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Let $A_i$ be a subset of a metric space for each $i\in \mathbb{N}$.

  1. Let $B_n := \bigcup_{i=1}^n A_i$. Prove (for any) $n \in \mathbb{N}$ that $\overline{B_n} = \bigcup_{i=1}^n \overline{A_i}$.
  2. If $B = \bigcup_{i=1}^\infty A_i$, prove that $\overline{B} \supseteq \bigcup_{i=1}^\infty \overline{A_i}$. Give an example to show that this containment might be proper.

If $A_i$ is closed then $A_i = \bar A_i$, but I'm stuck as to how to prove $B=\bar B_n$. If I prove the first statement for when $A_i$ is closed does that mean it is also true for when $A_i$ is open because I can construct a closed set containing $A_i$?

For the example, would constructing a sequence of closed segments between $0$ and $1$ that gets arbitrarily close to $1$ and taking the union of the segments be considered a proper containment?

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3 Answers 3

up vote 3 down vote accepted

To prove the true statements we'll make use of the following facts:

  • If $A \subseteq B$, then $\overline{A} \subseteq \overline{B}$.
  • $x \in \overline{A}$ iff $U \cap A \neq \varnothing$ for every open neighbourhood $U$ of $x$. (This is in the more general context of general topological spaces. For metric spaces it suffices to restrict the $U$s to the open $\varepsilon$-balls centred at $x$.)

To see that $\bigcup_{i=1}^n \overline{A_i} \subseteq \overline{B_n}$ observe that $A_i \subseteq B_n$ for each $i \leq n$, and apply the first fact above.

To see that $\overline{B_n} \subseteq \bigcup_{i=1}^n \overline{A}_i$ note that this is equivalent to showing $X \setminus \bigcup_{i=1}^n \overline{A}_i \subseteq X \setminus \overline{B_n}$. If $x \in X \setminus \bigcup_{i=1}^n \overline{A_i}$, then by the second fact above it follows that for each $i \leq n$ there is an open neighbourhood $U_i$ of $x$ which is disjoint from $A_i$. Now note that $U := U_1 \cap \cdots \cap U_n$ is an open neighbourhood of $x$ which is disjoint from $B_n$.


Note that the first fact above also shows that $\bigcup_{i=1}^\infty \overline{A_i} \subseteq \overline{\bigcup_{i=1}^\infty A_i} = \overline{B}$.

To give an example of where the equality does not hold, just enumerate the set $\mathbb{Q}$ of rational numbers as $\{ a_i : i \in \mathbb{N} \}$, and for each $i$ define $A_i = \{ a_i \}$. Working in $\mathbb{R}$ with the usual (metric/order) topology, we can show that $$ {\textstyle \bigcup_{i=1}^\infty} \overline{A_i} = \mathbb{Q} \neq \mathbb{R} = \overline { {\textstyle \bigcup_{i=1}^\infty} A_i }. $$

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Thanks! That helped me get started. –  Andrew C Sep 13 '12 at 17:49

Hint 1: for $\overline{A_1}\cup\overline{A_2}\subset\overline{A_1\cup A_2}$

Suppose that $x\in\overline{A_1}$. This means that for any open $U$ containing $x$, $U\cap A_1\not=\varnothing$. Obviously, for any $U$ containing $x$, $U\cap(A_1\cup A_2)\supset U\cap A_1\not=\varnothing$. Thus, if $x\in\overline{A_1}$, then $x\in\overline{A_1\cup A_2}$; that is, $\overline{A_1}\subset\overline{A_1\cup A_2}$.

Hint 2: for $\overline{A_1\cup A_2}\subset\overline{A_1}\cup\overline{A_2}$

Suppose that $x\in\overline{A_1\cup A_2}$. This mean that for any open $U$ containing $x$, $U\cap(A_1\cup A_2)\not=\varnothing$. Suppose that $x\not\in\overline{A_1}$; that is, for some open $U_1$ containing $x$, we have that $U_1\cap A_1=\varnothing$, and that $x\not\in\overline{A_2}$; that is, for some open $U_2$ containing $x$, we have $U_2\cap A_2=\varnothing$. Consider $U=U_1\cap U_2$. Obviously, $U$ is open and $x\in U$. Find a contradiction.

Hint 3: for $\bigcup\limits_{i=1}^n\overline{A_i}=\overline{\bigcup\limits_{i=1}^nA_i}$

Use induction on $\overline{A_1}\cup\overline{A_2}=\overline{A_1\cup A_2}$.

Hint 4: for $\bigcup\limits_{i=1}^\infty\overline{A_i}\subset\overline{\bigcup\limits_{i=1}^\infty A_i}$

If $x\in\bigcup\limits_{i=1}^\infty\overline{A_i}$ then there is some $n$ so that $x\in\overline{A_n}$. Use Hint 3 to show $\overline{A_n}\subset\overline{\bigcup\limits_{i=1}^nA_i}$ and Hint 1 to show that $\overline{\bigcup\limits_{i=1}^nA_i}\subset\overline{\bigcup\limits_{i=1}^\infty A_i}$

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Almost $2$ years but anyway I am reading (studying) your answer (+1) : is the last hint (4), I think it's the union of the closure, right? –  MarcGato Nov 4 at 0:10
    
Do you have a question about Hint 4? –  robjohn Nov 4 at 0:23
    
Yes, you said use Hint 1 to show that $\overline{\bigcup\limits_{i=1}^nA_i}\subset\overline{\bigcup\limits_{i=1}^{\inf‌​ty} A_i}$, why it's not $\bigcup\limits_{i=1}^n\overline{A_i}\subset\overline{\bigcup\limits_{i=1}^{\inf‌​ty} A_i}$? –  MarcGato Nov 4 at 0:25
1  
Hint 1 says that $\overline{\bigcup\limits_{i=1}^nA_i}\cup\overline{\bigcup\limits_{i=n+1}^\infty A_i}\subset\overline{\bigcup\limits_{i=1}^\infty A_i}$. Therefore, $\overline{\bigcup\limits_{i=1}^nA_i}\subset \overline{\bigcup\limits_{i=1}^\infty A_i}$ –  robjohn Nov 4 at 1:03

Notice that $A_i \subset B_n \subset \overline{B_n}$. Since $\overline{B_n}$ is closed it follows that $\overline{A_i}\subset \overline{B_n}$, hence it follows that $\cup_{i=1}^n \overline{A_i}\subset \overline{B_n}$. This is true even with arbitrary unions.

Furthermore, the union of a finite number of closed sets is closed, and we have $B_n \subset \cup_{i=1}^n \overline{A_i}$, so it follows that $\overline{B_n} \subset \cup_{i=1}^n \overline{A_i}$. This is not necessarily true with infinite unions.

Another example illustrating the latter point is $A_i = \frac{1}{i} = \overline{A_i}$, then $0 \in \overline{B} \setminus \cup_{i=1}^\infty \overline{A_i}$.

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