If $X$ is Regular space and $A$ is an infinite subset of $X$, then there is a sequence $U_1,U_2,\dots$ of open subsets of $X$ such that

  1. $\overline{U_n}\cap \overline{U_m}=\emptyset$ if $n\ne m$

  2. the set $A$ meets every open subset of the sequence $(U_n)$ for all $n$

    [Hint: Use induction]

  • $\begingroup$ Take $U_i$ to be empty for all $i$. Perhaps you intended to ask something else? $\endgroup$ May 3, 2012 at 14:48
  • $\begingroup$ thank you Chris, but the question exactly like this If X is Regular space and A is an infinite subset of X,then there is a sequence U1,U2,... of open subsets of X such that 1. the intersection between Cl(Un) and Cl(Um) is empty if n not equals m (Notation:I denote the closure of the open subset by Cl(Un) ) 2. the closed set A meets every open subset of the sequence (Un) for all n $\endgroup$
    – fatma
    May 3, 2012 at 15:05
  • $\begingroup$ I think you need to assume $X$ is $T_1$ (singleton sets are closed). $\endgroup$ May 3, 2012 at 15:22
  • $\begingroup$ thx David , but the definition of regular includes that the space is T1. $\endgroup$
    – fatma
    May 3, 2012 at 15:28
  • $\begingroup$ You should mention that then. Many authors (Kelly, Willard, e.g.) do not include $T_1$ in the definition of a regular space. $\endgroup$ May 3, 2012 at 15:30

2 Answers 2


Here, "by induction" means define the sets inductively. That is

1) Show that the open set $U_1$ can be constructed so that it meets $A$.

2) Assume that the open sets $U_1$, $U_2$, $\ldots\,$, $U_n$, have been constructed so that each $U_i$ meets $A$ and that $\overline {U_i}\cap \overline{U_j}=\emptyset$ whenever $i\ne j$ and $1\le i,j\le n$.

3) Then show that $U_{n+1}$ can be constructed to have the appropriate properties.

In order to construct the required infinite sequence of open sets, you will need to impose the additional property that for each $j$ the complement of $\overline{U_1}\cup\overline{U_2}\cup\cdots\cup\overline{U_j}$ contains infinitely many points of $A$. (Without this requirement, the induction may stop after a finite number of steps.)

So, you need to do the following:

1) Show that the open set $U_1$ can be constructed so that it meets $A$ and $\overline{U_1}^C\cap A$ is infinite.

2) Assume that the open sets $U_1$, $U_2$, $\ldots\,$, $U_n$, have been constructed so that

$\ \ \ \ $a) each $U_i$ meets $A$,

$\ \ \ \ $b) $\overline {U_i}\cap \overline{U_j}=\emptyset$ whenever $i\ne j$ and $1\le i,j\le n$,

$\ \ \ \ $c) $ (\overline{U_1}\cup\overline{U_2}\cup\cdots\cup\overline{U_n} )^C\cap A$ is infinite.

3) Then show that an open set $U_{n+1}$ can be constructed so that it meets $A$, its closure is disjoint from the closure of each previously defined set, and that $ (\overline{U_1}\cup\overline{U_2}\cup\cdots\cup\overline{U_n}\cup\overline{U_{n+1}})^C\cap A$ is infinite.

To establish 1), you could argue as Martin does in his answer.

Assuming that 2) holds, to establish 3), you would use regularity:

$F_n=\overline{U_1}\cup\overline{U_2}\cup\cdots\cup\overline{U_n}$ is closed and by 2c) its complement contains infinitely many points of $A$. Given $a\in F_n^C\cap A$, by regularity, you can find an open set $U_{n+1}$ containing $a$ whose closure is disjoint from $F_n$. But, you need to be a bit more careful, you need to select $a$ so that $ (\overline{U_1}\cup\overline{U_2}\cup\cdots\cup\overline{U_n}\cup\overline{U_{n+1}})^C\cap A$ is infinite. I'll leave this for you...

  • $\begingroup$ thank you very much, really it was a useful answer for me :)) $\endgroup$
    – fatma
    May 3, 2012 at 17:23

I am not entirely sure how to understand the condition 2 in the (present version of the) post, but I'll assume it is supposed to be formulated as in Problem 14D from Willard, i.e. we want that $U_n\cap A\ne\emptyset$ for each $n$.

If we choose arbitrary $a, b\in A$, then by Hausdorffness we can find and open neighborhoods $U_a$ of $a$ and $U_b$ of $b$ such that $U_a\cap U_b=\emptyset$. Using regularity we can get a new neighborhood $V_a$ of $a$ such that $a\in V_a\subseteq \overline{V_a}\subseteq U_a$.

  • If $A\setminus \overline V_a$ is infinite, we put $a_1=a$ and $U_1=V_a$.

  • If $A\setminus\overline V_a$ is finite, then we have $\overline U_b \subseteq X\setminus U_a \subseteq X\setminus \overline{V_a}$, which gives $A\setminus\overline U_b\supseteq A\setminus(A\setminus\overline V_a)$, hence $A\setminus\overline U_b$ is infinite. We put $a_1=b$ and $U_1=U_b$

In either case we have found a set $U_1$ such that $\overline U_1\cap A\ne\emptyset$ and $A\setminus\overline{U_1}$ is non-empty.

In the next step we will repeat the same thing with the infinite subset $A\setminus\overline{U_1}$ of the space $X\setminus\overline{U_1}$.

If we try to continue by induction, our goal is to have $A\setminus (\overline{U_1} \cup \dots \cup \overline{U_n})$ infinite after each step. (And, of course, the sets $\overline{U_1},\dots, \overline{U_n}$ should be disjoint, as this is one of the requirements of the given problem.)

  • $\begingroup$ Slezia, thank you very much, unfortunately I solved it wrong :(( thx again. $\endgroup$
    – fatma
    May 3, 2012 at 17:03

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