- #1

fishturtle1

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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##.

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