Find the infimum and/or supremum and see if the set is bounded

In summary: The x-values would be (##-∞,∞##). Then it would be correct to say that there will not be a minimum and maximum, but there would be an infimum and a supremum?Yes, that is correct. The only thing I would add is that the infimum and supremum in this case are both actually minimum and maximum, since there are no values of x that make the inequality false. So the set is bounded below and above, with the minimum and maximum values being -∞ and ∞, respectively.In summary, the set ##S_3## of real numbers defined by ##x^2+x+1≥0## is bounded below and above, with the infimum and suprem
  • #1
Nicci
23
0
Homework Statement
Determine whether the following set is bounded (from below, above, or both). If so,
determine the infimum and/or supremum and find out whether these infima/suprema
are actually minima/maxima.
Relevant Equations
##S_3 = \left\{ \ x∈ℝ : x^2+x+1≥0 \right\}##
##S_3 = \left\{ \ x∈ℝ : x^2+x+1≥0 \right\}##

I am not sure if I have done this correctly. The infimum/supremum and maximum/minimum are confusing me a bit.

This is how I started:

##x^2+x+1=0##
##x^2+x+ \frac1 4\ =\frac{-3} {4}\ ##
## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ = 0##
Therefore: ## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ ≥ 0##
Thus ##S_3 = ℝ##

Since## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ ## will never be negative, ##S_2## is not bounded above or below.
The interval of ##S_2## is ## \left ( -∞,∞ \right ) ##
Therefore: ##inf S_2 = -∞## and ##sup S_2 = ∞ ##

I do not think there will be any minima or maxima since it is an open interval.
Can someone please give me some advice or hints on this problem. I am very confused at this moment and I'm not really sure if I have done this correctly.
Thank you very much in advance.
 
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  • #2
Nicci said:
Homework Statement:: Determine whether the following set is bounded (from below, above, or both). If so,
determine the infimum and/or supremum and find out whether these infima/suprema
are actually minima/maxima.
Relevant Equations:: ##S_3 = \left\{ \ x∈ℝ : x^2+x+1≥0 \right\}##

##S_3 = \left\{ \ x∈ℝ : x^2+x+1≥0 \right\}##

I am not sure if I have done this correctly. The infimum/supremum and maximum/minimum are confusing me a bit.

This is how I started:

##x^2+x+1=0##
##x^2+x+ \frac1 4\ =\frac{-3} {4}\ ##
So far, so good, but the line below is no help, and you have a mistake in the line beyond that.
From the above, ##(x + \frac 1 2)^2 = -\frac 3 4##
The left side is always greater than or equal to zero, so can the left side possibly be equal to -3/4?

Also, in your factorization, you have ##(x^2 + \frac 1 2)^2##, which is incorrect.

One more thing. What you started with is an inequality -- you should be working with inequalities, not equations. It might be helpful to sketch a graph of ##y = x^2 + x + 1##.
Nicci said:
## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ = 0##
Therefore: ## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ ≥ 0##
Thus ##S_3 = ℝ##

Since## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ ## will never be negative, ##S_2## is not bounded above or below.
The interval of ##S_2## is ## \left ( -∞,∞ \right ) ##
Therefore: ##inf S_2 = -∞## and ##sup S_2 = ∞ ##

I do not think there will be any minima or maxima since it is an open interval.
Can someone please give me some advice or hints on this problem. I am very confused at this moment and I'm not really sure if I have done this correctly.
Thank you very much in advance.
 
  • #3
Mark44 said:
So far, so good, but the line below is no help, and you have a mistake in the line beyond that.
From the above, ##(x + \frac 1 2)^2 = -\frac 3 4##
The left side is always greater than or equal to zero, so can the left side possibly be equal to -3/4?

Also, in your factorization, you have ##(x^2 + \frac 1 2)^2##, which is incorrect.

One more thing. What you started with is an inequality -- you should be working with inequalities, not equations. It might be helpful to sketch a graph of ##y = x^2 + x + 1##.
I'm sorry about the ##x^2##, that was a typo. It should be ##(x + \frac 1 2)^2 ≥ -\frac 3 4##
If I look at the graph of ##y = x^2 +x +1##, I notice that the turning point of the parabola is at (-1/2 , 3/4).
Does this mean that ##S_3## is bounded from below? If it is bounded from below, does that mean that the min##S_3## = 3/4?
 
