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I have to prove that:

By adding a finite number of sets to the set sequence $(A_n)_{n\in\mathbb{N}}$ the limit superior doesn't change.

I constructed a new sequence $(E_n)_{n \in \mathbb{N}}$ by adding the set $A$ to the sequnce $(A_n)$. That is $$E_1=A,\; E_2=A_1,\; \ldots,\; E_n=A_{n-1},\; \ldots$$ I have to prove that $$\lim\sup E_n=\lim\sup A_n$$ By definition: $$\lim\sup E_n= \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k$$ $$\lim\sup A_n= \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$$

We have that: \begin{align*} &\bigcup_{k=n}^\infty E_k=E_n\cup E_{n+1}\cup \ldots=A_{n-1}\cup A_n\cup\ldots \\ & \bigcup_{k=n}^\infty A_k=A_n\cup A_{n+1}\cup \ldots \\ \Longrightarrow & \bigcup_{k=n}^\infty A_k \subseteq \bigcup_{k=n}^\infty E_k\\ \Longrightarrow& \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\subseteq \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k\\ \Longrightarrow & \lim\sup A_n \subseteq \limsup E_n \end{align*} Now I want to prove that $$\lim\sup E_n \subseteq \limsup A_n$$ So I took a random $x\in \lim\sup E_n$. That means \begin{align*} (\forall n\in \mathbb{N}) \; x\in \bigcup_{k=n}^\infty E_k & \Leftrightarrow x\in E_n \cup E_{n+1} \cup \ldots \\ & \Leftrightarrow x\in A_{n-1}\cup A_n \cup \ldots \\ & \Leftrightarrow x\in \bigcup_{k=n-1}^\infty A_k \end{align*}

Now I have somehow to imply from here that $x\in \bigcup_{k=n}^\infty A_k$, but that's where I have a problem, because for $n=1$ I have that $$x\in \bigcup_{k=0}^\infty A_k=A_0\cup A_1 \cup \ldots$$ What if $x\in A_0$, then I can't say that $x\in \bigcup_{k=n}^\infty A_k$. So if anyone has a good idea what to do, please.

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Assume $x\in\limsup E_n$. Let's prove that for every $n$ there exists $k>n$ such that $x\in A_k$. Given $n$, there exists $k>n+1$ such that $x\in E_k=A_{k-1}$ (because $x\in\limsup E_n$). Since $k-1>n$, we have that $x\in\bigcup_{k=n}^{\infty}A_k$. Since $n$ was arbitrary, we deduce that $x\in\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$. This proves that $\limsup E_n\subset \limsup A_n$.

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  • $\begingroup$ Why is $k>n$, and not $k\geq n$? $\endgroup$ – zermelovac Oct 29 '15 at 20:37
  • $\begingroup$ Because $k>n+1$. If you give me $n$, I use the fact that $x$ belongs to the union of $A_k$'s when $k$ starts from any $n$, and in particular from $n+1$. $\endgroup$ – uniquesolution Oct 29 '15 at 20:39
  • $\begingroup$ So, the situation that $x\in A_0$ that I came across isn't possible? $\endgroup$ – zermelovac Oct 29 '15 at 20:41
  • $\begingroup$ It is possible, but $A_0$ would not be the only $A_k$ that has $x$ in it. $\endgroup$ – uniquesolution Oct 29 '15 at 20:42
  • $\begingroup$ In your previous comment, did you mean that $x$ belongs to the union of $E_k$'s or? $\endgroup$ – zermelovac Oct 29 '15 at 20:45
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The limsup of the sequence of sets $A_n$ is the set of all $x$ in $X$ such that there are infinitely many $n$ with $x \in A_n$. This is often an easier way of thinking about it. This follows directly from the formulaic definition if you think about it a little.

Adding finitely many sets to the sequence doesn't change those $x$. That's the whole proof in a nutshell.

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