every bounded sequence has limit along an ultrafilter

Theorem 1.

Let $\mathrm{F}$ be an ultrafilter on $\mathrm{N}$ and $\mathrm{(}{x}_n\mathrm{)}$ be a real boundedsequence. Then $\mathrm{F}\mathit{}\mathit{\text{-}}\mathit{}\lim \mathit{}{x}_n$ exists.

Proof.

Let $(x_n)$ be a bounded sequence. Choose $a_0$ and $b_0$ suchthat $a_0\le x_n\le b_0$. Put $c_0:=\frac{a_0+b_0}{2}$. Thenprecisely one of the sets $\{n\in \mathbb{N};x_n\in ⟨a_0,c_0⟩\}$,$\{n\in \mathbb{N};x_n\in ⟨c_0,b_0⟩\}$ belongs to the filter $ℱ$.(Their union is $\mathbb{N}$ and the filter $ℱ$ is an ultrafilter.) Wechoose $⟨a_1,b_1⟩$ as that subinterval from$⟨a_0,c_0⟩$ and $⟨c_0,b_0⟩$ for which $C:=\{n\in \mathbb{N};x_n\in ⟨a_1,b_1⟩\}$ belongs to $ℱ$.

Now we again bisect the interval $⟨a_1,b_1⟩$ by putting$c_1=\frac{a_1+b_1}{2}$. Denote $A:=\{n\in \mathbb{N};x_n\in ⟨a_1,c_1⟩\}$, $B:=\{n\in \mathbb{N};x_n\in ⟨c_1,b_1⟩\}$. Itholds $B\cup A\cup (\mathbb{N}\setminus C)=\mathbb{N}$. By the alternativecharacterization of ultrafilters we get that one of these sets isin $ℱ$. The set $\mathbb{N}\setminus C$ doesn’t belong to $ℱ$, thereforeit must be one of the sets $A$ and $B$. We choose thecorresponding interval for $⟨a_2,b_2⟩$.

By induction we obtain the monotonous sequences $(a_n)$, $(b_n)$with the same limit $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}{b}_n:=L$ such thatfor any $n\in \mathbb{N}$ it holds $\{n\in \mathbb{N};x_n\in ⟨a_1,b_1⟩\}\in ℱ$.

We claim that $ℱ\text{-}lim{x}_n=L$. Indeed, for any $\epsilon >0$ there is$n\in \mathbb{N}$ such that $⟨a_n,b_n⟩\subseteq (L-\epsilon ,L+\epsilon )$, thus$\{n\in \mathbb{N};x_n\in ⟨a_n,b_n⟩\}\subseteq A(\epsilon )$. The set$\{n\in \mathbb{N};x_n\in ⟨a_1,b_1⟩\}$ belongs to $ℱ$, hence$A(\epsilon )\in ℱ$ as well.∎

Note that, if we modify the definition of $ℱ$-limit in a such waythat we admit the values $\pm \mathrm{\infty }$, then every sequence has$ℱ$-limit along an ultrafilter $ℱ$. (The limit is $+\mathrm{\infty }$ iffor each neighborhood $V$ of infinity, the set $\{n\in \mathbb{N};x_n\in V\}$ belongs to $ℱ$. Similarly for $-\mathrm{\infty }$.)