Proposition 1.Suppose $M$, $N\subseteq\mathbb{R}$ have minima and maxima. If $\max(M\cap N)$ exists, then $\max(M\cap N)\leq\min\{\max M,\max N\}$.
Proof.Since $\max(M\cap N)$ exists, let $x\in M\cap N$, then $x\in M$ and $x\in N$. So $x\leq\max M$ and $x\leq\max N$. Hence $\max(M\cap N)\leq\min\{\max M,\max N\}$. $\square$
Notice that $\max(M\cap N)$ does not always exist even if $M\cap N\neq\varnothing$: Take $M=[0,1]$ and $N=[0,1)\cup\{2\}$, then $M\cap N=[0,1)$, which has no maximum.
Proposition 2.If $A_n$ is countable for every $n\in\mathbb{N}$, then $\bigcup_{n\in\mathbb{N}}A_n$ is countable.
Proof.Since $A_n$ is countable for every $n\in\mathbb{N}$, there is a bijection $f_n:\mathbb{N}\to A_n$ for every $n$. Define
\[h : \mathbb{N}\times\mathbb{N}\to\bigcup_{n\in\mathbb{N}}A_n,(n,m)\mapsto f_n(m).\]Let $x\in\bigcup_{n\in\mathbb{N}}A_n$, then there is an $n\in\mathbb{N}$ such that $x\in A_n$. Since $f_n$ is bijective, there is an $m\in\mathbb{N}$ with $f_n(m)=x$. We have $x=f_n(m)=h(n,m)$, which shows $h$ is surjective. Since $\mathbb{N}\times\mathbb{N}$ is countable, $\bigcup_{n\in\mathbb{N}}A_n$ is countable. $\square$
Proposition 3.For any set $X$ there is no surjection $X\to\mathcal{P}(X)$.
Proof.Let $X$ be a set. Assume there is a surjection $f:X\to\mathcal{P}(X)$. Let
\[A = \{x\in X : x\notin f(x)\},\]then $A\in\mathcal{P}(X)$, so there is an $a\in X$ such that $f(a)=A$. Now if $a\in f(a)$, then $a\notin f(a)$; if $a\notin f(a)$, then $a\in f(a)$. A contradiction. $\square$
Theorem 1.The limit of a sequence does not change after its members are rearranged. That means, let $a:\mathbb{N}\to\mathbb{R}$ be a convergent sequence and $f:\mathbb{N}\to\mathbb{N}$ be a bijection, then
\[\lim_{n\to\infty} a_{f(n)} = \lim_{n\to\infty} a_n.\]We prove a stronger result:
Proposition 4.Let $a:\mathbb{N}\to\mathbb{R}$ be a convergent sequence and $f:\mathbb{N}\to\mathbb{N}$ be a mapping such that $\lim_{n\to\infty} f(n) = \infty$, then
\[\lim_{n\to\infty}a_{f(n)} = \lim_{n\to\infty} a_n.\]Proof.Let $\varepsilon>0$. Put $\lim_{n\to\infty} a_n = a^*$, then there is an $M\in\mathbb{N}$ such that $\lvert a_m-a^*\rvert<\varepsilon$ for all $m\geq M$. Since $\lim_{n\to\infty} f(n) = \infty$, there is an $N\in\mathbb{N}$ such that $f(n)\geq M$ for all $n\geq N$. It follows that $\lvert a_{f(n)}-a^*\rvert<\varepsilon$, which means $\lim_{n\to\infty} a_{f(n)}=a^*$. $\square$
Proposition 4 is stronger than Theorem 1 as every injection in $\mathbb{N}$ diverges to $\infty$.