A proper open subset of $\mathbb{R}$ is never homeomorphic to a proper open subset of $\mathbb{R^n}$

I am reading a proof of the statement:

A proper open subset $V_{1}$ of $\mathbb{R}$ is never homeomorphic to a proper open subset $V_{2}$ of $\mathbb{R^n}$ for $n>1$.

Proof: Suppose by contradiction $V_{1} \simeq V_{2}.$ Then there exists a connected component $W_{1} \subset V_1$ homeomorphic to a connected subset $W_2 \subset V_2$. But connected components of $V_1$ are intervals. So for any $x \in W_1$, $W_1$ \ $ \{x\}$ is disconnected. This implies that $W_{2} \backslash \{y\}$ is disconnected for any $y \in W_2$ which is impossible since $W_2$ contains a ball centred at $y$.

It is this last part in bold that I don't understand. How does the fact that $W_{2}$ contains a ball centred at $y$ imply $W_2$ can't be disconnected?

Solution 1:

For $n\geq 2$, a ball $B$ remains connected, in fact, it remains path connected, even if you remove a single point $y$ from the ball.

This can quite easily be proven, since for any two points $x_1,x_2\in B\setminus \{y\}$, you have two options:

1. If the straight line from $x_1$, $x_2$ does not path through $y$, then that straight line is a path from $x_1$ to $x_2$ in $B\setminus \{y\}$.
2. If the straight line from $x_1$ to $x_2$ does pass through $y$, then there exists some point $z$ such that $z$ is not on the line from $x_1$ to $x_2$. In this case, there exists a path from $x_1$ to $z$ and from $z$ to $x_1$, so there also exists a path from $x_1$ to $x_2$.

EDIT:

I now see what you mean by your question, and I agree that the proof you cite is a little handwavy.

The point of the proof is that the following three things cannot happen all at once:

• $W_2$ is connected.
• $W_2$ contains a ball, centered at $y$.
• $W_2\setminus \{y\}$ is not connected.

You can prove this the three points above form a contradiction by proving that if the first two points hold, then the third must not hold, and you do this like so:

Take any $u,v\in W_2\setminus y$. Then, because $\mathbb R^n$ is connected iff it is path connected, there exists a path $\gamma$ from $u$ to $v$ in $W_2$. Now,

• if $\gamma$ does not pass through $y$, then $\gamma$ is a path from $u$ to $v$ in $W_2\setminus\{y\}$.
• Otherwise, $B$ be the closed ball centered at $y$ that is contained in $W_2$. Let $u'$ be the point at which $\gamma$ first enters $B$, and $v'$ be the point at which $\gamma$ leaves $B$ for the last time. Then, clearly, there exists a path from $u$ to $u'$, and from $v$ to $v'$, because you can just take the part of $\gamma$ that suits you. Also, from what I wrote above, there also exists a path in $B$ from $u'$ to $v'$. So, concatenating the three paths, there exists a path from $u$ to $v$, this $W_2\setminus \{y\}$ is path connected.

Solution 2:

Suppose $y \in W_2$ and that $U,V$ are disjoint open sets with $U \cup V = W_2 \setminus \{y\}$. Let $B$ be an open ball with $y \in B \subseteq W_2$. Then $B \setminus \{y\}$ is connected as we are in $>1$ dimension. Note that $U \cap B$ and $V \cap B$ are disjoint open sets with $(U \cap B) \cup (V \cap B) = B \setminus \{y\}$. By connectedness we can WLOG conclude that $U \cap B = B \setminus \{y\}$ and $V \cap B = \emptyset$. Let $U' = U \cup \{y\}$. Since $B \setminus \{y\} = U \cap B \subseteq U$ we also have $U' = U \cup B$, so $U'$ is open. Note that $U'$ is disjoint from $V$. Moreover $U' \cup V = U \cup \{y\}\cup V = W_2$. Since $U' \neq \emptyset$ and $W_2$ is connected, we conclude that $V = \emptyset$ and $U' = W_2$, which implies $U = W_2 \setminus \{y\}$. Thus $W_2 \setminus \{y\}$ is connected.