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Can anyone show me how to prove that: In a paracompact space: given a family of sets $A_i$ where $i \in I$ (some set of indexes) then: $\overline{\cup{A_i}} = \cup \overline{A_i}$ Thanx!

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2 Answers 2

$\newcommand{\cl}{\operatorname{cl}}$The statement is false in general, even if the sets $A_i$ are open. Let $S$ be the Sorgenfrey line, i.e, the real line topologized by taking $\big\{[a,b):a<b\big\}$ as a base. $S$ is $T_3$ and Lindelöf, so it’s paracompact. For $x\in(0,1)$ let $U_x=[x,1)$, and let $\mathscr{U}=\{U_x:x\in(0,1)\}$. $\mathscr{U}$ is a family of clopen sets in $S$, so $$\bigcup_{x\in(0,1)}\cl_SU_x=\bigcup_{x\in(0,1)}U_x=(0,1)\;,$$ but $$\cl_S\bigcup_{x\in(0,1)}U_x=\cl_S(0,1)=[0,1)\;.$$

The closest result that I can think of is this one:

Theorem. Let $X$ be a space, and let $\mathscr{A}$ be any locally finite family of subsets of $X$. Then $$\cl_X\bigcup\mathscr{A}=\bigcup\big\{\cl_XA:A\in\mathscr{A}\big\}\;.$$

Proof. Clearly it suffices to show that $\cl_X\bigcup\mathscr{A}\subseteq\bigcup\big\{\cl_XA:A\in\mathscr{A}\big\}$, so suppose that $x\in\cl_X\bigcup\mathscr{A}$. Let $V$ be an open nbhd of $x$ meeting only finitely many members of $\mathscr{A}$, and let $\mathscr{A}_0=\{A\in\mathscr{A}:V\cap A\ne\varnothing\}$. If $U$ is any open nbhd of $x$, so is $U\cap V$, so $(U\cap V)\cap\bigcup\mathscr{A}\ne\varnothing$. But $V\cap\bigcup\mathscr{A}=V\cap\bigcup\mathscr{A}_0$, so $(U\cap V)\cap\bigcup\mathscr{A}=(U\cap V)\cap\bigcup\mathscr{A}_0$ and it follows that $x\in\cl_X\bigcup\mathscr{A}_0$. But $\mathscr{A}_0$ is finite, so $\cl_X\bigcup\mathscr{A}_0=\bigcup\big\{\cl_XA:A\in\mathscr{A}_0\big\}$, and there is therefore some $A_x\in\mathscr{A}_0$ such that $x\in\cl_XA_x\subseteq\bigcup\big\{\cl_XA:A\in\mathscr{A}\big\}$. $\dashv$

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Thanks a lot! Yes, this is exactly the statement I needed. Since I was tring to show that a paracompact Hausdorff space is Nornal. So I was using a locally finite reinfinment of some cover of X. –  Shir Sivroni Jan 30 '13 at 11:01
    
@user60164: You’re welcome! –  Brian M. Scott Jan 30 '13 at 18:34

This is true, given that all $A_i$ are open.

First off, we get $\cup \overline{A_i} \subseteq \overline{\cup{A_i}}$ due to the nature of the closure operator.

Now let $x \in \overline{\cup A_i}$. Since the space is paracompact, and since $\cup A_i$ is an open cover, $\cup A_i$has some locally finite refinement $\cup B_i$, so there is some neighbourhood $U$ of $x$ such that $U$ intersects finitely many $B_i$.

Suppose towards contradiction that $x \notin \cup \overline{A_i}$. $\cup \overline{B_i} \subseteq \cup \overline{A_i}$, so $x \notin \cup \overline{B_i}$, hence for each $B_i$, $x \notin \overline{B_i}$.

Now let $i_1, i_2, ..., i_n$ be the indices such that $U \cap B_{i_1}, U \cap B_{i_2}, ..., U \cap B_{i_n}$ are all non-empty, and let $V = U \cap (\displaystyle \cup_{k=1}^n \overline{B_{i_k}}')$ ($U$ subtract the compliments of the closures). Notice $V$ is an open neighbourhood of $x$ that does not intersect any $B_i$.

But that means $V$ does not intersect $\cup B_i$, so $V'$ is a closed set containing $\cup B_i$ such that $x \notin V$, which means $x \notin \overline{\cup B_i}$. $\cup B_i$ is a refinement of $\cup A_i$, so $x \notin \overline{\cup A_i}$, contrary to our assumption.

That means we must in fact have $x \in \cup \overline{A_i}$, giving $\overline{\cup A_i} \subseteq \cup \overline{A_i}$, which combines with our first inequality for the equality $\overline{\cup A_i} = \cup \overline{A_i}$.

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I almost lost my head...$\mathbb R$ is paracompact, yet the property asked by the OP is not true for arbitrary sets. The members $A_i$ must be open sets in order that ferson2020's argument works (second line, concretely). –  Matemáticos Chibchas Jan 30 '13 at 3:44
    
@MatemáticosChibchas Thank you for pointing that out! For example, if we enumerate $\mathbb{Q}$ as $\{q_1, q_2, ...\}$, and let $A_i = \{q_i\}$, then $\cup \overline{A_i} = \mathbb{Q}$ but $\overline{\cup A_i} = \mathbb{R}$. –  ferson2020 Jan 30 '13 at 3:52
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It is not necessarily true even if all of the $A_i$ are open. $\{A_i:i\in I\}$ is an open cover of $\bigcup_iA_i$, but $\bigcup_iA_i$ need not be paracompact. Your argument requires that $X$ be hereditarily paracompact. This is the case, for instance, if $X$ is metrizable. –  Brian M. Scott Jan 30 '13 at 5:04
    
@BrianM.Scott My interpretation of the question was that the space itself is paracompact, and so any open cover admits a locally finite refinement. Given that, is there anything wrong with my proof? –  ferson2020 Jan 30 '13 at 14:11
    
Yes: you don’t have an open cover of the space. The sets $A_i$ are not assumed to cover the space. –  Brian M. Scott Jan 30 '13 at 18:35

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