Every compact Hausdorff space can be expressed as a disjoint union of finitely many open sets.

Question

Let $X$ be a compact Hausdorff space. Can we express as disjoint union of open sets ?

I got a proof for that.

Since $X$ is compact there exist a finite sub collection of open sets $\{W_{i}\}_{i=1}^{n}$ that covers $X$. Construct disjoint open sets $W_{i}'$ for $i=1, \cdots,n$ as follows : \begin{eqnarray*} W_{1}'= & W_{1} \\ W_{k}'= & W_{k}-\displaystyle \bigcup_{j=1}^{k-1}\overline{W_{j}'}' \end{eqnarray*}

Then by the construction $W_{i}'$ are disjoint. Suppose $x \in W_{i}$ only then $x$ belongs to $W_{i}'$. Now let $x$ be an element of more than one $W_{i}$. Let $i$ be the least index for which $x \in W_{i}$ then $x $ belongs to that $W_{i}'$ and does not belongs to any other $W_{j}'$. Thus the collection $\{ W_{i}' \}_{i=1}^{n}$ covers $X$.

Is this proof correct ? If it is other questions are immaterial. I know that I have not used Hausdorff condition in the proof.

Can we have a counter example ?

Can we apply some conditions on compact Hausdorff space so that it can be expressed as disjoint union of open sets ?

Answer 1

The result is far from being true.

No connected (compact Hausdorff) space can be expressed in this form (except in the trivial way in which there is only one non-empty open set in the collection).

Answer 2

You can always do it with $n=1$ and $W_1=X$: if $X$ is connected, this is the one and only possibility. Otherwise, your procedure fails in general, your error lying in the fact that $\{W_i',\:\, i\in\Bbb N\}$ may not be a covering. Also, the entire idea of a non-trivial case $n\ge2$ fails catastrophically: again, see the concept of connected space, and related concepts such as connected components of a topological space.

Answer 3

If the unit disc, say, were a disjoint union of open sets, it would be disconnected. Just split the open sets into two groups. But it is not

Answer 4

Apart from the trivial case $n=1$, your new set isn't a covering. Take $X=[0,1]$, $W_1=[0,3/4)$ and $W_2=(1/4,1]$. Then $W_2'=[1/4,1]\setminus[0,3/4]=(3/4,]$ so $3/4$ does not belong either to $W_1'$ or $W_2'$.

In general your result is false if the space is connected. Even if it is not connected, you cannot chose $n$ arbitrarily, it depends on the number of connected components of the space.