Exercise: Upper limit of a sequence of sets.

Question

Let $\langle A_n:n\in\Bbb N\rangle$ be a sequence in $X$ ,$\lim\sup A_n=\bigcap_{n\in N}\bigcup_{k>n} A_k$.$$$$ Is the solution of the following exercise correct?I'm looking for the upper limit. $$ A_n= \begin{cases} (\frac{1}{n}, \frac{2}{3}-\frac{1}{n})\text{if}\,n=1,3,5...\\ (\frac{1}{3}-\frac{1}{n},1+\frac{1}{n})&\text{if}\,n=2,4,6..\\ \end{cases} $$

$\bigcup_{k>n}(\frac{1}{n}, \frac{2}{3}-\frac{1}{n})=(\frac{1}{n},\frac{2}{3})$ si $n=1,3,5...$

$\bigcup_{k>n}(\frac{1}{3}-\frac{1}{n},1+\frac{1}{n})=(\frac{1}{3}-\frac{1}{n},1+\frac{1}{n})$ si $n=2,4,6...$

$\bigcap_{n\in N}(\frac{1}{n},\frac{2}{3})=(0,\frac{2}{3})$ si $n=1,3,5...$

$\bigcap_{n\in N}(\frac{1}{3}-\frac{1}{n},1+\frac{1}{n})=(\frac{1}{3},1]$

$(0,\frac{2}{3})\cup(\frac{1}{3},1]=(0,1]$

Is it correct to evaluate the case of even numbers and odd numbers separately in order to finally union both results?

My reasoning is as follows:

Let $A_e$ the sequence of subsets for even numbers and $A_o$ the sequence of subsets for odd numbers.

$\bigcup_{k>n}A_e\cup A_o=\bigcup_{k>n}A_e\cup \bigcup_{k>n}A_o$

$\bigcap_{n\in N}(\bigcup_{k>n}A_e\cup \bigcup_{k>n}A_o)=\bigcap_{n\in N}(\bigcup_{k>n}A_e)\cup\bigcap_{n\in N}(\bigcup_{k>n}A_o)$

Is it correct?

Answer 1

The final answer is correct, but there are intermediate steps which do not look right.

• Firstly, you should say to which space/set $X$ the elements in your sequence $\langle A_n : n\in \mathbb N \rangle$ belong to; I'm assuming that in this case $X$ is either $\mathbb Q$ or $\mathbb R$.
• You can indeed evaluate the expression by separating the even and odd cases, but the notation in your argument to support this claim is slightly misleading, since it seems as if for each $n \in \mathbb N$, $A_k = A_e \cup A_o$ for all $k > n$, and this is not true. A better way of writing this would be as follows: for each $n \in \mathbb N$ we have that $$\begin{align} \bigcup_{k > n} A_k &= \bigcup_{k \text{ even } > n} A_k \cup \bigcup_{k \text{ odd } > n}A_k \\ &=\bigcup_{e > n} A_e \cup \bigcup_{o > n}A_o \end{align},$$ where $e \in \{k \in \mathbb N : k \text{ is even}\}$ and $o \in \{k \in \mathbb N : k \text{ is odd}\}$.
• It also seems as if you've confused the $k$ and $n$ subscripts when computing the unions. Note that we have $\lim\sup A_n=\bigcap_{n\in N}\bigcup_{k>n} A_k$, so for $n \in \mathbb N$ we have:

$$ \bigcup_{k \text{ odd } > n}A_k = \bigcup_{k \text{ odd }>n}\Big(\frac{1}{\color{red} k}, \frac{2}{3}-\frac{1}{\color{red} k}\Big)=\Big(0,\frac{2}{3}\Big) \ \ $$

and

$$\bigcup_{k \text{ even } > n}A_k = \bigcup_{k \text{ even }>n}\Big(\frac{1}{3}-\frac{1}{\color{red}k},1+\frac{1}{\color{red}k}\Big)= \begin{cases} \Big(\frac{1}{3} - \frac{1}{\color{red}{n+2}}, 1 + \frac{1}{\color{red}{n+2}}\Big) & \text{ if } n \text{ is even} \\ \Big(\frac{1}{3} - \frac{1}{\color{red}{n+1}}, 1 + \frac{1}{\color{red}{n+1}}\Big) & \text{ if } n \text{ is odd}\end{cases}$$ Therefore the corresponding intersections are computed as follows:

$$\bigcap_{n \in \mathbb N} \bigcup_{k \text{ odd } > n}A_k = \bigcap_{n \in \mathbb N} \Big(0,\frac{2}{3}\Big) = \Big(0,\frac{2}{3}\Big)$$

and

$$\bigcap_{n \in \mathbb N}\bigcup_{k \text{ even } > n}A_k = \Big[\frac{1}{3}, 1\Big]; $$ note that $\frac{1}{3}$ is included in all sets $A_e$, so the interval resulting from this last intersection is closed. Therefore:

$$\begin{align} \lim\sup_{n \to \infty} A_k &= \bigcap_{n \in \mathbb N} \bigcup_{k > n} A_k \\ &= \bigcap_{n \in \mathbb N} \Big( \bigcup_{k \text{ even } > n} A_k \cup \bigcup_{k \text{ odd } > n}A_k \Big) \\ &= \bigcap_{n \in \mathbb N} \Big( \bigcup_{k \text{ even } > n} A_k \cup \Big(0,\frac{2}{3}\Big) \Big) \\ &= \Big( \bigcap_{n \in \mathbb N} \bigcup_{k \text{ even } > n} A_k\Big) \cup \Big(0,\frac{2}{3}\Big) \\ &= \Big[ \frac{1}{3}, 1\Big] \cup \Big(0,\frac{2}{3}\Big) \\ &= \Big( 0, 1\Big] \end{align}$$