# Is there a neat way to prove that $R^2$ is not homeomorphic to $R^2$\{(0,0)}

Although similar questions has been asked, I still didn't find a clear and simple answer only involved with connectedness and restriction function.

In particular, to prove $$[0,1)$$ is not homeomorphic to $$(0,1)$$, we only need to consider the restriction function $$f: [0,1)\setminus \{0\}$$ to $$(0,1)\setminus \{f(0)\}$$, obviously the former is connected and the latter disconnected, contradicting $$f$$ being a homeomorphism.

However, I have tried $$\Bbb R^2 \setminus \{p\}$$ and $$\Bbb R^2\setminus \{L\}$$ where $$p$$ is a point and $$L$$ a line, none of them provide a satisfactory counterexample. Is this possible to construct a contradiction?

• All proofs of this fact will involve some non-trivial fact (it doesn't simply follow from a straightforward connectedness argument). Nov 20, 2021 at 14:39
• @HennoBrandsma Noted with thanks! Nov 20, 2021 at 14:56
• A couple of extra ways to prove this without Algebraic Topology. You may use the different compactifications: Alexander's one point, and Freudenthal's by ends. In the first case their compactifications yield $\mathbb{S}^2$, and $\mathbb{S}^2$ with two points identified. Alternatively, $\mathbb{R}^2$ has one end, while $\mathbb{R}^2\setminus\{0\}$ has two.
– Laz
Oct 12, 2022 at 17:04

Here is a rather simple proof, which does not involve algebraic topology: Call a space $$X$$ compactly connected, iff for each compact subset $$A$$ of $$X$$ there is compact subset $$B$$ of $$X$$, such that $$A \subseteq B$$ and $$X \setminus B$$ is connected.
It is easy to see that R$$^2$$ is compactly connected.
But $$X :=$$ R$$^2 \setminus \{(0,0)\}$$ is not:
Let $$A:= S^1$$ be the unit circle in $$X$$ and assume there is a compact $$B$$ in $$X$$, such that $$A \subset B$$ and $$X \setminus B$$ is connected. Let D be the open unit ball in $$X$$ and $$E := X \setminus (D \cup A)$$, which is open in $$X$$. $$X \setminus B$$ is the disjoint union of $$D \cap X \setminus B$$ and $$E \cap X \setminus B$$, hence one of these sets must be empty. If the second one is empty, $$E \subseteq B$$, hence $$E$$ is bounded. Contradiction. Therefore $$D \subseteq B$$. But then $$D^\prime := \{(x,y) \in R^2: ||(x,y)|| \le \frac{1}{2}\} \setminus \{(0,0)\}$$ is a closed subset of $$B$$, hence compact. Contradiction!
• Very nice (+1). Only one remark: If the second one is empty, then $E \subset B$. Therefore $D \cap X \subset B$. Nov 22, 2021 at 17:58