Computing $\liminf A_{n}$ and $\limsup A_{n}$ for a given sequences of sets For each $n \in \mathbb{N}$, let
$$A_{2n-1}=\left\lbrace (x,y) \in \mathbb{R}^{2} \: | \: y\geq 0, x\geq 1 +\frac{1}{n} \right\rbrace$$
and 
$$A_{2n}=\left\lbrace (x,y) \in \mathbb{R}^{2} \: | \: x \geq 0, 0 \leq y \leq 
 \frac{1}{n} \right\rbrace$$
Calculate  $\liminf A_{n}$ and $\limsup A_{n}$. I don't know how to calculate the mentioned limits manually but my intuition says this sequence converges and it converges to 
$$A= \lbrace (x,y) \in \mathbb{R}^{2} \: | \: x \geq 1, y=0\rbrace.$$ So if this is true then 
$$\liminf A_{n}=A=\limsup A_{n}?$$
I need help with the technicalities of the proof as I've never have worked with limits of sequences involving sets. Thanks!
 A: As @Marios Gretsas indicated in a comment, $\liminf$ is the set $\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k$, 
and $\limsup$ is the set $\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$. 
In other words, $x$ belongs to $\limsup$ if $x$ belongs to infinitely 
many $A_k$, and $x$ belongs to $\liminf$ if $x$ belongs to all but finitely 
many $A_k$. See also https://en.wikipedia.org/wiki/Set-theoretic_limit 
For the sets in the question we do not have that $\limsup=\liminf$. 
It is not clear how you have used your intuition to come up with the result you state in the question. Perhaps you could have provided more details. 
$\limsup$ might be easier to consider first. From the definition of $A_{2n-1}$ 
it is easily seen that $\limsup A_n\supseteq O:=\{(x,y)|y\ge0,x>1\}$. In fact, 
$\limsup A_{2n-1}=O$ since every point $x$ in $O$ belongs to all but finitely many 
$A_{2n-1}$ (and only points $x$ in $O$ have this property). Since $A_{2n-1}$ is 
an increasing sequence (in particular monotone), we have moreover that 
$\limsup A_{2n-1}=\liminf A_{2n-1}=\lim A_{2n-1}=O$. For an increasing sequence of sets we always have $\limsup =\liminf=\lim=$ the union of the sets, so 
$\lim A_{2n-1}=\cup_n A_{2n-1}=O$ (explaining why $O$ looks the way it does, as defined earlier). 
Similarly, $A_{2n}$ is a decreasing sequence and 
$\limsup A_{2n}=\liminf A_{2n}=\lim A_{2n}=\cap_n A_{2n}=E:=\{(x,y)|x\ge0,y=0\}$. 
This is the positive part of the $x$-axis, including the origin. 
It follows that $\limsup A_n=O\cup E$. 
On the other hand, $\liminf A_n=O\cap E=\{(x,y)|x>1,y=0\}$. 
