Find the $\sup$ of the set.

Question

part of an analysis hw is to find the $\sup$ of the following set:

$E = \{3^n + 1 | n ∈ N\}$ The hint is to prove that $3^n>n$

What I did so far is this:

1)Proof by induction:

$3^0>0$ assume true for $3^k>k$ for any $k = n$

therefore $3^{k+1} = 3 \cdot 3^k > 3k > k+1$

so $3^n > n$. since the set $\{n\}$ is not bounded above then $\{3^n + 1\}$ is not bounded above. so $\sup (E)$ does not exist

Is this nearly correct??

Answer 1

I have a few comments about the layout of your proof. The first is not so essential, but makes it a little easier to read: state the claim you aim to prove before the proof, if the claim is not the full statement. For example "we prove that $3^n>n$ for all $n\in\mathbb N$. Now for the main comments.

Break your induction into clear segments. State first the base case (or anchor step, or whatever you call the starting point of the induction): "For $n=1$ we have $3^1=3>1$ (for example)". Then state the inductive assumption: "Assume now that the claim holds for some (not any or all, these are incorrect) $k\in\mathbb N$". Now prove that the inductive assumption holds for $k+1$ given that it holds for $k$:$$ 3^{k+1}=3\cdot 3^k>3k=2k+k>2k+1>k+1 $$ (I have inserted one or two steps that you omitted. These may or may not be necessary, depending on the complexity of the calculation). It is generally good practice to state where the inductive assumption has been used (the first inequality above). Don't be afraid to add some words, but not too many. Finally, state the conclusion of your induction: "Hence, by induction, we have $3^n>n$ for all $n\in\mathbb N$.

For the final assertion, we see that, for each $n\in\mathbb N$, we can find an element $x\in E$ such that $x>n$ (taking $x=3^n+1$ will do by our induction), and so $\sup E>n$ for all $n\in\mathbb N$. Thus $\sup E=\infty$ (or it doesn't exist, depending on your definition of the supremum.

Answer 2

You are correct that the set $E$ is unbounded and hence has no supremum.

For your induction proof, your base case should be $n=1$, as $0$ is not a natural number (some might debate this). Then you should assume that for some natural number $k \gt 1, 3^k \gt k$. The wording is important.

From there you aim to prove that this assumption holds for $k+1$. So we have now $3^{k+1} = 3 * 3^k \gt 3 * k \gt k+1$, where our assumption comes into play in the first inequality. From there we can conclude that $3^n \gt n$ for all natural numbers $n$.

Finally {n} is ill-defined. You should write your set as $\{n: n \in \mathbb N\}$, and then instead of writing $E$ as $\{3^n+1\}$ (similarly ill-defined), just say $E$.