Can a set include negative infinity and be bounded below

  • #1
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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##.
 
Last edited:

Answers and Replies

  • #2
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Are you considering the negative numbers x where x2 ≥ 1? How low can those go?
 
  • #3
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Are you considering the negative numbers x where x2 ≥ 1? How low can those go?
To negative infinity
 
  • #4
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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.
 
  • #5
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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...
 
  • #6
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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.
 
  • #7
SammyS
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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 −∞ .
 
  • #8
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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.
 

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