Compute $\lim_{n\to \infty} \sup A_{n}$ and $\lim_{n\to \infty} \inf A_{n}$, where $A_{n} = [0,a_{n}]$

Question

I want to compute the limit (if exists) of the sequence of sets defined by $A_{n} = [0,a_{n}]$. Where $\{a_{n}\}$ is a sequence of nonnegative numbers that converges to $a\geq 0$.

I know that I have to compute $ \lim_{n\to \infty } \inf A_{n} $ and $ \lim_{n\to \infty } \sup A_{n} $.

Here's my attempt:

$ \lim_{n\to \infty } \inf A_{n} := \cup_{n=1}^{\infty} \cap_{k=n}^{\infty} A_{k} $.

Let $T_{n} := \{a_{n},a_{n+1},... \} $

I proved that $\cap_{k=n}^{\infty} A_{k} = [0,\inf T_{n}] $, and note that $\{ \inf T_{n} \}_{n \geq 1}$ is a non decreasing sequence.

Then $\lim_{n\to \infty } \inf A_{n} = \cup_{n=1}^{\infty} \cap_{k=n}^{\infty} A_{k} = \cup_{n=1}^{\infty} [0,\inf T_{n}] = [0,\sup_{n\geq 1}\{\inf T_{n}\}] $

since $ \lim_{n\to \infty} a_{n} = a \implies \lim_{n \to \infty} \inf a_{n} :=\sup_{n\geq 1}\{\inf Tn\} = a $. Then

$\lim_{n\to \infty } \inf A_{n} = [0,a]$

Now, I have a problem when I try to compute $ \lim_{n\to \infty } \sup A_{n}$

$ \lim_{n\to \infty } \sup A_{n} := \cap_{n=1}^{\infty} \cup_{k=n}^{\infty} A_{k} $.

I tried to prove that $\cup_{k=n}^{\infty} A_{k} = [0,\sup{T_{n}}]$ but I was not able to prove that $ [0,\sup{T_{n}}] \subset \cup_{k=n}^{\infty} A_{k} $.

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Proof that $\cap_{k=n}^{\infty} A_{k} = [0,\inf T_{n}] $

$\subset:$ Let $x \in \cap_{k=n}^{\infty} A_{k} \implies \forall k \geq n \ \ x \in A_{k} \implies \forall k \geq n \ \ 0\leq x \leq a_{k} \implies x \leq \inf T_{n} \implies x \in [0,\inf T_{n}] $. Then $\cap_{k=n}^{\infty} A_{k} \subset [0,\inf T_{n}] $

$\supset:$ Let $x \in [0,\inf T_{n}] \implies 0 \leq x \leq \inf T_{n} \leq a_{k} \ \ \ \forall k \geq n \implies x \in \cap_{k=n}^{\infty} [0,a_{k}] = \cap_{k=n}^{\infty} A_{k} $. Then $\cap_{k=n}^{\infty} A_{k} \supset [0,\inf T_{n}] $

which implies $\cap_{k=n}^{\infty} A_{k} = [0,\inf T_{n}] $

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Questions:

1. What do you think about the computation of $\lim_{n\to \infty} \inf A_{n} $, is it correct?

2. Do you know how to compute $\lim_{n\to \infty} \sup A_{n}$?

Thanks in advance.

Answer 1

Take $\epsilon>0$. By definition of limit, there exists $n_{\epsilon}: k>n_{\epsilon} \rightarrow |a_k-a|<\epsilon$. We can then say $[0,a-\epsilon] \subset \cap_{k=n_{\epsilon}}^{\infty} A_{k} $. Since we can take $\epsilon$ arbitrarily small, it follows that $[0,a) \subset \lim_{n\to \infty } \inf A_{n} $. The limsup will be either $[0,a)$ or $[0,a]$; if there are an infinite $a_n$ less than $a$, then limsup =$[0,a)$, otherwise it is $[0,a]$.