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Let X be a topological space. Define a binary relation $\sim$ in $X$ as follows: $x \sim y$ if there exists a connected subspace $C$ included in $X$ such that $x,y$ belong to $C$. Show the following.

(i) $\sim$ is an equivalence relation.

(ii) Each equivalence class is a maximal connected subspace of $X$. These equivalence classes are called the connected components of $X$.

(iii) Each connected component is a closed subset of $X$. To this end, show that the closure of a connected set is connected.

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    $\begingroup$ What did you try?... (i) is pretty straight forward. $\endgroup$
    – Ludolila
    Feb 25, 2013 at 15:32
  • $\begingroup$ Why is no answer accepted? $\endgroup$ Feb 28, 2022 at 18:13

3 Answers 3

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Hint: (i) I guess you're ok with $x \sim x$ and $x\sim y \Rightarrow y \sim x$. For transitivity, recall that the union of two connected sets with nonempty intersection is also a connected set.

(ii) Use the same fact of (i) (possibly with infinite elements) to check that the equivalence classes are connected. If C is a connected set in $X$, note that any two points in $C$ are equivalent, so they all must be contained in an equivalence class.

(iii) Closure of a connected subset of $\mathbb{R}$ is connected?

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    $\begingroup$ In (ii), we can argue the connectivity of equiv classes alternately as well: Let $C$ be an equiv class and let $C$ be a disjoint union of the sets $U\cap C$ and $V\cap C$ ($U$, $V$ open in $X$). We need to show that one of these is empty. If possible, choose $x\in U\cap C$ and $y\in V\cap C$. Now, take a connected subspace $Y$ that contains both $x$ and $y$, and observe that $Y\subseteq C$ so that $Y = Y\cap C$, getting that $Y$ is a disjoint union of nonempty opens $U\cap Y$ and $V\cap Y$, contradicting the connectivity of $Y$. $\endgroup$
    – Atom
    Oct 19, 2023 at 12:06
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$\textbf{Hint:}$ (i) is straightforward.

(ii) If $A$ is an equivalence class and $A \subseteq B$ where $B$ is connected, show that $B \subseteq A$ (note that $\forall x \in B$, $\forall a \in A$ we have $x$~ $a$).

(iii) If $A$ is a connected component, note that $A$ is dense in $cl(A)$ and apply (ii) to get $A=cl(A)$.

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You can prove the following: If $A$ is connected in $X$, then $A\subseteq B\subseteq \bar A$ implies $B$ is connected. Argue that if $B$ is not connected, then neither is $A$.

Thus, the closure of a connected set is connected. In particular, $\overline{\operatorname{Cmp}(a)}\ni a$ is connected, so $\overline{\operatorname{Cmp}(a)}\subseteq {\operatorname{Cmp}(a)}$ and the reverse inclusion always holds, so $$\overline{\operatorname{Cmp}(a)}={\operatorname{Cmp}(a)}$$

and the component is closed.

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