Infimum and Supremum of a set 2^k

Question

I am trying to find the infimum and supremum of a set 2^k where k is an integer. I have determined that as k gets larger, so does 2^k so it is not bounded above and therefore there is no supremum. Then, as k get smaller (More negative), 2^k will become 1/(2^k) and so it will trend to 0, which will be the infimum.

While I understand this, I have no idea how to formally answer this question beyond the fact that I got these answers intuitively.

Thank you in advance!

Answer 1

To show that 0 is the infimum, you can first show that $0<2^k$ for all $k\in\mathbb{Z}$. Then, you can show that if $\epsilon\in\mathbb{R}$ is greater than 0, then there is some $k_0\in\mathbb{Z}$ such that $2^{k_0}\leq\epsilon$.

To show that $\infty$ is the supremum, you can show that no matter what upper bound you pick, say $M$, that there is some $k_1\in\mathbb{Z}$ such that $2^{k_1}\geq M$.

Hint: The Archimedean property will be very helpful!

Answer 2

Good job on the intuition!

Now formal:

To show $\{2^k\}$ is not bounded above you must show there is no $M$ so that $M\ge 2^k$ for all $k\in \mathbb Z$.

If $\{2^k\}$ were bounded above then $\sup 2^k$ exists. We can assume $2^k > 1$ for obvious reasons (because $\sup 2^k \ge 2^1 > 1$) But then for any $1 < w < \sup 2^k$, $w$ is not an upper bound. So there will be a $2^m$ so that $w < 2^m \le \sup 2^k$. But that would mean $1< w < 2^m < 2^{m+1} \le \sup 2^k$.

But let's not forget that $w$ can be any number less than $\sup 2^k$. That is to say we can have $w$ be as arbitrarily close to $\sup 2^k$ as we like. Bue $\sup 2^k - w > 2^{m+1} - 2^{m} = 2^m(2-1) = 2^m > 1$. Bu this would be true for $w = \sup 2^k -1$. If $w=\sup 2^k-1$ we must have $2^m$ and $2^{m+1}$ so that $w= \sup 2^k - 1 < 2^m < 2^{m+1} \le \sup 2^k$ which is impossible.

So we can't have $\{2^k\}$ bounded above.

Now to prove $\inf 2^k = 0$.

Note $2^k > 0$ so $0$ is a lower bound. So $\inf 2^k$ exists and $\inf 2^k \ge 0$.

Let's suppose $\inf 2^k > 0$. then means for any $w > \inf 2^k$ then there exist an $2^m$ so that $w > 2^m \le \inf 2^k$.

But what if we make $w = 2\cdot \inf 2^k$. If $\inf 2^k > 0$ then $w =2\cdot \inf 2^k> \inf 2^k$. But that means $2\cdot \inf 2^k > 2^m > 2^{m-1} \ge \inf 2^k$.

Do you see how that is a contradiction? There's "no room" for those powers of $2$. If we divide each term but $\frac 12$ we get $\frac w2 = \inf 2^k > 2^{m-1} > 2^{m-2}\ge \frac 12\cdot \inf 2^{k}$ contradicting both that $\inf 2^k$ is a lower bound, and the $2^{m-1} \ge \inf 2^k$.

So we can't have $\inf 2^k > 0$. So $\inf 2^k \ge 0$ but $\inf 2^k > 0$ is impossible... so $\inf 2^k = 0$.