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How do I go about proving that the set $A=\{ (-2)^n : n \in \mathbb{N} \}$ is unbounded?

I started by showing (using induction) that if $n$ is even then $(-2)^n=2^n$, and since $\mathbb{N}$ is unbounded in $\mathbb{R}$ oubviously $A$ is unbounded. Am I in the right direction? Is it enought to say that for every $n \in \mathbb{N}$ there exist $k \in \mathbb{N}$ such that $(-2)^n=2^n \lt 2^k=(-2)^k$?

If I'm write then showing that $A$ is unbounded from beloew is almot similar.

Any thoughts? Thanks!

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You may want to show that $2^n\ge n$. Then you'd be done. I'm not sure what you're asking in your second to last question. – David Mitra Mar 26 '12 at 12:53
One way to rephrase the statement "$A$ is unbounded" is "there is no positive real number $r$ such that $A$ is contained in the interval $[-r, r]$." That is, there is no $r>0$ such that $|x|\leq r$ for all $x\in A$. Do you see why this is true for your particular set $A$? – Brad Mar 26 '12 at 12:53
I can see that, but how do I prove it? Also, it is enough to say that for every element in the set there exist a greater element? Thanks! – yotamoo Mar 26 '12 at 12:59
Arguing directly, you want to show that given any integer $n$, there is an element in the set greater than $n$ (here, it would be either $(-2)^n$ or $(-2)^{n+1}$). – David Mitra Mar 26 '12 at 13:01
up vote 2 down vote accepted

The argument in your first sentence is correct. But you need to justify the statement "obviously $A$ is unbounded". Why is this so? Well, of course, because $A$ contains all even powers of $2$ and the set of all even powers of 2 is unbounded from above. So, what you need to show is that:

given any positive integer $n$, there is an even positive integer $k$ so that $2^{ k}\ge n$.

The above claim would follow from a more general statement:

given any positive integer $n$, the inequality $2^n\ge n$ holds.

You can prove this statement by induction.

So, I would start of by proving the Lemma: $2^n\ge n$ for any positive integer $n$. Then to show $A$ is unbounded, argue as follows:

Let $n$ be a positive integer.

If $n$ is even, then $2^n=(-2)^n$ is in $A$ and from the lemma we have $(-2)^n=2^n\ge n$.

If $n$ is odd, we have $2^{n+1}=(-2)^{n+1}$ is in $A$ and, again from the lemma, $(-2)^{n+1}=2^{n+1}\ge n+1>n$.

So in either case, we have an element in $A$ greater than or equal to $n$. Since $n$ was an arbitrary positive integer, it follows that $A$ is not bounded from above.

Note that, to show that your $A$ is not bounded, you need only show that $A$ is not bounded from above (or show that it is not bounded from below). Of course, if you were required to show that $A$ is unbounded in both directions, you'd use a similar argument to show that $A$ is not bounded from below.

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