If every Component in a compact space is open then the number of components is finite.

Question

Let $X$ a compact set. Prove that if every connected component is open then the number of components is finite.

Ok, $X = \bigcup C(x)$ where $C(x)$ is the connected component of $x \in X.$ I know that $X \subset UA_\lambda$, where $A_\lambda$ is a finite family of opens set but how can i conclude that the components are finite??

Answer 1

$\textbf{Hint:}$ Argue by contradiction: assume $X$ has infinite $\textit{open}$ connected components and prove this implies that $X$ is not compact.

Recall that $X$ being non-compact is equivalent to the existence of an open cover of $X$ that has no finite subcover. Can you think of such a cover in this case?

Answer 2

Hint. Since $X$ is compact and since, as you observed, the connected component form an open cover of $X$, one can extract from them a finite subcover. On the other hand, the connected component form a partition of $X$. Thus...

Answer 3

Under the assumption that components are open, they form an open cover of the compact space $X$. They form a partition of non-empty sets (always), so we cannot omit any one of them and still have a cover.

So $X$ has a finite subcover of components which is not really a proper subcover, but just the original cover of components. Which is thus finite.