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

[Nicci]

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.

 

[Mark44]

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 $y = x^2 + x + 1$.

$\left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ = 0$
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.

 

[Nicci]

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?

 

[Mark44]

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?

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 $S_2$.

 

[Nicci]

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.

 

[Eclair_de_XII]

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.

 

[PeroK]

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.