If $\langle a_n \rangle_{n=0}^\infty$ is convergent and $\lim_{n \to \infty}a_n=a$, then for every nonprincipal ultrafilter $U$, $\lim_U a_n=a$

Question

My textbook Introduction to Set Theory 3rd by Hrbacek and Jech introduces a definition related to ultrafilter as follows:

2.7. Definition Let $U$ be an ultrafilter on $\mathbb{N}$, and let $\langle a_n\rangle_{n=0}^{\infty}$ be a bounded sequence of real numbers. We say that a real number $a$ is the $U$-limit of the sequence,

$$ a = \lim_{U} a_n, $$

if for every $\epsilon > 0$, $\{n \mid |a_n - a| < \epsilon\} \in U$.

First, we observe that if $U$-limit exists then it is unique. For, assume that $a< b$ and both $a$ and $b$ are $U$-limits of $\langle a_n\rangle_{n=0}^{\infty}$. Let $\epsilon = (b-a)/2$. Then the sets $\{n \mid |a_n - a| < \epsilon\}$ and $\{n \mid |a_n - b| < \epsilon\}$ are disjoint and so cannot both be in $U$.

If $U$ is a principal ultrafilter, $U = \{ A \mid n_0 \in A\}$ for some $n_0$, then $\lim_U a_n = a_{n_0}$, for any sequence $\langle a_n\rangle_{n=0}^{\infty}$. This is because for every $\epsilon > 0$, $\{ n \mid |a_n - a_{n_0}| < \epsilon\} \supseteq \{n_0\} \in U$.

If $\langle a_n\rangle_{n=0}^{\infty}$ is convergent and $\lim_{n\to\infty} a_n = a$, then for every nonprincipal ultrafilter $U$, $\lim_U a_n = a$. This is because for every $\epsilon > 0$, there exists some $k$ such that $\{n \mid |a_n - a| < \epsilon\} \subseteq \{ n \mid n \geq k \}$, and $\{n \mid n \geq k\} \in U$.

The important property of ultrafilter limits is that the $U$-limit exists for every bounded sequence:

There is a statement in the paragraph:

If $\langle a_n \rangle_{n=0}^\infty$ is convergent and $\lim_{n \to \infty}a_n=a$, then for every nonprincipal ultrafilter $U$, $\lim_U a_n=a$. This is because for every $\epsilon > 0$, there exists some $k$ such that $\{n \mid |a_n - a| < \epsilon\} \supseteq \{n \mid n \ge k\}$, and $\{n \mid n \ge k\} \in U$.

Please confirm if my understanding of this paragraph is correct or not! Thank you so much!

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Lemma: An ultrafilter which contains a finite set is principal.

Proof: Let $U$ be an ultrafilter on $X$ and $A\in U$ be finite. Then either there exists some $a \in A$ such that $\{a\} \in U$, implying that $U$ is principal, or else $\{a\} \notin U$ for all $a \in A$ and thus $X- \{a\} \in U$ for all $a \in A$, so $U \ni A \cap \bigcap_{a\in A}(X - \{a\}) =\emptyset$, leading to a contradiction.

It follows from $\lim_{n \to \infty}a_n=a$ that $\forall \epsilon > 0, \exists N \in \Bbb N, \forall n \ge N: |a_n - a| < \epsilon$ and thus $\{n \in \Bbb N \mid |a_n - a| < \epsilon\} \supseteq \{n \in \Bbb N \mid n \ge N\}$.

I claim that there exists $M \in \Bbb N$ such that $M \ge N$ and $\{n \in \Bbb N \mid n \ge M\} \in U$. If not, $\{n \in \Bbb N \mid n \ge M\} \notin U$ for all $M \ge N$. This combining with the fact that $U$ is a ultrafilter implies $\Bbb N - \{n \in \Bbb N \mid n \ge M\} \in U$ or $\{n \in \Bbb N \mid n < M\} \in U$ for all $M \ge N$. As a result, there exists at least a finite set of natural numbers in $U$. This clearly contradicts to our Lemma.

We have $\{n \in \Bbb N \mid |a_n - a| < \epsilon\} \supseteq \{n \in \Bbb N \mid n \ge N\} \supseteq \{n \in \Bbb N \mid n \ge M\}$ where $\{n \in \Bbb N \mid n \ge M\} \in U$. It follows that $\{n \in \Bbb N \mid |a_n - a| < \epsilon\} \in U$.

Answer 1

Your understanding is good. If you want to utilize the lemma, however, it would be better to know a more direct argument:

• Assume that $\lim a_n = a$ and $U$ is a non-principal ultrafilter. Our goal is to show that, for each $\epsilon > 0$, the set $A_{\epsilon} = \{ n : |a_n - a| < \epsilon \}$ lies in $U$.

  Since $U$ is an ultrafilter, either $A_{\epsilon}\in U$ or $\mathbb{N}\setminus A_{\epsilon}\in U$. But since $\lim a_n = a$, we know that $\mathbb{N}\setminus A_{\epsilon}$ is finite, and so, cannot lie in $U$ by the lemma. Therefore $A_{\epsilon} \in U$ as required.

Alternatively, we may combine the following several independent facts:

1. If $F \subseteq F'$ are filters and $\lim_F a_n = a$, then $\lim_{F'} a_n = a$.

2. Let $F_{\mathrm{co}} = \{ A \subseteq \mathbb{N} : \mathbb{N}\setminus A\text{ is finite} \}$ be the Fréchet filter (a.k.a. cofinite filter). Then $\lim_n a_n = a$ if and only if $\lim_{F_{\mathrm{co}}} a_n = a$.

3. Fréchet filter is contained in every non-principal ultrafilter. (This is a direct consequence of the lemma in OP.)