as n->infinity)...but if you look at the domain, the infinum is actually a minimum and equals 1...

help me clear up this ambiguity?

 

[gb7nash]

When we talk about supremum/infinum/etc, does it mean the largest/smallest number on the x-axis or y axis??

Ok so first my professor was explaining how when x^2 < 2 , the supremum is root of 2 (so he was talking about how the domain has a least upper bound).

But then he said that when we look at 1/n, where n is 1,2,3,... , the infinum is 0...so in this case i guess he meant that the range is bounded from below (since 1/n approaches 0 as n->infinity)...but if you look at the domain, the infinum is actually a minimum and equals 1...

help me clear up this ambiguity?

An infimum is the greatest lower bound and the supremum is the least upper bound. When you're looking at the set:

A = {1,1/2,1/3,1/4,...}

sup(A) = 1. This is because 1 is an upper bound for A (every element is less than 1) and there is no upper bound that is lower than 1.

inf(A) = 0. This is because 0 is a lower bound for A (every element in A is greater than 0) and there is no lower bound that is greater than 0. If there was, call it a, you could always find an n such that 1/n < a.

I think the confusion you were having is that the infimum would be the first element in the set, but this isn't true. You need to consider all elements of the set and see what's going on.

 

[kahwawashay1]

An infimum is the greatest lower bound and the supremum is the least upper bound. When you're looking at the set:

A = {1,1/2,1/3,1/4,...}

sup(A) = 1. This is because 1 is an upper bound for A (every element is less than 1) and there is no upper bound that is lower than 1.

inf(A) = 0. This is because 0 is a lower bound for A (every element in A is greater than 0) and there is no lower bound that is greater than 0. If there was, call it a, you could always find an n such that 1/n < a.

I think the confusion you were having is that the infimum would be the first element in the set, but this isn't true. You need to consider all elements of the set and see what's going on.

Oh ok so i understand about the 1/n...but what my professor wrote exactly was that supremum of
\left\{x\in Q: x^2 &lt; 2\right\}
is root of two...but shouldn't it be just 2?

 

[kahwawashay1]

Oh ok so i understand about the 1/n...but what my professor wrote exactly was that supremum of
\left\{x\in Q: x^2 &lt; 2\right\}
is root of two...but shouldn't it be just 2?

ohh nvm the supremum is root of two in this case because we are talking about x in Q that satisfy the condition x^2 < 2...but if we just ask for supremum of {x^2 < 2}, it would be 2?

 

[gb7nash]

Oh ok so i understand about the 1/n...but what my professor wrote exactly was that supremum of
\left\{x\in Q: x^2 &lt; 2\right\}
is root of two...but shouldn't it be just 2?

Think about it. \left\{x\in Q: x^2 &lt; 2\right\} means the set of rational numbers such that each rational number squared is less than 2. 2² = 4. 2 is definitely an upper bound for the set, but it's not the least upper bound. (note that the supremum does not need to be in the set. In this case, the supremum doesn't need to be a rational number. It only needs to be a least upper bound) What's something else you could plug into x such that there it is an upper bound for the set, and also that it is the least upper bound?

 

[kahwawashay1]

Think about it. \left\{x\in Q: x^2 &lt; 2\right\} means the set of rational numbers such that each rational number squared is less than 2. 2² = 4. 2 is definitely an upper bound for the set, but it's not the least upper bound. (note that the supremum does not need to be in the set. In this case, the supremum doesn't need to be a rational number. It only needs to be a least upper bound) What's something else you could plug into x such that there it is an upper bound for the set, and also that it is the least upper bound?

kk thanks i get that,

so as another example just so i kno for sure i get it, if

f: (-1, 3)->R
f(x)=x^2

then the supremum is 9 and minimum is 0 ?

 

[gb7nash]

kk thanks i get that,

so as another example just so i kno for sure i get it, if

f: (-1, 3)->R
f(x)=x^2

then the supremum is 9 and minimum is 0 ?

If the set you're considering is f(x) = x² for x in (-1,3), then yes.