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For my course in real analysis I'm working on the problem:

Let $A_1,A_2,A_3,...$ be subsets of a metric space and $B_n=\cup^{n}_{i=1}A_i$, prove that $$\overline{B_n}=\cup^{n}_{i=1}\overline{A_i},\quad n=1,2,3,...$$

My attempt: (1) Let $x\in\overline{A_i}$ a limit point of $A_i$, then for any neighbourhood $N$ of $x$, $A_i\cap N\neq \emptyset$ so $N\cap B_n$ is not empty. Therefore $x\in B_n'$ and $x\in \overline{B}_n$.

(2) Let $x\in B_n'$, then for any neighbourhood $N$ of $x$, $N\cap(A_1\cup ... \cup A_n)\neq\emptyset$. So for any neighbourhood, $N\cap A_i\neq\emptyset$ for some $A_i$. Therefore $x$ is a limit point of this $A_i$.

Now I talked about the proof with one of the teachers why said that the second argument was invalid; but I didn't manage to understand why that would be the case. So I'm hoping someone here can explain the error better.

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  • $\begingroup$ The last sentence is wrong because another nhood may intersect with some A_j /= A_i but not with A_i. $\endgroup$ – William Elliot May 11 '17 at 9:59
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The problem lies in the detail here: To show that $x$ is a limit point of $A_j$, you have to show that $N \cap A_j \neq \emptyset$ for every $N$. What you have shown, however, is that for every $N$ there exists an $i$ such that $N \cap A_i \neq \emptyset$. Do you see the problem? You can't show that $j = i$ holds for all $N$, the $i$ might differ between $N$.

To repair your proof, you might want to show/use that a finite union of closed sets is closed. Then use that the closure is the minimal closed set, such that...

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  • $\begingroup$ I understand. Is it actually impossible to show that there is at least one such $A_i $ for which all neighbourhoods are non-empty? Since that is the argument I'm trying to make here. Or does that sort of argument run in to problems when you're working outside the reals? (I don't see a problem in the reals because any open set in the reals contains infinitely many elements (Q is dense in R)) admittedly, I don't entirely know how you'd formalise that argument though. $\endgroup$ – Mitchell Faas May 11 '17 at 11:47
  • $\begingroup$ Of course there is always one such $A_i$, namely the one that satisfies $x \in \overline{A_i}$. But then you would already use what you are trying to show... $\endgroup$ – Dirk May 11 '17 at 12:54
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Proceed by induction. The equality is trivially true for $n= 1.$ For $n=2$ we have:

$\overline{B_2} = \overline{A_{1}\cup A_{2}}$

$= (A_{1}\cup A_{2})\cup (A_{1}\cup A_{2})'$

$= (A_{1}\cup A_{2})\cup (A_{1}'\cup A_{2}')$

$= (A_{1}\cup A_{1}')\cup (A_{2}\cup A_{2}')$

$=\overline{A_1}\cup\overline{A_2}$. Let the statement be true for some $n=k$. Now consider $B_{k+1} = \displaystyle \cup_{i=1}^{k+1}A_i$

$\Rightarrow \overline{B_{k+1}} = \overline{\displaystyle \cup_{i=1}^{k+1}A_i}$

$=\displaystyle\overline{\cup_{i=1}^{k}A_i\cup A _{k+1}}$

$=\displaystyle \overline{\cup_{i=1}^{k}A_i}\cup\overline{A_{k+1}}$

$= \overline{B_{k}}\cup\overline{A_{k+1}}$

$= \displaystyle\cup_{i=1}^{k}\overline{A_i}\cup\overline{A_{k+1}}$

$=\cup_{i=1}^{k+1}\overline{A_{i}}$

Note: for the case of $n=2$ we used the fact that $(A_{1}\cup A_{2})'= (A_{1}'\cup A_{2}')$ which can be easily proved as below:

Consider $x \in (A_{1}\cup A_{2})'$.

$\iff \forall \, \varepsilon >0 \, \, N_{\varepsilon}(x) \cap ((A_{1}\cup A_{2})\setminus \{x\}) \neq \emptyset$

$\iff \forall \, \varepsilon >0 \, \, N_{\varepsilon}(x) \cap (A_{1} \setminus \{x\})\cup(N_{\varepsilon}(x) \cap (A_{2} \setminus \{x\})\neq \emptyset$

$\iff x$ is a limit point of $A_{1}$ or $A_{2}$

$\iff x \in A_{1}'\cup A_{2}'$

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