shah_nosrat Posted March 12, 2012 Share Posted March 12, 2012 Hi, I came across this theorem and decided to prove it, as follows: Theorem: A set [math]A \subseteq R[/math] is bounded if and only if it is bounded from above and below. I would like the prove the converse of the above statement; If a set [math]A \subseteq R[/math]is bounded from above and below, then it is bounded. Let [math]M = |M_{1}| + |M_{2}|[/math]and using this preliminary result I proved earlier [math]-|a| \leq a \leq |a|[/math]. Now, [math]\forall a \in A[/math] we have [math] a \leq M_{1}[/math] ---> definition of bounded from above. and [math] M_{2} \leq a[/math] ---> definition of bounded from below. Using the result: [math]-|a| \leq a \leq |a|[/math]. Since [math]M = |M_{1}| + |M_{2}|[/math] is the sum of absolute value of [math]|M_{1}|[/math] and [math]|M_{2}|[/math], it is a big number. (trying to convince myself). Also, [math]-M = -(|M_{1}| + |M_{2}|)[/math], This is on the opposite of the spectrum. Now, [math]-M \leq -|M_{2}| \leq M_{2} \leq -|a| \leq a \leq |a| \leq M_{1} \leq |M_{1}| \leq M [/math]. [math]M = |M_{1}| + |M_{2}| \geq |M_{1} + M_{2}| \geq 0 [/math] ---> Which shows that M is positive by using the triangle inequality. Hence, [math]-M \leq a \leq M[/math]. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% I'm excited for this proof, hopefully it's correct. Your help is once again appreciated. Link to comment Share on other sites More sharing options...
kavlas Posted May 15, 2012 Share Posted May 15, 2012 (edited) Hi, I came across this theorem and decided to prove it, as follows: Theorem: A set [math]A \subseteq R[/math] is bounded if and only if it is bounded from above and below. I would like the prove the converse of the above statement; If a set [math]A \subseteq R[/math]is bounded from above and below, then it is bounded. Let [math]M = |M_{1}| + |M_{2}|[/math]and using this preliminary result I proved earlier [math]-|a| \leq a \leq |a|[/math]. Now, [math]\forall a \in A[/math] we have [math] a \leq M_{1}[/math] ---> definition of bounded from above. and [math] M_{2} \leq a[/math] ---> definition of bounded from below. Using the result: [math]-|a| \leq a \leq |a|[/math]. Since [math]M = |M_{1}| + |M_{2}|[/math] is the sum of absolute value of [math]|M_{1}|[/math] and [math]|M_{2}|[/math], it is a big number. (trying to convince myself). Also, [math]-M = -(|M_{1}| + |M_{2}|)[/math], This is on the opposite of the spectrum. Now, [math]-M \leq -|M_{2}| \leq M_{2} \leq -|a| \leq a \leq |a| \leq M_{1} \leq |M_{1}| \leq M [/math]. [math]M = |M_{1}| + |M_{2}| \geq |M_{1} + M_{2}| \geq 0 [/math] ---> Which shows that M is positive by using the triangle inequality. Hence, [math]-M \leq a \leq M[/math]. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% I'm excited for this proof, hopefully it's correct. Your help is once again appreciated. yes but you do not show how: [math]-M\leq -|M_{2}|[/math]? Also how do you know that:[math]M_{2} \leq -|a|[/math] ?? Edited May 15, 2012 by kavlas Link to comment Share on other sites More sharing options...
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