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Let X be an infinite Hausdorff space. Prove that there exist an infinite disjoint class of open subsets of X.

Ok the first time I tried to prove this I started by taking pairwise disjoint sets given by the Hausdorff property and then taking the intersection of all those sets for each point; but you can only prove the finite case by doing this, since one can only assert that the finite intersection of open set is an open set. So there must be a different reasoning for the infinite case (if it is in fact true).

Then I thought that the statement was false, since you can take a set, for example $\mathbb{R}$ which has a countable dense subset $\mathbb{Q}$, then it would be impossible to create such class taking the pairwise disjoints sets for every point in $\mathbb{R}$. But this was no counterexample since one can take for example all the balls $B_{\frac{1}{2}}(n)$ where $n \in \mathbb{Z}$ and you get and infinite family of open disjoint subsets.

At this point I think the assertion must be true, but I have no clue of how to prove it. Trying to construct this infinite family of disjoint open subsets for an arbitrary Hausdorff space has proven to be a difficult problem, so I also tried to prove the assertion by contradiction, but I haven't had any luck either by using this approach.

Do you have any suggestions? is the statement even true?

I know that there are other questions here which address this problem, but until now every one that I found gave the proof or the idea only for the finite case.

Thank you in advance for all your help.

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  • $\begingroup$ I'm starting to think that the best way to prove this might be by contradiction, yet I haven't been able to get it. There are not a lot of properties of infinite Hausdorff spaces in the most common topology books. $\endgroup$ – pjox Nov 2 '13 at 2:20
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Let $X$ be an infinite Hausdorff space, and let $A$ be the set of isolated points of $X$. If $A$ is infinite, then $\big\{\{x\}:x\in A\big\}$ is an infinite family of pairwise disjoint open sets in $X$. If $A$ is finite, let $Y=X\setminus A$: $Y$ is an infinite Hausdorff space with no isolated points, and $Y$ is open in $X$ (since $A$, being finite, is closed), so it suffices to find an infinite family of pairwise disjoint open subsets of $Y$: they will automatically be open in $X$ as well.

Let $x_0$ and $x_1$ be distinct points of $Y$; $Y$ is Hausdorff, so $x_0$ and $x_1$ have disjoint open nbhds $U_0$ and $V_1$, respectively. $Y$ has no isolated points, so there is a point $x_2\in V_1\setminus\{x_1\}$. The points $x_1$ and $x_2$ have disjoint open nbhds $U_1$ and $V_2$, respectively, and (by intersection them with $V_1$ if necessary) we may assume without loss of generality that $U_1\subseteq V_1$ and $V_2\subseteq V_1$. Continue in this fashion: given $x_n$ and its open nbhd $V_n$, choose $x_{n+1}\in V_n\setminus\{x_n\}$ and let $U_n$ and $V_{n+1}$ be disjoint open nbhds of $x_n$ and $x_{n+1}$, respectively, such that $U_n\subseteq V_n$ and $V_{n+1}\subseteq V_n$. Clearly the recursive construction goes through for all $n\in\Bbb N$, and $\{U_n:n\in\Bbb N\}$ is an infinite family of pairwise disjoint open sets in $Y$ (and hence in $X$).

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  • $\begingroup$ Thank you for your answer, but I was looking for an infinite family of disjoint open sets not of pairwise disjoint open sets. I mean, given an open set in the family, it must be disjoint from every other open set in the family. $\endgroup$ – pjox Nov 2 '13 at 0:15
  • $\begingroup$ @PJOX: That's exactly what pairwise disjoint means: no two of them intersect, and therefore each is disjoint from all the rest. And that’s exactly what I gave you, by what I consider the simplest argument around. I suggest that you make a sketch of the first few stages of the construction: it should very quickly become obvious what is going on and why it works. $\endgroup$ – Brian M. Scott Nov 2 '13 at 3:26
  • $\begingroup$ Oh I'm sorry, you are right, I was confused; also I was associating pairwise disjoint with the wrong concept in my own language (My native language is Spanish). Now I get it, thank you for answering. $\endgroup$ – pjox Nov 2 '13 at 6:55
  • $\begingroup$ @PJOX: You’re welcome. $\endgroup$ – Brian M. Scott Nov 2 '13 at 6:56
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The statement is true. Let $X$ be an infinite Hausdorff space. It will suffice to show that we can find a nonempty open set $V_1$ such that $X\setminus\overline V_1$ is infinite. (Then we find $V_2$ in the same way with the infinite Hausdorff space $X\setminus\overline V_1$ playing the role of $X$, and so on.)

At any rate, since $X$ is Hausdorff and has more than one point, we can find a nonempty open set $U$ such that $X\setminus\overline U\ne\emptyset$.

Case 1. If $X\setminus\overline U$ is infinite, let $V_1=U$.

Case 2. If $X\setminus\overline U$ is finite, let $V_1=X\setminus\overline U$. Then $V_1$ is a finite nonempty clopen set, and $X\setminus\overline V_1=X\setminus V_1$ is infinite.

Edit in response to original poster's comment:

If you just wanted to show that for every $n\in\mathbb N$ you can constuct a family of $n$ pairwise disjoint nonempty open sets, you wouldn't need a proof by induction. Just choose $n$ distinct points, choose disjoint neighborhoods for each pair of points, and intersect the (finitely many) chosen neighborhoods of each point. What I described above was a method for constructing an infinite sequence of pairwise disjoint nonempty open sets. I will try to describe the construction a little more formally.

Suppose that $V_1,V_2,\dots,V_n$ have already been defined, so that $V_1,V_2,\dots,V_n$ are pairwise disjoint nonempty open subsets of $X$, and the set $X_n=X\setminus\overline{V_1\cup V_2\cup\dots\cup V_n}=\mathrm{int}(X\setminus(V_1\cup V_2\cup\dots\cup V_n))$ is infinite.

Find a nonempty open set $U\subset X_n$ such that $X_n\setminus\overline U\ne\emptyset$.

If $X_n\setminus\overline U$ is infinite, let $V_{n+1}=U$; if $X_n\setminus\overline U$ is finite, let $V_{n+1}=X_n\setminus\overline U$.

Now $V_1,V_2,\dots,V_n,V_{n+1}$ are pairwise disjoint nonempty open sets, and $X\setminus\overline{V_1\cup V_2\cup\dots\cup V_n\cup V_{n+1}}$ is infinite.

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  • $\begingroup$ Your proof seems right, but there is a little detail that I don't quiet understand, and it is, how do you make sure that you can repeat this process infinitely many times and you will always get an open set. I mean, one can proof by induction that for every $n \in \mathbb{N}$ you can construct a family of $n$ disjoint sets, but that is different from actually exhibiting the infinite family of disjoint sets. By the way, thanks for answering! $\endgroup$ – pjox Nov 2 '13 at 0:26

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