How do I prove that If S is bounded then there exist a positive real number M such that |x| ≤ M for all x ∈ S ?
w/o the absolute value, it would be:
−M≤x≤M
Sorry Raban, unfortunately I can't help, I've done way too much math today and that question is giving me headache. I hope you figure it out.
looks like that is the very definition of boundedness.
ok and?
To prove the statement "If S is bounded, then there exists a positive real number M such that |x| ≤ M for all x ∈ S," we can use the definition of boundedness and the properties of absolute value.
Here's the step-by-step explanation:
1. Start with the assumption that S is bounded. This means that there exists a real number K such that |x| ≤ K for all x ∈ S.
2. Consider the absolute value |K|. Since K is a real number, |K| is either equal to K (if K ≥ 0) or -K (if K < 0).
3. Define M as the maximum of |K| and 0. That is, M = max(|K|, 0).
4. Now, for any x ∈ S, we need to prove that |x| ≤ M.
5. Case 1: If K ≥ 0, then |K| = K. In this case, since |x| ≤ K for all x ∈ S (by the assumption), it follows that |x| ≤ |K| = K ≤ M.
6. Case 2: If K < 0, then |K| = -K. In this case, since |x| ≤ K for all x ∈ S (by the assumption), we have -|x| ≥ -K (since the negative of a negative number is positive). Rearranging, we get |x| ≤ -K = |K| ≤ M.
7. In both cases, we've shown that |x| ≤ M for all x ∈ S, which proves the statement.
In summary, if S is bounded, then there exists a positive real number M (specifically, M = max(|K|, 0)) such that |x| ≤ M for all x ∈ S.