# How to prove that any ball minus a point is a path-connected set?

Let $B_1\subset\mathbb{R}^n$ be a open ball and $B_2\subset\mathbb{R}^n$ be a closed ball, where $n\geq2$. How to show that if $p_1\in B_1$ and $p_2\in B_2$, then $B_1\setminus \{p_1\}$ and $B_2\setminus \{p_2\}$ are path-connected? Is it necessary "two proofs" or there is a unique proof that works for the two balls?

Thanks.

• $n > 1$ is necessary. – Daniel Fischer Jul 18 '13 at 22:58
• Perguntador: show that for any two points, there are two paths connecting them that intersect only at those points. In fact there are infinitely many, but you only need two. – dfeuer Jul 18 '13 at 23:00
• I suggest drawing some pictures of the two-dimensional case. – dfeuer Jul 18 '13 at 23:05
• @dfeuer I've tried do this: given $x,y \in B_1$ consider the segment $[x,y]=\{(1-t)x+ty;\;0\leq t\leq 1\}$. If $p_1\notin[x,y~]$, then $[x,y]$ is a path from $x$ to $y$. If $p_1\in [x,y]$, then we take some point $z\in B_1$ such that $z\notin [x,y]$. So, we consider the segments $[x,z]$ and $[z,y]$ to get a path for $x$ to $y$. What do you thing about this? – Pedro Jul 18 '13 at 23:18
• That's a perfectly good approach. – dfeuer Jul 18 '13 at 23:46

Let $A$ be an open ball.
Let $B$ be the closure of $A$.
Let $A\subseteq C \subseteq B$.
Let $p \in C$.
Then you can show that $C \setminus \{p\}$ is path-connected.