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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?

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    $\begingroup$ All proofs of this fact will involve some non-trivial fact (it doesn't simply follow from a straightforward connectedness argument). $\endgroup$ Nov 20, 2021 at 14:39
  • $\begingroup$ @HennoBrandsma Noted with thanks! $\endgroup$
    – Dinoman
    Nov 20, 2021 at 14:56
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    $\begingroup$ 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. $\endgroup$
    – Laz
    Oct 12, 2022 at 17:04

1 Answer 1

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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!

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  • $\begingroup$ Very nice (+1). Only one remark: If the second one is empty, then $E \subset B$. Therefore $D \cap X \subset B$. $\endgroup$
    – Paul Frost
    Nov 22, 2021 at 17:58
  • $\begingroup$ @Paul: yes, of course. I corrected it. Thank you very much! $\endgroup$
    – Ulli
    Nov 22, 2021 at 18:07

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