A connected subset lies entirely in one set of partition of open sets.

Question

Let $X$ be a topological space and let $C$ be a connected subset of $X$. Let $\{X_\alpha\}$ be a collection of open sets that are disjoint one another, whose union is $X$. (That is, $X_\alpha$ form a partition of $X$ where $X_\alpha$ is open for each $\alpha$.)

Does it follow there exists $\alpha$ such that $C\subset X_\alpha$?

This is the end of the question, but the below is where the question come from.

[Munkres, Topology 2ed, Section 54, Lemma 54.2]

$C$ is a connected set, $\tilde F:I\times I\to E$ is a continuous function(it will be a lifting of $F$.). $\{V_\alpha\}$ is a partition of a subset $p^{-1}(U)$ of $E$, where $V_\alpha$ is open for all $\alpha$. The book says '$\tilde F(C)$ is connected and must lie entirely within one of the sets $V_\alpha$.'

[Added] Here is my attempt. Suppose not. Then there exists $X_0$ such that $C\cap X_0$ and $C\setminus X_0$ are both nonempty. How do they form a separation of $C$?

Answer 1

In the setting where $X$ is partitioned by the open sets $X_\alpha$, note that all $X_\alpha$ are also closed: each $X_\alpha = X\setminus \bigcup_{\beta \neq \alpha} A_\beta$, which is the complement of (a union of opens hence) open set and thus closed. So all sets $X \setminus X_\alpha$ are open too.

So if $C \subseteq X$ is connected, let $p \in C$ and there is some $\alpha$ such that $p \in X_\alpha$. If $C \nsubseteq X_\alpha$, this would then mean that both $(X_\alpha \cap C)$ and $C \cap (X\setminus X_\alpha)$ are open in $C$ and non-empty and they clearly partition $C$ and would make $C$ disconnected. So $C \subseteq X_\alpha$.

Answer 2

Yes.
Trivially if C is empty.
So assume exists x in C.
Let P be the open partition of X.
Exists U in P with x in U.
V = $\cup${ W : W in P - {U} } is open.
If C not subset V, then exists y in C $\cap$ V.
Thus U and V disconnect C, a contradiction.