# Homeomorphic characterization of the real line?

Let $$A$$ be a path-connected subset of $$\mathbb R^2$$ such that the removal of any singleton from $$A$$ splits $$A$$ into two open connected components, each of which is path-connected.

Is $$A$$ necessarily homeomorphic to $$\mathbb{R}$$?

• related post: mathoverflow.net/questions/76134/… Jan 1 '19 at 12:33
• I voted for reopening, because the question does not assume local connectedness while the answer at the other question does (and the purportedly "duplicate" question does not ask the same, being more open-ended).
– YCor
Sep 6 '19 at 19:07

Ward has given the following characterization of the real line: a connected, locally connected separable metric space in which each point is a cut point, i.e., its removal splits the space into two connected subsets (Proc. London Math. Soc. 1936). This implies a positive answer to your question, assuming the set has more than one point.

• In order to apply this result, we need to establish the local connectednes sof $A$. And how to prove this fact? Jan 5 '19 at 21:35

Theorem. Suppose $$X$$ is a connected separable metric space such that $$X\setminus \{x\}$$ has exactly two (non-empty) open path-connected components for every $$x\in X$$. Then $$X\simeq \mathbb R$$.

Proof. We claim that each point $$x\in X$$ is contained in an arc $$A$$ with endpoints $$a$$ and $$b$$ such that $$x\in A\setminus \{a,b\}$$ and the interval $$A\setminus \{a,b\}$$ is open in $$X$$. Clearly this would imply $$X$$ is locally connected, and then by Ward's theorem (see earlier answer) we get $$X\simeq \mathbb R$$.

Let $$x\in X$$. Let $$y\in X\setminus \{x\}$$. It is a simple exercise to prove that $$X\setminus \{x,y\}$$ is the union of three non-empty disjoint open sets $$U\sqcup V\sqcup W$$ such that $$U\cup \{x\}\cup V$$ is path-connected. Let $$A$$ be an arc with endpoints $$a\in U$$ and $$b\in V$$. Then we have $$x\in A\setminus \{a,b\}$$.

We need to two claims. The proofs are by contradiction.

Claim 1: $$X$$ contains no simple closed curve. Suppose $$S\subseteq X$$ is a simple closed curve. Then for every point $$x\in S$$ there are two non-empty open sets $$U_x$$ and $$V_x$$ such that $$X\setminus \{x\}=U_x\sqcup V_x$$ and $$S\subseteq U_x\cup\{x\}$$. Notice that the sets $$V_x$$, $$x\in S$$, are pairwise disjoint. This contradicts the fact that $$X$$ is separable.

Claim 2: $$X$$ contains no simple triod. Suppose $$T\subseteq X$$ is a simple triod which is the union of three arcs $$A_1,A_2,A_3$$ and has center vertex $$a$$. Write $$X\setminus \{a\}= U\sqcup V$$ so that $$U$$ and $$V$$ are path-connected. Without loss of generality $$A_1\cup A_2\subseteq U\cup \{a\}$$. Let $$B\subseteq U$$ be an arc meeting both $$A_1$$ and $$A_2$$. Then $$B\cup A_1\cup A_2$$ must contain a simple closed curve. This contradicts Claim 1.

By Claims 1 and 2 we have:

Claim 3. The union of any two intersecting arcs in $$X$$ is again an arc.

Finally we can show $$A\setminus \{a,b\}$$ is open in $$X$$. Suppose to the contrary that $$c\in A\setminus \{a,b\}$$ is the limit of a sequence of points $$x_n\in X\setminus A$$. By the path-connected property and Claim 3, there is an increasing collection of arcs $$A_0\subseteq A_1\subseteq ...$$ such that $$\{x_n\}\cup A\subseteq A_n$$ for every $$n<\omega$$. Then we get a one-to-one continuous image of the half-line $$[0,\infty)$$ which begins in $$A$$ and also limits to $$A$$ at the infinity end. This forms a circle-like configuration, which is impossible by the arguments in Claim 2 (the "ray limiting to itself" would remain connected upon the deletion of any point, and this leads to an uncountable collection of pairwise disjoint non-empty open sets). $$\;\blacksquare$$

Note 1: I used path-connectedness of $$X$$ near the end. This was a hypothesis in the original question, but it actually follows from my assumptions in the theorem statement. It's a simple exercise.

Note 2: Ward's theorem is probably overkill. We already see that $$X$$ is an increasing union of arcs and contains no ray limiting to itself.