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Prove that if $x \in \operatorname{cl}_X(A)$, where $A$ is a connected subspace of a topological space $X$, then $A \cup \{x\}$ is connected.

My attempt:

Suppose, in order to find a contradiction, that $B \cup C = A \cup \{x\}$ where $B,C$ are open in $A \cup \{x\}$ non empty and disjoint.

Then, we have either $A \subseteq B$ or $A \subseteq C$. Indeed, if this wouldn't be the case, then both $A \cap C$ and $A \cap B$ are non empty, and then

$A = A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ gives a union of disjoint non empty open (in A) sets, contradicting the connectedness of $A$.

Without loss of generality, we may assume that $A \subseteq B$. Then $x \in C$, or else $A \cup \{x\} = B$, meaning that $C = \emptyset$, which isn't possible.

Write $C = G \cap (A\cup \{x\})$ with $G$ open in $X$. Because $x \in \operatorname{cl}(A)$, and $x \in G$, it follows that $A \cap G \neq \emptyset$. Pick $y \in A \cap G$. Then $y \in A \subseteq B$ and $y \in A \cap G \subseteq C$, so $y \in B \cap C$. This is the desired contradiction.

Is this correct?

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  • $\begingroup$ More generally, if $A$ is connected and $A\subset B\subset Cl(A)$ then $B$ is connected. $\endgroup$ – DanielWainfleet Apr 18 '18 at 1:24
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The proof you gave is fine, I think. One can prove something slightly more general:

Let $A$ be a connected subspace of $X$ and suppose $A \subseteq D \subseteq \overline{A}$. Then $D$ is also connected.

My preferred proof uses the following well-known characterisation of connectedness:

$X$ is connected iff every continuous function from $X$ to $\mathbf{2}$ (the discrete space $\{0,1\}$) is constant.

Let $f: D \to \mathbf{2}$ be continuous. Then we know that $f|A$ is constant with value $i_A \in \{0,1\}$. So $f$ and the constant function with value $i_A$ agree on $A$, both are continuous, and as $A$ is dense in $D$ (from $D \subseteq \overline{A}$) and $\mathbf{2}$ is Hausdorff, another classic theorem says that $f$ agrees with the constant map with value $i_A$ on the whole of $D$. So, $f$ is constant, and $D$ is connected. In particular $\overline{A}$ is connected.

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  • $\begingroup$ Dumb me. I proved that theorem in class (with the exact same proof) but didn't realise this was a special case. Thanks $\endgroup$ – user370967 Apr 17 '18 at 21:19
  • $\begingroup$ @Math_QED Glad I could help your enlightenment.. $\endgroup$ – Henno Brandsma Apr 17 '18 at 21:20
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The OP's proof is correct, but by changing the last 2 paragraphs it can be nailed down in another way.

Suppose, in order to find a contradiction, that $B \cup C = A \cup \{x\}$ where $B,C$ are open in $A \cup \{x\}$ non empty and disjoint.

Then, we have either $A \subseteq B$ or $A \subseteq C$. Indeed, if this wouldn't be the case, then both $A \cap C$ and $A \cap B$ are non empty, and then

$A = A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ gives a union of disjoint non empty open (in A) sets, contradicting the connectedness of $A$.

Without loss of generality, we may assume that $A \subseteq B$. But then it must be that $B = A$ and $C$ is the open singleton $\{x\}$.

Since $x \in \operatorname{cl}(A)$, any open set in $A \cup \{x\}$ that contains $x$ can't be a singleton. This is the desired contradiction.

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  • $\begingroup$ $A \cap G \subseteq (A \cup \{x\}) \cap G = C$. Is it clear now? $\endgroup$ – user370967 Apr 18 '18 at 12:58
  • $\begingroup$ Yup. Thanks. Was lost in the weeds I guess. $\endgroup$ – CopyPasteIt Apr 18 '18 at 13:13

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