Closed ball minus a point is connected Could someone help me to prove that if $B\subset\mathbb{R}^n$ ($n>1$) is a closed ball, then $B\setminus \{p\}$ is a connected set for all $p\in B$? 
 A: HINT: Let $x$ and $y$ be any two points of $B\setminus\{p\}$. If $p$ is not on the line segment $\overline{xy}$, then that line segment is a connected set in $B\setminus\{p\}$ containing both $x$ and $y$. If $p$ does lie on the line segment $\overline{xy}$, let $z$ be any point of $B$ that is not on the line through $x$ and $y$, and show that $\overline{xz}\cup\overline{zy}$ is a connected subset of $B\setminus\{p\}$ that contains both $x$ and $y$. (This actually shows that $B\setminus\{p\}$ is not just connected, but path connected.)
A: You can solve this problem by reducing to dimension $2$. Firstly, any closed ball in $R^n$ is homeomorphic to the unit closed ball in $\mathbb R^n$, so it suffices to prove that the unit closed ball has the property you mention. Now, in the $\mathbb R^2$, you can draw the argument to see that the closed unit ball minus any point is path connected. 
Now, for the general case, given the closed unit ball $B$ in $\mathbb R^n$, with $n\ge 2$, a point $p$ that is removed from $B$, and two points $x,y\in B-\{p\}$, let $V$ be any plane that contains the origin as well as $x$ and $y$ (this is where $n\ge 2$ is used). The intersection $V\cap B$ is homeomorphic to the unit ball in $\mathbb R^2$. Thus, the intersection $V\cap (B-\{p\})$ is homeomorphic either to the unit ball in $\mathbb R^2$ or to the unit ball with a point removed. In either case, $x$ and $y$ can be connected by a path in the intersection (by the two-dimensional argument) and thus also in $B-\{p\}$. 
This argument shows that the property is essentially two-dimensional. 
A: Let $C = B \setminus \{p\}$. Let's asssume that $C$ is not connected, and show that this leads to a contradiction. 
If $C$ is not connected, then we can find two closed sets $X$ and $Y$ such that $X \cap Y = \emptyset$ and $X \cup Y = C$. Let $Z = Y \cup \{p\}$. Then $Z$ is closed, $X \cap Z = \emptyset$, and $X \cup Z = B$. But this implies that $B$ is not connected, which is false, so our original assumption must have been wrong.
Caveat: the last time I wrote anything like this was about 40 years ago, so it's quite possible that I've forgotten a few things about topology. As always, you shouldn't necessarily believe the things you read on the internet.