  • #4
Nicci said:
I'm sorry about the ##x^2##, that was a typo. It should be ##(x + \frac 1 2)^2 ≥ -\frac 3 4##
If I look at the graph of ##y = x^2 +x +1##, I notice that the turning point of the parabola is at (-1/2 , 3/4).
Does this mean that ##S_3## is bounded from below?
No, it doesn't. The y values on the graph of the parabola are bounded below, but ##S_3## is a set of values of x. What are the x-values associated with the graph of the parabola?
Nicci said:
If it is bounded from below, does that mean that the min##S_3## = 3/4?
No. Again, ##S_3## is a set of x values. 3/4 is a y value.

Also, you have another typo in your first post where you also talk about ##S_2##.
 
  • #5
Mark44 said:
No, it doesn't. The y values on the graph of the parabola are bounded below, but ##S_3## is a set of values of x. What are the x-values associated with the graph of the parabola?
The x-values would be (##-∞,∞##). Then it would be correct to say that there will not be a minimum and maximum, but there would be an infimum and supremum. ##inf S_3 = -∞## and ##sup S_3 = ∞##
##S_3## will then not be bounded above or below because the x-values are from negative infinity to positive infinity?

I hope I have it correct now.
Thank you so much.
 
  • #6
Nicci said:
Then it would be correct to say that there will not be a minimum and maximum, but there would be an infimum and supremum.

Just to clear things up (someone correct me if I mess this up):
(1) A set only has a supremum/infimum iff it is bounded from above/below, respectively.
(2) The minimum/maximum of a set is just another word for an infimum/supremum that is contained within the set. For example, ##0## is the infimum for ##[0,1]## and is contained within the set. So it is the minimum of the interval. On the other hand, ##0## is the infimum for ##(0,1]##, but is not contained within the interval, so it cannot be the minimum of the interval. Just some terminology to keep in mind.
 
  • #7
Nicci said:
The x-values would be (##-∞,∞##). Then it would be correct to say that there will not be a minimum and maximum, but there would be an infimum and supremum. ##inf S_3 = -∞## and ##sup S_3 = ∞##
##S_3## will then not be bounded above or below because the x-values are from negative infinity to positive infinity?

I hope I have it correct now.
Thank you so much.

One way to think about the supremum is a generalisation of concept of the maximum. If you have a set with a maximum value, then that is also the supremum. But, there are sets that do not have a maximum. One obvious example is the open interval ##(0, 1)##. The number ##1## is clearly something important for this set, and it is indeed the smallest number greater than all members of the set, which is the least upper bound or supremum. If you don't like the term supremum, use "least upper bound" instead.

There are also sets like the set of whole numbers, which are not bounded above. In that case we are going to define the supremum of such a set as ##+\infty##. Note that in this case the supremum is not a number. And sometimes you must be careful in proofs to distinguidsh the case where the set is bounded above and the supremum is a real number; and, the case where set is not bounded above and the supremum is ##+\infty##.

Likewise the infimum is the greatest lower bound and is a generalisation of the concept of a minimum.
 

FAQ: Find the infimum and/or supremum and see if the set is bounded

1. What is the definition of infimum and supremum?

The infimum of a set is the greatest lower bound, meaning it is the largest number that is less than or equal to all the elements in the set. The supremum is the least upper bound, meaning it is the smallest number that is greater than or equal to all the elements in the set.

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

To find the infimum, you must first list out all the elements in the set and then find the smallest number among them. To find the supremum, you must list out all the elements in the set and then find the largest number among them. It is important to note that the infimum and supremum may or may not be included in the set itself.

3. What does it mean for a set to be bounded?

A set is bounded if both the infimum and supremum exist and are finite numbers. This means that there is a clear upper and lower limit to the elements in the set. If a set is unbounded, it means that either the infimum or supremum (or both) do not exist, or they are infinite.

4. How do you determine if a set is bounded?

If a set is bounded, it means that both the infimum and supremum exist and are finite numbers. To determine if this is the case, you must first find the infimum and supremum of the set. If they are both finite numbers, then the set is bounded. If either one or both of them are infinite, then the set is unbounded.

5. Can a set be both bounded and unbounded?

No, a set cannot be both bounded and unbounded. By definition, a set is either bounded or unbounded. If a set is bounded, it means that both the infimum and supremum exist and are finite numbers. If a set is unbounded, it means that either the infimum or supremum (or both) do not exist, or they are infinite. These two conditions cannot coexist in the same set.

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