How to show that removing a point of an open and connected $U \subset \mathbb{R}^n$ such that $U\setminus \{p\}$ is no longer connected, then $n = 1$

Question

Basically title. Let $0 \in U \subset \mathbb{R}^n$ open and connected with $U \setminus \{0\}$ being no longer connected, then show that n = 1.

It's obvious that removing a point from an intervall in $\mathbb{R}$ makes the interval no longer connected, but it's difficult for me to actually prove rigoursly that this can't happen if n > 1. Is that even true?

Answer 1

I will try to answer what I guess you question is. Namely I will show that if $0 \in U \subseteq \mathbb R^k$ is an open, connected subset and $U \setminus \{0\}$ is not connected, then necessarily $k = 1$.

The first thing to note is that the open, punctured ball $B_\varepsilon(0) \setminus \{0\}\subseteq \mathbb R^k$ is connected for any $\varepsilon >0$ and $k > 1$, because it is even path connected.

Let $0 \subseteq U \subseteq \mathbb R^k$ be open and connected for some $k>1$. Choose some $\varepsilon > 0$, such that $B_\varepsilon(0) \subseteq U$. Now suppose that $U\setminus \{0\}$ was not connected. Then by definition, there exist two non-empty disjoint open sets $V_1, V_2$ such that $U \setminus \{0\} \subseteq V_1 \cup V_2$. If $(B_\varepsilon(0)\setminus \{0\}) \cap V_i$ are both non-empty, then we have written the connected set $B_\varepsilon(0)\setminus \{0\}$ as the union of two non-empty connected sets, which is a contradiction. So, w.l.o.g suppose that $B_\varepsilon(0) \subseteq V_1$. Now $W_1 := B_\varepsilon(0) \cup V_1$ and $W_2 := V_2$ are both open, disjoint (one needs to make sure that $0\notin V_2$, why is that so?) and cover $U$. But again, that contradicts the connectedness of $U$.

So $U \setminus \{0\}$ has to be connected as well, which establishes the claim.

Answer 2

Let $U$ an open connected set which is disconnected by removing the origin, clearly the real line is such a set, but the converse however is a bit more introcate.

Let the frontier of a set like $U$ be the set of points $p$ such that every set of which $p$ is an internal point meets both the interior and the exterior. My claim is that the origin is such a point iff $dim \geq 2$

But why is this? if the point disconnected a one dimensional line which was connected clearly it would still have been an internal point of the space. but the same is not the case in dim 2. Why?

Because one must in removing the point have a situation that any open neighbourhood about the origin contains infinitely many points both of the set $U$ and its complement. And if there in fact was a neighbourhood such that the origo was interior to the set then it couldn't disconnect it. since taking epsilon a little smaller for the ball about the origin and removing the whole ball would result in freedom to go around the origin (with path connectedness) since the bigger ball would have to touch on both pieces of the disconnected set

Answer 3

I'll show path-connectedness instead of connectedness because it is the stronger result, and in my opinion a bit more understandable by just drawing a picture. Note that we can use that $U$ is already path-connected since it is connected and we are in an open subset of euclidean space (where every point has path-connected open neighborhoods).

1. Case $n\geq 2$: Let $U\subseteq \mathbb{R}^n$ open and connected and such that $0\in U$. Because $U$ is open, there is some $B_\varepsilon(0)\subseteq U$. Now we will show that $X:=U\setminus\{0\}$ is path-connected and thus connected.

   Let $a,b\in X$ and let $p:[0,1]\to U$ be an $(a,b)$-path in $U$. If $p$ does not hit $0$, then $p$ is also an $(a,b)$-path in $X$. Else, let $t_0,t_1$ be the first/last occurence of $\overline{B_\varepsilon(0)}$ (you could define them as the $\inf,\sup$ of the preimage of the ball). We build an $(a,b)$-path $p'$ in X by modifying $p$: $$p'(t):=\begin{cases}p(t)&,t< t_0\\ \alpha(t)&,t_0\leq t\leq t_1\\ p(t)&,t>t_1.\end{cases}$$

   Where $\alpha:[t_0,t_1]\to \partial B_\varepsilon(0)$ is a $(p(t_0),p(t_1))$-path along the boundary of your ball. Quickly check the continuousness of $p'$ at the gluing points, and note that $\alpha$ exists because the boundary of $\varepsilon$-balls is path-connected for $n\geq 2$. Also, $p'$ never enters the interior ball, so it is a path in $X$, hence $X$ is path-connected.

2. Case $n=1$: Let $X_-,X_+$ be the negative and positive parts of $X$ (note that neither is empty because $U$ contains a ball around $0$). For each $x\in X$, define the "open ball" $B_x:=B_{\frac{|x|}{2}}(x)\cap X$. For $x>0$, $B_x$ lies in $X_+$ and for $x<0$, $B_x$ lies in $X_-$, and $X_+=\bigcup\limits_{x>0}B_x$ as well as $X_-=\bigcup\limits_{x<0}B_x$. As unions of open sets, $X$ is divided into the non-trivial open sets $X_-,X_+$ and is thus not connected.