Defining convergence in terms of the set of rationals in the gap between $\liminf$ and $\limsup$

Question

Consider a real sequence $x_n$ and the set $A = \{ r | \liminf x_n < r < \limsup x_n \, , \, r \in \mathbb Q\}$. Is it true that $|A| = 0$ if and only if $x_n$ converges to a real limit?

My attempt:

Note that if $\liminf x_n < r < \limsup x_n$ for some real $r$ then there exists a $q \in \mathbb Q$ close enough to $r$ such that $\liminf x_n < q < \limsup x_n$ since the rationals are dense in the reals. Also, if there exists a rational $q$ such that $\liminf x_n < q < \limsup x_n$ then the inequality is satisfied for a real value since $\mathbb Q \subset \mathbb R$.

So, the set $A$ above has cardinality $0$ if and only if the set $B$ has cardinality $0$ where $B = \{r | \liminf x_n < r < \limsup x_n, r \in \mathbb R\}$.

But $B$ has cardinality $0$ only when $\liminf x_n = \limsup x_n$ which implies that $x_n$ converges to a real limit.

I could use some feedback on my reasoning.

Answer 1

It is correct that $A$ is empty exactly if $B$ is empty.

It is also correct that both sets are empty exactly if $\liminf x_n = \limsup x_n$.

That is the case if the sequence is convergent to a real number, or if it diverges to $+\infty$ or to $-\infty$, i.e. if it converges in the extended real line.

Answer 2

If $|A|$ means measure of $A$ then $|A|=0 $ for any sequence $x_n .$

If $|A|$ means the cardinality of $A$ then $|A|=0$ if and only if $x_n$ converges to finite or infinite limit.

Answer 3

We need the fact: $\lim_{n}x_{n}$ converges to $x\in[-\infty,\infty]$ iff $\liminf_{n}x_{n}=\limsup_{n}x_{n}=x$. Once we have that fact, it is trivial that $A=\emptyset$ iff $\lim_{n}x_{n}$ converges in $[-\infty,\infty]$. To prove the fact, we observe the following:

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Lemma 1: Let $l=\limsup_{n}x_{n}$. Consider the following cases. Case 1: $l\in\mathbb{R}$. In this case, for each $\varepsilon>0$, $\{n\mid x_{n}>l+\varepsilon\}$ is a finite set and $\{n\mid x_{n}>l-\varepsilon\}$ is an infinite set. Case 2: $l=\infty$. In this case, there exists a subsequence $(x_{n_{k}})$ such that $x_{n_{k}}\rightarrow\infty$. Case 3: $l=-\infty$. In this case, $\lim_{n}x_{n}\rightarrow-\infty$.

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Proof of Lemma 1: Case 1: Let $\varepsilon>0$ be given. Since $\lim_{n}(\sup_{k\geq n}x_{k})=l<l+\varepsilon$, there exists $N$ such that $(\sup_{k\geq n}x_{k})<l+\varepsilon$ whenever $n\geq N$. In particular, for any $n\geq N$, $x_{n}\leq(\sup_{k\geq n}x_{k})<l+\varepsilon$. Hence, $\{n\mid x_{n}>l+\varepsilon\}\subseteq\{1,\ldots,N\}$. This shows that $\{n\mid x_{n}>l+\varepsilon\}$ is a finite set. Note that $(\sup_{k\geq n}x_{k})_{n}$ is a decreasing sequence and $\lim_{n}(\sup_{k\geq n}x_{k})=l>l-\varepsilon$. Therefore, $\sup_{k\geq n}x_{k}>l-\varepsilon$ for each $n$. Put $n=1,$ there exists $k\geq1$ such that $x_{k}>l-\varepsilon$. Denote such $k$ by $n_{1}$. Suppose that $n_{1}<n_{2}<\ldots<n_{k}$ have been chosen such that $x_{n_{k}}>l-\varepsilon$. Put $N=n_{k}+1$, then $\sup_{i\geq N}x_{i}>l-\varepsilon$. Choose $i\geq N$ such that $x_{i}>l-\varepsilon$. Denote such $i$ by $n_{k+1}$. By recursion theorem, we obtain a sequence $(n_{k})$ such that $n_{1}<n_{2}<\ldots$ and $x_{n_{k}}>l-\varepsilon$ for each $k$. This shows that $\{n\mid x_{n}>l-\varepsilon\}$ is an infinite set.

Case 2 and Case 3 are easy and are left to you.

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Similarly, we have:

Lemma 2: Let $l=\liminf_{n}x_{n}$. Consider the following cases. Case 1: $l\in\mathbb{R}$. In this case, for each $\varepsilon>0$, $\{n\mid x_{n}<l-\varepsilon\}$ is a finite set and $\{n\mid x_{n}<l+\varepsilon\}$ is an infinite set. Case 2: $l=-\infty$. In this case, there exists a subsequence $(x_{n_{k}})$ such that $x_{n_{k}}\rightarrow-\infty$. Case 3: $l=\infty$. In this case, $\lim_{n}x_{n}\rightarrow\infty$.

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Lemma 1 indicates that $\limsup_{n}x_{n}$ is the largest accumulation point of $(x_{n})$ while Lemma 2 indicates that $\liminf_{n}x_{n}$ is the smallest accumulation point of $(x_{n})$.

Now, we are able to prove the fact at the beginning: Let $h=\liminf_{n}x_{n}$ and $k=\limsup_{n}x_{n}$. We have that $h\leq k$.

We go to show that $h<k\Rightarrow\lim_{n}x_{n}$ does not exist in $[-\infty,\infty]$. Firstly, consider the case that $h,k\in\mathbb{R}$. Choose $a,b\in\mathbb{R}$ such that $h<a<b<k$. Then $A=\{n\mid x_{n}<a\}$ and $B=\{n\mid x_{n}>b\}$ are infinite sets. If $\lim_{n}x_{n}=l$ exists in $[-\infty,\infty]$, by letting $n\rightarrow\infty$, $n\in A$, we have $l\leq a$. Similarly, by letting $n\rightarrow\infty$, $n\in B$, we have $l\geq b$. This is a contradiction. The cases that $h=-\infty$, $k=\infty$ are left to you.

We go to show that $h=k\Rightarrow\lim_{n}x_{n}$ exists in $[-\infty,\infty]$. Denote $l=h=k$. Firstly we consider the case that $l\in\mathbb{R}$. Let $\varepsilon>0$ be arbitrary. By Claim 1, $\{n\mid x_{n}>l+\varepsilon\}$ is a finite set, so there exists $N_{1}$ such that $x_{n}\leq l+\varepsilon$ whenever $n\geq N_{1}$. By Claim 2, $\{n\mid x_{n}<l-\varepsilon\}$ is a finite set, so there exists $N_{2}$ such that $x_{n}\geq l-\varepsilon$ whenever $n\geq N_{2}$. Let $N=\max(N_{1},N_{2})$. If $n\geq N$, then $l-\varepsilon\leq x_{n}\leq l+\varepsilon$, i.e., $|x_{n}-l|\leq\varepsilon$. This shows that $\lim_{n}x_{n}=l$. The cases that $l=\infty$ and $l=-\infty$ are left to you.