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.
Best Answer
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.