prove a set is bounded: Proof

[shah_nosrat]

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].

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I'm excited for this proof, hopefully it's correct.

 

Your help is once again appreciated.

[kavlas]

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