Cofinal set is uncountable
Consider sequences of natural numbers. We say that a sequence $a$ majorizes $b$, notation $a\succ b$, if $$\lim_{n\to \infty}\frac{b_n}{a_n}=0.$$ We call a set of sequences $C$ cofinal if for any sequence $v$ there is $u\in C$ such that $u$ majorizes $v$. $\textbf{Statement}.$ Every cofinal set is uncountable. $\textbf{Proof}.$ The proof is utilization of Cantor's diagonal argument. Assume $C$ is a cofinal countable set, denote its elements as $a^k$ for $k=1,2,3,\ldots$. First of all observe that $\succ$ is a transitive relation and for any finite set of sequences, there is a sequence $u$ which majorizes all of them (consider $n$ times the pointwise maximum). Construct a sequence of sequences $b^k$ as follows - take $b^1_n=\max\{a^1_1,\ldots,a^1_n\}$. Suppose we have constructed $b^j$ for $j=1,\ldots,k$. Construct $b^{k+1}$ as an nondecreasing sequence which majorizes all of $a^i$ for $i=1,\ldots,k+1$ and $b^j$ for $j=1,\ldots,k$. Thus $b^j\succ a^j$ and $b^{j+1}\succ b^j$ a