Showing that the one-point compactification of an open set in the Euclidean plane is path-connected

Question

Problem: Let $U\subseteq \mathbb{R}^2$ be any open set in the Euclidean plane $(\mathbb{R}^2$ with the standard topology) and consider $(U,T)$ as a topological space with the subspace topology inherited from the plane. Prove or disprove that $(U^*, T^*)$ is path-connected (where $(X^*,T^*$) denotes the one-point compactification/Alexandroff Extension of a topological space $(X,T)$).

I sketched out a proof as follows: Let $U$ be any open set in $\mathbb{R}^2$. Then we know that we can write $U= \cup_{\alpha}B_\alpha$ where $B_\alpha\subseteq U$ is an open ball. Then, \begin{align*} U^*&=(\cup_\alpha B_\alpha)^*\\ &=(\cup_\alpha B_\alpha)\cup \{\infty\}\\ &=\cup_\alpha (B_\alpha\cup \{\infty\}) \end{align*} Since $B_\alpha\cup\{\infty\}$ is homeomorphic to $S^2$ and $S^2$ is path-connected we have that $B_\alpha \cup \{\infty\}$ is path-connected for all $\alpha$ (path-connectedness is preserved under homeomorphisms).

Then, I want to use a result I've proved in the past: "Prove that if $A_i\subset X$ is path-connected for every $i\in I$, and $A_i\cap A_j\neq \emptyset$ for all $i,j\in I$, then $A=\cup_{i\in I} A_i$ is path-connected. " Applying this result then gives that $A^*$ is path-connected.

My concern with this proof is if I skipped too many steps for the homeomorphism part. However, is this sketch in the right direction?

Answer 1

For $U = \mathbb R^2$ it is trivial. So consider a proper subset $U$ of $\mathbb R^2$. To prove that $U^*$ is path connected, it suffices to show that for each $x \in U$ there exists a path $v :[0,1] \to U^*$ from $x$ to $\infty$.

Choose $y \in \mathbb R^2 \setminus U$ and a path $u : [0,1] \to \mathbb R^2$ such that $u(0) = x$ and $u(1) = y$. The set $u^{-1}(\mathbb R^2 \setminus U)$ is closed in $[0,1]$, hence compact with a minimum $m > 0$. By definition $u([0,m)) \subset U$ and $u(m) \in \mathbb R^2 \setminus U$.

We may assume w.l.o.g. that $m = 1$. Define $$v : [0,1] \to U^*, v(t) = u(t) \text{ for } t < 1, v(1) = \infty .$$ This map is continuous at all $\tau < 1$ because $v \mid_{[0,1)} = i \circ \tilde u$ with inclusion $i : U \to U^*$ and restriction $\tilde u : [0,1) \to U$ of $u$.

Continuity at $1$:

Let $V$ be an open neighborhood of $\infty$ in $U^*$. This means that $C = U^* \setminus V$ is a compact subset of $U$, thus a closed subset of $\mathbb R^2$. Therefore $u^{-1}(C)$ is closed in $[0,1]$ and does not contain $1$ (because $u(1) \in \mathbb R^2 \setminus U$ and $C \subset U$).

Therefore $W = [0,1] \setminus u^{-1}(C)$ is an open neighborhood of $1$ in $[0,1]$. To show that $v(W) \subset V$, it suffices to verify that $v(t) = u(t) \in V$ for $t \in W \setminus \{1\}$.

$t \in W$ means $t \notin u^{-1}(C)$, that is $u(t) \notin C$ . Therefore $v(t) = u(t) \in U^* \setminus C = V$.

By the way, all that was needed in the above proof is that $U$ is a proper open subset of a path connected Hausdorff space $X$.

Answer 2

You cannot expect that each $B_\alpha \cup \{\infty\}$ is path connected. As an example take $U = \mathbb R^n$. Let $B_n$ be the open ball with center $0$ and radius $n$. Then $U = \bigcup B_n$. Identifying $U^*$ with $S^2$ (where $\infty$ is the north pole $N$), you see that no $B_n \cup \{\infty\}$ is not path connected.

So what can we do?

Identifying $\mathbb R^2$ with $S^2 \setminus \{N\}$, we see that $A = S^2 \setminus U$ is a compact non-empty subset of $S^2$. Consider the quotient map $p : S^2 \to S^2/A$. It restricts to a continuous bijection $p' : U \to p(U) = (S^2/A) \setminus \{*\}$, where $*$ is the point obtained by collapsing $A$. Note that $\{*\}$ is closed in $S^2/A$ since $p^{-1}(\{*\}) = A$. Hence $p(U)$ is an open subspace of the compact space $S^2/A$.

We have $p^{-1}(p(U)) = U$, thus $U$ is a saturated subset of $S^2$ and $p'$ is a quotient map. Hence $p'$ is an open map (because it is bijective) and therefore a homeomorphism.

Let us show that $S^2/A$ "is" the one-point compactification of $U$. Indeed, define $$\phi : U^* \to S^2/A, \phi(x)= \begin{cases} p'(x) & x \in U \\ * & x = \infty \end{cases}$$ which is clearly a bijection. We claim that $\phi$ is a homeomorphism.

1. $\phi$ is continuous.

It is clearly continuous in all $x \in U$ since $p(U)$ is open is $S^2/A$. Let us show continuity in $\infty$. Let $V$ be an open neighborhood of $*$ in $S^2/A$. Its complement $V' = (S^2/A) \setminus V$ is a compact subset of $(S^2/A) \setminus \{*\}$, thus $W'= (p')^{-1}(V)$ is a compact subset of $U$ and $W = U^* \setminus W'$ is an open neigborhood of $\infty$ in $U^*$. By construction, $\phi(W) = V$.

2. $\phi^{-1}$ is continuous.

The proof is analogous to that of 1. Simply use $$\phi^{-1}(y)= \begin{cases} (p')^{-1}(y) & y \in (S^2/A) \setminus \{*\} \\ \infty & y = * \end{cases}$$

It therefore suffices to prove that $S^2/A$ is path connected. Let us show that for each $y \in (S^2/A) \setminus \{*\}$ there exists a path from $y$ to $*$.

Let $x = (p')^{-1}(y) \in U$. Choose a path $\gamma : I \to S^2$ such that $\gamma(0) = x$ and $\gamma(1) = N$. Then $\hat \gamma = p \circ \gamma : I \to S^2/A$ satisfies $\hat \gamma(0) = y$ and $\hat \gamma(1) = *$ (since $N \in A$).

Remark.

The proof can easily be generalized to open subsets of $\mathbb R^n$.