[proofplan] We verify each property directly from the definition $C_x = \bigcup \{A \subseteq X : A \text{ is connected and } x \in A\}$. Non-emptiness follows because $\{x\}$ is connected. Connectedness of the union follows from the fact that all sets in the union share the common point $x$, so any two overlap, and the [union of overlapping connected sets is connected](/theorems/298). Maximality is immediate from the definition. The partition property follows from maximality and the union-of-overlapping-connected-sets argument. Closedness follows because the closure of a connected set is connected, so $\overline{C_x}$ is a connected set containing $x$ and must be absorbed by $C_x$. [/proofplan] [step:Verify that $C_x$ is non-empty] The singleton $\{x\}$ is connected (it has no non-trivial separation) and contains $x$, so $\{x\} \subseteq C_x$. In particular, $C_x \neq \varnothing$. [/step] [step:Show $C_x$ is connected using the common-point union principle] Every connected set $A$ appearing in the union $C_x = \bigcup \{A \subseteq X : A \text{ is connected and } x \in A\}$ contains the point $x$. Therefore any two such sets intersect (at $x$). By the [union of overlapping connected sets theorem](/theorems/298): if $\{A_i\}_{i \in I}$ is a family of connected subsets of $X$ with $\bigcap_{i \in I} A_i \neq \varnothing$, then $\bigcup_{i \in I} A_i$ is connected. The common point $x$ lies in every $A_i$, so $\bigcap A_i \supseteq \{x\} \neq \varnothing$, and the theorem applies. Therefore $C_x$ is connected. [guided] Why is the union of connected sets connected here, when unions of connected sets are not connected in general? The key condition is the shared point $x$. In general, $A \cup B$ can be disconnected even if $A$ and $B$ are each connected (e.g., $(0,1) \cup (2,3)$ in $\mathbb{R}$). The issue is that separated components cannot communicate. But when all the sets share a common point, this cannot happen. Suppose for contradiction that $C_x = U \cup V$ where $U, V$ are open in $C_x$, disjoint, and non-empty. The common point $x$ must lie in one of them --- say $x \in U$. Then every set $A_i$ in the defining collection satisfies $x \in A_i \cap U$, so $A_i \cap U \neq \varnothing$. Since each $A_i$ is connected, the sets $A_i \cap U$ and $A_i \cap V$ would form a separation of $A_i$ if both were non-empty. Since $A_i \cap U \neq \varnothing$ and $A_i$ is connected, we must have $A_i \cap V = \varnothing$, hence $A_i \subseteq U$. This holds for every $A_i$, so $C_x = \bigcup A_i \subseteq U$, giving $V = \varnothing$ --- contradicting the assumption that $V$ is non-empty. This is precisely the content of the [union of overlapping connected sets theorem](/theorems/298). [/guided] [/step] [step:Show $C_x$ is a maximal connected subset of $X$] Suppose $A$ is a connected subset of $X$ with $C_x \subseteq A$. Then $x \in C_x \subseteq A$, so $A$ is one of the connected sets containing $x$ that appears in the defining union. Therefore $A \subseteq C_x$. Combined with $C_x \subseteq A$, this gives $A = C_x$. [/step] [step:Show $\{C_x : x \in X\}$ partitions $X$ by proving $C_x \cap C_y \neq \varnothing$ implies $C_x = C_y$] Suppose $C_x \cap C_y \neq \varnothing$. Then $C_x$ and $C_y$ are both connected (by the previous step) and they overlap. By the [union of overlapping connected sets theorem](/theorems/298), $C_x \cup C_y$ is connected. Since $x \in C_x \subseteq C_x \cup C_y$ and $C_x \cup C_y$ is connected, the definition of $C_x$ gives $C_x \cup C_y \subseteq C_x$, hence $C_y \subseteq C_x$. By symmetry (swapping the roles of $x$ and $y$), $C_x \subseteq C_y$. Therefore $C_x = C_y$. Since every $x \in X$ belongs to $C_x$, the collection $\{C_x : x \in X\}$ covers $X$. Together with the disjointness property, this is a partition. [guided] The partition property requires two things: (i) the components cover $X$, and (ii) any two distinct components are disjoint. **Covering:** Every point $x \in X$ belongs to its own component $C_x$ (since $\{x\} \subseteq C_x$), so $X = \bigcup_{x \in X} C_x$. **Disjointness:** Suppose $C_x \cap C_y \neq \varnothing$. Both $C_x$ and $C_y$ are connected subsets that share some point $z \in C_x \cap C_y$. Since they overlap, the [union of overlapping connected sets theorem](/theorems/298) gives that $C_x \cup C_y$ is connected. Now $x \in C_x \subseteq C_x \cup C_y$, and $C_x \cup C_y$ is a connected set containing $x$. By the definition of $C_x$ as the union of all connected sets containing $x$, we have $C_x \cup C_y \subseteq C_x$. In particular, $C_y \subseteq C_x$. By the symmetric argument (replacing $x$ with $y$): $y \in C_y \subseteq C_x \cup C_y$, so $C_x \cup C_y \subseteq C_y$, giving $C_x \subseteq C_y$. Therefore $C_x = C_y$. The contrapositive: $C_x \neq C_y$ implies $C_x \cap C_y = \varnothing$. [/guided] [/step] [step:Show $C_x$ is closed using the closure-of-connected-sets theorem] By the [closure of connected sets theorem](/theorems/297), if $A$ is a connected subset of $X$, then $\overline{A}$ is connected. Applying this to $A = C_x$: the closure $\overline{C_x}$ is connected in $X$. Since $x \in C_x \subseteq \overline{C_x}$, the set $\overline{C_x}$ is a connected subset of $X$ containing $x$, so by the definition of $C_x$ as the union of all such sets, $\overline{C_x} \subseteq C_x$. The reverse inclusion $C_x \subseteq \overline{C_x}$ always holds. Therefore $C_x = \overline{C_x}$, and $C_x$ is closed. [guided] The closedness of connected components is a consequence of a general fact: the [closure of a connected set is connected](/theorems/297). The connected component $C_x$ is connected (proved in the second step). By the closure-of-connected-sets theorem, $\overline{C_x}$ is also connected. Moreover, $x \in C_x \subseteq \overline{C_x}$, so $\overline{C_x}$ is a connected subset of $X$ containing $x$. By the definition of $C_x$ as the union of all connected sets containing $x$, every such set is absorbed: $\overline{C_x} \subseteq C_x$. The reverse inclusion $C_x \subseteq \overline{C_x}$ always holds (every set is contained in its closure). Combining: $C_x = \overline{C_x}$. A set equal to its own closure is closed. Note: connected components need not be open in general. For example, in $\mathbb{Q}$ with the subspace topology from $\mathbb{R}$, the connected component of any point $q$ is $\{q\}$, which is closed but not open (since $\mathbb{Q}$ has no isolated points in the subspace topology inherited from $\mathbb{R}$). [/guided] [/step]