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A fun problem: show that for any sequence $A_{1},A_{2},...$ of subsets of a topological space we have:

$$\overline{\bigcup_{i=1}^{\infty} A_{i}} = \bigcup_{i=1}^{\infty} \overline{A_{i}} \cup \bigcap_{i=1}^{\infty} \overline{\bigcup_{j=0}^{\infty}A_{i+j}}.$$

So I think I have the inclusion $\supseteq$. Let's see:

First note that for each $i$ we have $\displaystyle A_{i} \subseteq \bigcup_{i=1}^{\infty} A_{i}$. Taking closure on both sides gives:

$$\overline{A_{i}} \subseteq \overline{\bigcup_{i=1}^{\infty} A_{i}}.$$

Taking the union on both sides from $i=1$ to $\infty$ yields:

$$\bigcup_{i=1}^{\infty} \overline{A_{i}} \subseteq \overline{\bigcup_{i=1}^{\infty} A_{i}}.$$ Call this (*). Now note that (taking $i=1$)

$$\bigcap_{i=1}^{\infty} \overline{\bigcup_{j=0}^{\infty}A_{i+j}} \subseteq \overline{\bigcup_{j=0}^{\infty} A_{j+1}}.$$

Thus:

$$\bigcup_{i=1}^{\infty} \overline{A_{i}} \cup \bigcap_{i=1}^{\infty} \overline{\bigcup_{j=0}^{\infty}A_{i+j}} \subseteq \bigcup_{i=1}^{\infty} \overline{A_{i}} \cup \overline{\bigcup_{j=0}^{\infty} A_{j+1}}.$$

Now by (*) it follows that the above set is a subset of:

$$\overline{\bigcup_{i=1}^{\infty} A_{i}} \cup \overline{\bigcup_{j=0}^{\infty} A_{j+1}}.$$

But the latter set is equal to $\displaystyle \overline{\bigcup_{i=1}^{\infty} A_{i}}$, so we have the inclusion $\supseteq$.

Is this OK? Now, how to prove the other inclusion? I tried contradiction but gets messy.

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What you did for this inclusion seems ok. –  Davide Giraudo Jul 7 '11 at 10:04
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This is an interesting exercise: we know that in general $ \bigcup_{i\in\mathbb{N}}\overline{A_i}\subset \overline{\bigcup_{i\in\mathbb{N}}A_i}$ but we don't have the equality (for example taking $\left\{r_n\right\}$ an enumeration of $\mathbb Q$ and $A_i =\left\{r_i\right\}$). The question we can ask is: what do have to add to $\bigcup_{i\in\mathbb{N}}\overline{A_i}$ to have an equality? And this exercise gives us the answer. –  Davide Giraudo Jul 7 '11 at 12:35
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up vote 2 down vote accepted

Let $\displaystyle x\in\overline{\bigcup_{i\in\mathbb{N}^*}A_i}$. We assume that $\displaystyle x\notin\bigcup_{i\in\mathbb N^*}\overline{A_i}$ (if it's not the case we are done).We have to show that for all $i\geq 1$ we have $\displaystyle x\in\overline{\bigcup_{j\geq i}A_j}$. Let $V$ a neighborhood of $x$. Since $\displaystyle x\notin\bigcup_{i\in\mathbb N^*}\overline{A_i}$ we can find for all $i\geq 1$ an open set $U_i$ which contains $x$ and such that $U_i\cap A_i=\emptyset$. For $i\geq 1$, let $\displaystyle V_i=\bigcap_{k=1}^iU_k$: this is an open set which contains $x$. Since $V\cap V_i$ is still a neighborhood of $x$ and $\displaystyle x\in\overline{\bigcup_{i\in\mathbb{N}^*}A_i}$ we have $\displaystyle V\cap V_i\cap\bigcup_{k\in\mathbb{N}}A_k \neq \emptyset$. Now we can conclude since $\displaystyle V_i\cap\bigcup_{k\in\mathbb{N}}A_k =V_i\cap \bigcup_{k\geq i+1}A_k$.

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