MHB Finding Supremum and Infimum of Sets with Inequalities

[evinda]

Hello! (Wave)

I want to find the supremum, infimum of the following sets:

$$\{ x \in \mathbb{R}: 0<x^2-1<3\}, \{1+\frac{(-1)^n}{n}: n=1,2, \dots \}$$

For the first set I have thought the following:

$$ 0<x^2-1<3 \Rightarrow 1<x^2<4 \Rightarrow x^2>1 \text{ and } x^2 <4 \Rightarrow (x>1 \text{ or } x<-1) \text{ and } -2<x<2 \Rightarrow (-2<x<-1) \text{ or } (1<x<2)$$

Can we say that $-2$ is the infimum of the set and $2$ the supremum? Or not, since only one of these two inequalities hold? (Thinking)

For the second set, I have thought the following:$1+\frac{(-1)^n}{n}=\left\{\begin{matrix}
1+\frac{1}{n}, &\text{ if n is even} \\ \\
1-\frac{1}{n}, & \text{ if n is odd.}
\end{matrix}\right.$

We have that for all $n \in \mathbb{N}$, $1+\frac{1}{n} \geq 1$ and $1+\frac{1}{n} \leq 2$.

Also, we have that $1-\frac{1}{n} \leq 1$ and $1-\frac{1}{n} \geq 0$.

So is $0$ the infimum and $2$ the supremum? (Thinking)

 

[S.G. Janssens]

Hello! (Wave)

I want to find the supremum, infimum of the following sets:

$$\{ x \in \mathbb{R}: 0<x^2-1<3\}, \{1+\frac{(-1)^n}{n}: n=1,2, \dots \}$$

For the first set I have thought the following:

$$ 0<x^2-1<3 \Rightarrow 1<x^2<4 \Rightarrow x^2>1 \text{ and } x^2 <4 \Rightarrow (x>1 \text{ or } x<-1) \text{ and } -2<x<2 \Rightarrow (-2<x<-1) \text{ or } (1<x<2)$$

Can we say that $-2$ is the infimum of the set and $2$ the supremum? Or not, since only one of these two inequalities hold? (Thinking)

You can replace your $\Rightarrow$ with $\Leftrightarrow$ because you are actually describing the set in a number of ways that are all equivalent. Yes, you are correct. (In this case, it may also help to draw the graph of the standard parabola and see for which values of $x$ it lies strictly between $1$ and $4$.)

For the second set, I have thought the following:$1+\frac{(-1)^n}{n}=\left\{\begin{matrix}
1+\frac{1}{n}, &\text{ if n is even} \\ \\
1-\frac{1}{n}, & \text{ if n is odd.}
\end{matrix}\right.$

We have that for all $n \in \mathbb{N}$, $1+\frac{1}{n} \geq 1$ and $1+\frac{1}{n} \leq 2$.

Also, we have that $1-\frac{1}{n} \leq 1$ and $1-\frac{1}{n} \geq 0$.

So is $0$ the infimum and $2$ the supremum? (Thinking)

I agree with the inf but not with the sup.

 

[evinda]

You can replace your $\Rightarrow$ with $\Leftrightarrow$ because you are actually describing the set in a number of ways that are all equivalent. Yes, you are correct. (In this case, it may also help to draw the graph of the standard parabola and see for which values of $x$ it lies strictly between $1$ and $4$.)

Looking at this again, since the infimum is the greatest lower bound for the set, shouldn't it be equal to $1$ ?

More specifically, the lower bounds of the set are $1,-2$ and since $1$ is greater, this is the infimum.

Similarly, the upper bounds are $-1$ and $2$ and the smaller of them is $-1$. Thus, this is the supremum.

Or am I wrong? (Thinking)

 

[S.G. Janssens]

Looking at this again, since the infimum is the greatest lower bound for the set, shouldn't it be equal to $1$ ?

More specifically, the lower bounds of the set are $1,-2$ and since $1$ is greater, this is the infimum.

Similarly, the upper bounds are $-1$ and $2$ and the smaller of them is $-1$. Thus, this is the supremum.

Or am I wrong? (Thinking)

Now you are wrong and initially you were right. (Wink)

In interval notation, the set in question is: $(-2, -1) \cup (1, 2)$.
A lower bound has to be a lower bound for the whole set (i.e. the union of the two open intervals), not just for one of the intervals.
So, $-\pi$ is a lower bound, as is $-2$, but $1$ is not a lower bound.
For the upper bounds a similar comment applies.

 

[evinda]

I agree with the inf but not with the sup.

Why isn't $2$ the supremum? (Thinking)

- - - Updated - - -

Now you are wrong and initially you were right. (Wink)

In interval notation, the set in question is: $(-2, -1) \cup (1, 2)$.
A lower bound has to be a lower bound for the whole set (i.e. the union of the two open intervals), not just for one of the intervals.
So, $-\pi$ is a lower bound, as is $-2$, but $1$ is not a lower bound.
For the upper bounds a similar comment applies.

Ah I see... (Nod)

 

[evinda]

I agree with the inf but not with the sup.

