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We are given a compact metric topological space $X$ and a base $\beta$ of its topology. Is it always possible to find a subset of the basis, say $\beta_0 \subset \beta$, such that each point of the space is contained in a closure of a set from $\beta_0$, i.e. $\bigcup_{U \in \beta_0} \overline{U} = X$, and the sets from $\beta_0$ are pairwise disjoint, i.e. $\forall U,V \in \beta_0 : \ U = V \vee U \cap V = \emptyset$.

My guess is that it is impossible in general, because of two arguments. Firstly, there exist dense open subsets of $[0,1]$ with arbitralily small measure, so we need to be careful choosing sets for $\beta_0$ (making it any maximal subset of $\beta$ with pairwise disjoint will not suffice). Secondly, it would just be too good to be true, and a proof of a certain measure-theoretical problem would become too simple.

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For your example of the unit interval: Take $\beta_0 = (0,\frac{1}{4})$ and $\beta_1 = (\frac{1}{4},\frac{1}{2})$ and $\beta_3 = (\frac{1}{2},1)$. Those are all basis elements (if you consider only open interval with rational end points) and you have that $[0,\frac{1}{4}]\cup [\frac{1}{4},\frac{1}{2}]\cup [\frac{1}{2},1] = [0,1]$. – Asaf Karagila May 20 '11 at 9:24
@Asaf Karagila: I don't think you get to choose the original basis $\beta$, it's given to you. $\beta$ could be something other than the open intervals with rational endpoints, and you can't be sure the sets you chose are in $\beta$. – Nate Eldredge May 20 '11 at 15:07
@Nate: Of course it could be the discrete topology, which means $[0,1]$ is not compact at all :-) I just gave an example. – Asaf Karagila May 20 '11 at 15:15
@Asaf: Your example works for the standard basis for the topology of $[0,1]$, but the given basis $\beta$ might not contain the sets you have listed. Any construction of an example of this being true needs to start with an arbitrary basis for the space $X$. – wckronholm May 20 '11 at 15:15
I think you should use "base", not "basis": a vector space has a basis, a topological space has a base. – Henno Brandsma May 21 '11 at 8:10
up vote 2 down vote accepted

Firstly, sorry for my own stupidity. The counterexample is surprisingly simple and elementary.

It is sufficient to take $X = [0,1]$ and compose a countable basis $\beta$ of such intervals that no two have the same endpoint and $X \not \in \beta$ and $\beta$ contains no intervals of the form $(0,x)$ nor $(x,1)$. Then, for any two sets $U,V \in \beta$ with $U \cap V = \emptyset$ we can squeze an (nondegenetare) interval between $U$ and $V$ and for any $U \in \beta$ we have $\# \partial U \leq 2$.

Now, suppose that we manage to select $\beta_0$ such that $\bigcup_{U \in \beta_0} \overline{U} = X$ and sets from $\beta_0$ are pairwise disjoint. Since $\overline{U} \setminus U$ has at most two elements, the set $X \setminus \bigcup_{U \in \beta_0} U \subset \bigcup_{U \in \beta_0} \overline{U} \setminus U$ is at most countable. I will show that $X \setminus \bigcup_{U \in \beta_0} U$ has no isolated points, and it is easy to check that it is closed and nonempty. But a complete space with no isolated points has to have a power of continuum, which leads to contradiction.

Suppose $X \setminus \bigcup_{U \in \beta_0} U$ has an isolated point $x \neq 0,1$. We know that $x \in \partial U$ for some $U \in \beta$. Without loss of generality assume that $x$ is the left endpoint of the interval $U$ so that $U = (x,y)$ for some $y$. Since $x$ is an isolated point of $X \setminus \bigcup_{U \in \beta_0} U$ there has to exist an interval $(z,x)$ which is disjoint from $X \setminus \bigcup_{U \in \beta_0} U$ or, in other words, contained in $\bigcup_{U \in \beta_0} U$. If we select any $t \in (z,x)$ it belongs to some $U = (a,b) \in \beta_0$ with $b \leq x$. But $x$ is already an endpoint of $(x,y) \in \beta$ so, by choice of $\beta$ it can't also be the endpoint of $(a,b)$, thus $b < x$. Note that $b \not \in V$ for $V \in \beta$, since sets in $\beta_0$ are pairwise disjoint, so $X \not \in \bigcup_{U \in \beta_0} U$. At the same time, $b \in (z,x) \subset \bigcup_{U \in \beta_0} U$ which is a contradiction.

The above is a little technical and lacks a few points, but I don't think it's worth to go into more detail here. In fact, one can see that $X \setminus \bigcup_{U \in \beta_0} U$ is homeomorphic to the Cantor set, and thus uncountable.

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Very nice example! – Nate Eldredge May 26 '11 at 12:35

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