Prove the sequence $\{ 2^n \}$ is unbounded using Russell's paradox I am really stuck on how to make use of Russel's Paradox.
This is where I started.
Proof by contradiction. Suppose $\{ 2^n \}$ converges then it is bounded.
Let $M$ be an element of the reals and a bound for $\{ 2^n \}$. Then there exists an $n$-th element of the naturals such that $n > M$ by the Archimedean property. Thus $2^n > n > M$ but $2^n < M$, arriving at a contradiction.
How do I use Russel's paradox in this proof. Does it have to do with there not being a bijection from the naturals to the powerset of naturals? How would I show this in my proof?
 A: This question isn't completely off base. Your problem can be solved by applying Cantor's Theorem, whose proof relies on an argument that is a generalization of the argument used to demonstrate Russel's paradox. Cantor's theorem is used to prove the inequality $n < 2^n$ which you used in your argument.
Cantor's Theorem: If $f : A \rightarrow \mathcal{P}(A)$ is a function from a set $A$ to it's power set $\mathcal{P}(A)$, then $f$ is not surjective. Consequently $|A| < |\mathcal{P}(A)|$. That is the cardinality of $A$ is strictly less than the cardinality of $\mathcal{P}(A)$.
For finite sets, the cardinality is the number of elements in the set. Using elementary combinatorics, you can prove that if $A$ is a set with $n$ elements then $\mathcal{P}(A)$ has $2^n$ elements. Applying Cantor's Theorem gives the equation
$$n < 2^n$$
for all natural numbers $n$. Since the natural numbers are unbounded this implies that the sequence $\{2^n\}_{n=0}^{\infty}$ is unbounded.
Now for the connection with Russel's Paradox. The proof of Cantor's theorem runs as follows. 
Suppose $f$ is a surjection between $A$ and $\mathcal{P}(A)$. For any $x \in A$, $f(x)$ will be a subset of $A$ and we can as if $x \in f(x)$. Consider the set
$B = \{x \in A:\: x \notin f(x)\}$. If $f$ is surjective than there must be some $y \in A$ such that $f(y) = B$, but then $y \in B$ if and only if $y \notin f(y) = B$. This is a contradiction.
To demonstrate Russel's paradox, suppose $A$ is the hypothetical set of all sets. Then let $f:A \rightarrow \mathcal{P}(A)$ be given by the identity function. Then $B = \{x \in A :\: x \notin x\}$, the set of all sets that are not an element of themselves. As before, this is a contradiction.
So while the problem statement is not technically correct, it's understandable where the confusion came from for the one who created it.
