Path connectedness of one point compactification

Question

Let $X$ be a locally compact, metrizable, path-connected space which is not compact. Is the one-point compactification of $X$ also path-connected?

Answer 1

The answer is "no"'. We shall construct a path-connected closed $X \subset \mathbb{R}^3$ (which is a fortiori locally compact and $\sigma$-compact) such that the one-point compactification is not path-connected.

Let $$\Sigma = \lbrace (x,\sin(1/x)) \mid x \in (0,1] \rbrace \subset \mathbb{R}^2 ,$$ $$S = \lbrace 0 \rbrace \times [-1,1] \cup \Sigma \subset \mathbb{R}^2,$$ $$S_n = [0,2^{-n}] \times [-1,1] \cup \Sigma \subset \mathbb{R}^2, n \in \mathbb{N}.$$ $S$ is the closed topologist's sine curve which has the two path-components $\lbrace 0 \rbrace \times [-1,1]$ and $\Sigma$. The $S_n$ are path-connected. Define $$T_n = S \times \lbrace 0 \rbrace \cup \bigcup_{i=n}^\infty S_n \times \lbrace 2^{-n} \rbrace\subset \mathbb{R}^3 .$$ All $T_n$ are compact. Next define translations

$$t_n : \mathbb{R}^3 \to \mathbb{R}^3, t_n(x_1,x_2,x_3) = (n + x_1,x_2,x_3)$$ and set $$T = \bigcup_{n=1}^\infty t_{n-1}(T_n) .$$

This is a closed subspace of $\mathbb{R}^3$ which has infinitely many path components. Its intersection with $\mathbb{R}^2 \times \lbrace 2^{-n} \rbrace$ is $$R_n = \bigcup_{i=1}^n t_{i-1}(S_n \times \lbrace 2^{-n} \rbrace)$$ which is path-connected. Its intersection with $\mathbb{R}^2 \times \lbrace 0 \rbrace$ is $$R = \bigcup_{i=1}^\infty t_{i-1}(S \times \lbrace 0 \rbrace)$$ which splits into infinitely many path components (these are $C_0 = \lbrace 0 \rbrace \times [-1,1] \times \lbrace 0 \rbrace$ and $C_n = t_{n-1}((\Sigma \cup (\lbrace 1 \rbrace \times [-1,1])) \times \lbrace 0 \rbrace)$).

Define $$K_0 = \lbrace (0,0) \rbrace \times [0,1/2],$$ $$K_n = \lbrace n \rbrace \times [1,2] \times \lbrace 0, 2^{-n} \rbrace \cup \lbrace (n,2) \rbrace \times [0,2^{-n}], n \in \mathbb{N}.$$ Then $$K = K_0 \cup \bigcup_{n=1}^\infty K_n$$ is closed in $\mathbb{R}^3$ and $$X = T \cup K$$ is a closed path-connected subset of $\mathbb{R}^3$ ($K_0$ connects $C_0$ and the $R_n$, $K_n$ connects $C_n$ and $R_n$). Let $$A_n = X \cap (\lbrace n \rbrace \times \mathbb{R}^2).$$ Let $X^+ = X \cup \lbrace \infty \rbrace$ be the one-point compactification of $X$. Any path in $X^+$ from $A_0$ to $\infty$ must go through all $A_n$ since $X^+ \backslash A_n$ is not connected.

Assume $X^+$ is pathwise connected. Let $u : I \to X^+$ be a path with $u(0) \in A_0$ and $u(1) = \infty$. Let $a = sup \lbrace t \in I \mid u(t) \in A_0 \rbrace$ and $b = inf \lbrace t \in I \mid u(t) = \infty \rbrace$. Then $u(a) \in A_0, u(b) = \infty$ and $u((a,b)) \subset X \backslash A_0$. Moreover, $u((a,b))$ must intersect all $A_n$ with $n \ge 1$.

On the other hand, $u((a,b))$ must be contained in a path component of $X \backslash A_0$. But none of these intersects all $A_n$ which is a contradiction.

Answer 2

My example can be generalized as follows. Let $Q$ denote the Hilbert cube, $Q' = [0,\infty) \times Q$, $Q_n = [n,n+1] \times Q$, $n \ge 0$, and $a_n = (n,q_0)$ with $q_0 = (0,0,0,...) \in Q$.

Take any sequence of connected but not path-connected compact metric spaces $X_n$ together with points $x_n \in X_n$ and closed $Y_n \subset X_n$ such that $x_n \notin Y_n$ and $\lbrace x_n \rbrace \cup Y_n$ contains at least one point from each path component of $X_n$. It may happen that already $Y_n$ has this property, but we do not exclude the case that $x_n$ is not contained in any path component meeting $Y_n$.

Via a suitable embedding we may assume that the $X_n$ are subsets of $Q_n$ with the following properties:

(1) $x_n = a_n$

(2) $X_n \cap (\lbrace n \rbrace \times Q) = \lbrace a_n \rbrace$

(3) $a_{n+1} \in X_n$

(4) $a_n$ and $a_{n+1}$ belong to distinct path components of $X_n$.

We can write $X_n = \bigcap_{i=0}^\infty X_n^i$ with a sequence of path-connected closed $X_n^i \subset Q_n$ such that $X_n^{i+1} \subset int(X_n^i)$. They all contain $a_n, a_{n+1}$. We define subspaces of $Q'' = I \times Q' \times I$ as follows (brackets $\lbrace$ and $\rbrace$ are omitted for sets containing only a single point):

$$T_n = 0 \times X_n \times 0 \cup \bigcup_{i=n}^\infty 2^{-i} \times X_n^i \times 0 \subset I \times Q' \times 0 .$$

These are compact and we set

$$T = \bigcup_{n=0}^\infty T_n $$

which is closed in $Q''$ and has infinitely many path components. Its intersection with $2^{-m} \times Q' \times 0$ is $$R_m = \bigcup_{n=0}^m 2^{-m} \times X_n^m \times 0$$ which is path-connected. Its intersection with $0 \times Q' \times 0$ is $$R = \bigcup_{n=0}^\infty 0 \times X_n \times 0$$ which splits into infinitely many path components.

The set $$K_{-1} = I \times (a_0,0)$$ connects all $R_m$. To connect the path components of $R$ with $T \backslash R$ we introduce

$$K_n = 2^{-n} \times Y_n \times I \cup [0,2^{-n}] \times Y_n \times 1 \cup 0 \times Y_n \times I $$ They are pairwise disjoint since $Y_n \subset (n,+1] \times Q$. The set $$K = \bigcup_{n=0}^\infty K_n$$ is closed in $Q''$ and $$X = K_{-1} \cup K \cup T$$ is a closed path-connected subset of $Q''$. This is due to the fact that each path component $P$ of $X_n$ either meets $Y_n$ or contains $a_n$. Then $0 \times P \times 0$ is connected to $R_n$ via $K_n$ in the first case and to $R_{n-1}$ via $K_{n-1}$ in the second case since $a_n$ is contained in a path component $P'$ of $X_{n-1}$ meeting $Y_{n-1}$ (so that $P \cup P'$ is path-connected).

Define $A_n = X \cap (n \times Q \times I)$. Then the rest of the proof goes as in my first answer.

Of course we can replace $Q$ by any other compact metric space in which case only very special $X_n$ will be allowed.