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 $C$?

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    $\begingroup$ I think this is answered in the chapter about connectedness. The answer is yes, just look at the intersection with one and union that with the intersection with the union of the others; then $C$ is not connected if both intersections are nonempty. $\endgroup$
    – user722227
    Commented Feb 13, 2020 at 1:53
  • $\begingroup$ @user722227 I added my attempt and it is just the way you said. But I can't proceed anymore. I have to show that both of the two sets are open. Or that both of them are closed. Or that a closure of the former doesn't intersect the latter and vice versa. $\endgroup$
    – shyzealot
    Commented Feb 13, 2020 at 2:45

2 Answers 2


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 $C \subseteq X_\alpha$.


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.

  • $\begingroup$ Thanks for your reply, but I can't catch it. Specifically, what is the meaning of "If $C$ not subset $V$, then exists $y$ in $C\cap V$"? Does it mean "If $C\not\subset V$, then there exists $y\in C\cap V$"? $\endgroup$
    – shyzealot
    Commented Feb 13, 2020 at 3:31
  • $\begingroup$ I have corrected the mistake (V should be U) you pointed out. V$^n$ is not ever used in the proof. $\endgroup$ Commented Feb 13, 2020 at 9:57

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