Can a set include negative infinity and be bounded below

[fishturtle1]

Homework Statement

Prove that {$x \epsilon \mathbb{R} : x^2 \ge 1$} is "not" bounded below.
EDIT: I Looked closely and realized there is a "not" that we all had to write in...sorry if you lost some time..

Homework Equations

Defintion: We say a nonempty subset $A$ of $\mathbb{R}$ is bounded below, if there is a real number $m$ such that $m \le x$ for all $x \epsilon A$. We call such an $m$ an lower bound of $A$.

The Attempt at a Solution

I don't think this is bounded below.

Proof: This will be a proof by contradiction. Suppose $m$ is a lower bound of the set A ={$x \epsilon \mathbb{R} : x^2 \ge 1$}. Then $m \le x$ for all $x \epsilon A$. But $m - 1 < m$ and $m - 1 \epsilon A$.(Not sure how to prove that last statement). Therefore $m > (m-1)$ and $(m-1) \epsilon A$. Therefore m is not a lower bound, a contradiction. We conclude there does not exist a lower bound of A and A is not bounded below. []

Edit: in class we proved that the statement was True and had 3 cases but I don't see why the proof makes sense.

Consider 3 cases:

Case 1: Let $m < -1$ Let $x = m - 1$. Then $x = m - 1 < m$.

Case 2: Let $m \epsilon [-1,1]$. Let $x = m-4$. Then $x = m - 4 < m$.

Case 3: Let $m > 1$. Then $x = -m$. Then $x = -m < m$.

So we showed for any real number m, there exists an x such that x < m. Which I guess fulfills the definition. But I'm saying since the $A$ is going toward negative infinity, then I can always find a $k$ such that $k < x$.

 

[FactChecker]

Are you considering the negative numbers x where x² ≥ 1? How low can those go?

 

[fishturtle1]

Are you considering the negative numbers x where x² ≥ 1? How low can those go?

To negative infinity

 

[FactChecker]

To negative infinity

Sorry, I didn't realize why there were three cases in the proof and didn't notice exactly what you were asking. Yes, your intuition is right that you can get as great a negative number as you want, but the complication of which region m is in was handled in 3 cases.
He could have just said something like a = -|m|-200 ∈ A and a < m. That would have taken care of all cases.

 

[fishturtle1]

Sorry, I didn't realize why there were three cases in the proof and didn't notice exactly what you were asking. Yes, your intuition is right that you can get as great a negative number as you want, but the complication of which region m is in was handled in 3 cases.
He could have just said something like a = -|m|-200 ∈ A and a < m. That would have taken care of all cases.

So basically, if a = -|m| - 200 $\epsilon$ A and a < m, it follows that there is no lower bound, because for all elements m $\epsilon$ A, there exists an a $\epsilon$ A, such that a < m.

edit: had some questions.. but rereading your post a few times answered them...

 

[FactChecker]

So basically, if a = -|m| - 200 $\epsilon$ A and a < m, it follows that there is no lower bound, because for all elements m $\epsilon$ A, there exists an a $\epsilon$ A, such that a < m.

edit: had some questions.. but rereading your post a few times answered them...

Yes. And a = -|m| - 200 << -1 is always such a large negative number that it is guaranteed to be in A and lower than m. So there is no need to have 3 cases.

 

[SammyS]

Homework Statement

Prove that {$x \epsilon \mathbb{R} : x^2 \ge 1$} is "not" bounded below.
EDIT: I Looked closely and realized there is a "not" that we all had to write in...sorry if you lost some time..

Homework Equations

Defintion: We say a nonempty subset $A$ of $\mathbb{R}$ is bounded below, if there is a real number $m$ such that $m \le x$ for all $x \epsilon A$. We call such an $m$ an lower bound of $A$.

The Attempt at a Solution

I don't think this is bounded below.

Proof: This will be a proof by contradiction. Suppose $m$ is a lower bound of the set A ={$x \epsilon \mathbb{R} : x^2 \ge 1$}. Then $m \le x$ for all $x \epsilon A$. But $m - 1 < m$ and $m - 1 \epsilon A$.(Not sure how to prove that last statement). Therefore $m > (m-1)$ and $(m-1) \epsilon A$. Therefore m is not a lower bound, a contradiction. We conclude there does not exist a lower bound of A and A is not bounded below. []

Edit: in class we proved that the statement was True and had 3 cases but I don't see why the proof makes sense.

Consider 3 cases:

Case 1: Let $m < -1$ Let $x = m - 1$. Then $x = m - 1 < m$.

Case 2: Let $m \epsilon [-1,1]$. Let $x = m-4$. Then $x = m - 4 < m$.

Case 3: Let $m > 1$. Then $x = -m$. Then $x = -m < m$.

So we showed for any real number m, there exists an x such that x < m. Which I guess fulfills the definition. But I'm saying since the $A$ is going toward negative infinity, then I can always find a $k$ such that $k < x$.

The title of your thread is misleading - or worse.

"Can a set include negative infinity and be bounded below" ?​

The set you describe is a subset of ℝ, the real numbers.

Neither −∞ , nor ∞ is a real number, so your set does not include −∞ .

 

[fishturtle1]

The title of your thread is misleading - or worse.

"Can a set include negative infinity and be bounded below" ?​

The set you describe is a subset of ℝ, the real numbers.

Neither −∞ , nor ∞ is a real number, so your set does not include −∞ .

That is my mistake, I see that −∞ is not in (−∞, ∞). What I meant was "If an interval is approaching −∞, can it also be bounded below?". I'm sorry for the carelessness.