Clopen subsets of a topological space don't form a complete lattice

Question

I am trying to find a counter-example to show that the clopen subsets of a topological space $(X,\tau)$ don't form a complete lattice. I have tried all the common example, i.e. Sorgenfrey , discrete topology etc, but could not get a counter-example.

Answer 1

Consider the rationals with the standard topology. Then $(-\infty,a)\cap\mathbb Q$ is a clopen subset of $\mathbb Q$ for any irrational $a$.

Now take the set $$S=\{(-\infty,a)\cap\mathbb Q:a\text{ is irrational and }a<0\}.$$

In the lattice of clopen sets, the meet of that subset is the smallest clopen set which contains every set in $S$, or equivalently, the smallest clopen set containing the union of the sets in $S$, which is $(-\infty,0)\cap\mathbb Q$.

But no such smallest clopen set exists. For if it did, then it would be equal to the intersection of all clopen sets containing $(-\infty,0)\cap\mathbb Q$. But the intersection of all clopen sets containing $(-\infty,0)\cap\mathbb Q$ is just $(-\infty,0)\cap\mathbb Q$, which is not clopen (in particular, it's not closed). Thus the set $S$ lacks a meet in the lattice of clopen sets, and so this lattice is not complete.

Answer 2

Take the topology $\mathcal{T}$ on $\mathbb{R}$ where $\mathcal{T}=\{[a,b)|a<b,a,b \in \mathbb{R}\}$

If i remember correctly this is called the lower limit topology.

Every interval of the form $[a,b)$ is clopen in this topology.

Take the collection $\{[\frac{1}{n}, +\infty)|n \in \mathbb{N}\}$

Prove that the union is not a clopen set.

Answer 3

Stone's representation theorem says that every Boolean algebra $B$ is isomorphic to the algebra of clopen sets of a certain compact Hausdorff space called its Stone space. To get the counterexample you want, take any non-complete Boolean algebra and consider its Stone space.

Perhaps the simplest concrete example is the Cantor set, which is the Stone space of the countable atomless Boolean algebra (aka the free Boolean algebra on $\aleph_0$ generators).