We have that $n \geq 1$ and so $\frac{1}{n} \leq 1 \Rightarrow 1+\frac{1}{n} \leq 2$. So why isn't $2$ the supremum? (Thinking)

 

[S.G. Janssens]

We have that $n \geq 1$ and so $\frac{1}{n} \leq 1 \Rightarrow 1+\frac{1}{n} \leq 2$. So why isn't $2$ the supremum? (Thinking)

It is an upper bound, yes, but not the sup.
Try writing down the first few elements of the set, say for $n = 1, 2, 3, 4$.
(Don't bother simplifying, just leave the terms as $1 \pm \frac{1}{n}$.)

 

[evinda]

It is an upper bound, yes, but not the sup.
Try writing down the first few elements of the set, say for $n = 1, 2, 3, 4$.
(Don't bother simplifying, just leave the terms as $1 \pm \frac{1}{n}$.)

We have that $1 \pm \frac{1}{n} \leq 1+\frac{1}{n}$.

We notice that $1+\frac{1}{n} \to 1$, while $n \to +\infty$.

Thus $\sup\{1+\frac{(-1)^n}{n}: n=1,2, \dots \}=1$.

Right? (Thinking)

 

[S.G. Janssens]

We have that $1 \pm \frac{1}{n} \leq 1+\frac{1}{n}$.

We notice that $1+\frac{1}{n} \to 1$, while $n \to +\infty$.

Thus $\sup\{1+\frac{(-1)^n}{n}: n=1,2, \dots \}=1$.

Right? (Thinking)

No, go back for a moment. $1$ is not an upper bound, so it cannot be the sup.
What value occurs for $n=2$?
What other values occur? Do any of these exceed the value for $n = 2$?

 

[evinda]

No, go back for a moment. $1$ is not an upper bound, so it cannot be the sup.
What value occurs for $n=2$?
What other values occur? Do any of these exceed the value for $n = 2$?

For $n=1$ we get that $1+\frac{1}{n}=2$, for $n=2$ we get that $1+\frac{1}{n}=1.5$, for $n=3$ we get that $1+\frac{1}{n}=1.333$ and for $n=4$ we get that $1+\frac{1}{n}=1.25$.

Since $\frac{1}{n}$ is a decreasing function, we will always get smaller values, won't we?

Why isn't the supremum the minimum value of $1+\frac{1}{n}$, which we get when $n \to +\infty$? (Wondering)

 

[S.G. Janssens]

For $n=1$ we get that $1+\frac{1}{n}=2$, for $n=2$ we get that $1+\frac{1}{n}=1.5$, for $n=3$ we get that $1+\frac{1}{n}=1.333$ and for $n=4$ we get that $1+\frac{1}{n}=1.25$.

Since $\frac{1}{n}$ is a decreasing function, we will always get smaller values, won't we?

Why isn't the supremum the minimum value of $1+\frac{1}{n}$, which we get when $n \to +\infty$? (Wondering)

Careful, for $n = 1$ we get $1 + \frac{(-1)^1}{1} = 0$. Generally, for odd $n$ we subtract $\frac{1}{n}$ and for even $n$ we add $\frac{1}{n}$.
Also, the supremum must be an upper bound. The set looks like
$$
\left\{0, 1 + \frac{1}{2}, 1 - \frac{1}{3}, 1 + \frac{1}{4}, \ldots.\right\},
$$
and while the associated sequence indeed converges to $1$, it is clear that $1$ is not an upper bound for the set.

 

[evinda]

Careful, for $n = 1$ we get $1 + \frac{(-1)^1}{1} = 0$. Generally, for odd $n$ we subtract $\frac{1}{n}$ and for even $n$ we add $\frac{1}{n}$.
Also, the supremum must be an upper bound. The set looks like
$$
\left\{0, 1 + \frac{1}{2}, 1 - \frac{1}{3}, 1 + \frac{1}{4}, \ldots.\right\},
$$
and while the associated sequence indeed converges to $1$, it is clear that $1$ is not an upper bound for the set.

Ah I see... But then how can we find the least upper bound? (Thinking)

 

[S.G. Janssens]

Ah I see... But then how can we find the least upper bound? (Thinking)

I am sure you can do that.

What would be a good candidate for the sup? Consider how the set looks in my previous post.
The candidate is a member of the set.

Once found, prove that the candidate is an upper bound.
Since it is also a member of the set, there cannot be a smaller upper bound.

 

[evinda]

I am sure you can do that.

What would be a good candidate for the sup? Consider how the set looks in my previous post.
The candidate is a member of the set.

Once found, prove that the candidate is an upper bound.
Since it is also a member of the set, there cannot be a smaller upper bound.

We have that $1 \pm \frac{1}{n} \leq 1+\frac{1}{n}\leq 1+\frac{1}{2}=\frac{3}{2}$.

Thus $\frac{3}{2}$ is the supremum of the set. Right? (Thinking)

 

[S.G. Janssens]

We have that $1 \pm \frac{1}{n} \leq 1+\frac{1}{n}\leq 1+\frac{1}{2}=\frac{3}{2}$.

Thus $\frac{3}{2}$ is the supremum of the set. Right? (Thinking)

Right!

 

[evinda]

Right!

Nice, thank you! (Happy)