Is $X-\{p\}$ connected if $p$ has a ball around it?

Question

Let $X$ be a path-connected topological space, and $p \in X$ be such that there exists a neighbourhood $B \subset X$ of $p$ such that $B$ is homeomorphic to an open ball of dimension greater than $1$. We can prove that $X-\{p\}$ is path-connected by using the long exact sequence of $(X-\{p\},D-\{p\})$ together with the excision and the fact that $X$ is path-connected. (We can also prove this directly with some work).

This made me wonder: Does this preservation hold if $X$ is connected? That is,

Let $X$ be a connected topological space, $p \in X$ be such that there exists a neighbourhood $B \subset X$ of $p$ such that $B$ is homeomorphic to an open ball of dimension greater than $1$. Is is true that $X -\{p\}$ is connected?

Things like the Kuratowski fan make me a little uneasy to approach the problem, although it is clearly not a counter-example.

Answer 1

Let $f:X-\{p\}\rightarrow \{0,1\}$ be a continuous function. Since $B-\{p\}$ is connected, $f(B-\{p\})=0$ or $1$, without restricting the generality, suppose that $f(B-\{p\})=0$. Then you can extend $f$ to $X$ by setting $f(p)=0$. Since $f$ is continue and $X$ connected, $f$ is constant. We deduce that $X-\{p\}$ is connected.

Answer 2

Yes.

Let $\phi\colon B\to B_R(0)\subset \Bbb R^n$ be the assumed homeomorphism of a neighbourhood of $p$ with an open $n$-dimensional ball (where we assume $\phi(p)=0$). Then $\phi^{-1}(B_{R/2})(0))$ is an open neighbourhood of $p$ that is homeomorphic to a ball. Hence we may assume wlog that $B$ is already an open neighbourhood.

Assume $X-\{p\}\subseteq U\cup V$ where $U,V$ are disjoint open subsets of $X$. Then $\phi(U\cap B)-\{0\}$ and $\phi(V\cap B)-\{0\}$ are disjoint open subsets of $B_R(0)$ that cover $B_R(0)-\{0\}$. By connectedness of $B_R(0)-\{0\}$, one of them is empty. So say $\phi(V\cap B)-\{0\}=\emptyset$. Then also $\phi(V\cap B)=\emptyset$. Then we have $X=(U\cup B)\cup V$ with disjoint open sets $U\cup B$ and $V$. As the first contains $p$, we conclude from connectedness of $X$ that $V=\emptyset$.

Answer 3

OK, let $U\sqcup V$ be a separation of $X- \{p\}$. Then one of these sets, say, $V$, contains $B- \{p\}$ since the latter is connected. Now, $U \sqcup (V\cup B)$ is a separation. since $U\cap B$ is empty. As I said, just follow the definition.