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why is $\Bbb R\times[0,1]\not \cong \Bbb R^2$? we can't use the popular argument of deleting a point and finding that one has more path components than the other here.

So my idea is to delete a strip $\{0\}\times[0,1]$ from $\Bbb R\times[0,1]$.

But is $\Bbb R^2-f(\{0\}\times[0,1])$ always path-conneted when $f:\Bbb R\times[0,1]\to \Bbb R^2$ is a homeo?

Pindakaas1998
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    If you know some algebraic topology, then you could delete the point $(0,0)$ from $\mathbb{R}\times[0,1]$; the result is contractible, but $\mathbb{R}^2$ minus a point retracts to a circle and is not contractible. – Michael Burr Apr 26 '16 at 14:55
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    @MichaelBurr $\mathbb{R}^2$ minus a point is not homeomorphic to a circle (think topological dimension, e.g.). – Noah Schweber Apr 26 '16 at 14:56
  • @NoahSchweber Yeah, I meant retractible. Thanks. – Michael Burr Apr 26 '16 at 14:56

5 Answers5

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That spaces have different one-point (Alexandroff) compactifications, hence they cannot be homeomorphic.

The one-point compactification of $\mathbb{R}^2$ is the sphere $S^2$, while the one-point compactification of $\mathbb{R}\times[0,1]$ is a closed disc in $\mathbb{R}^2$ with a pair of its boundary points identified. In $S^2$, two open neighbourhoods of two different points are always isomorphic, and by removing any point in a neighbourhood it stays connected. In the last space, there is a point $u$ with an unusual behaviour: by taking an open neighbourhood $U$ of $u$, $U\setminus\{u\}$ is disconnected.

Jack D'Aurizio
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  • In the sense of Stone-Čech (for the strip)? Is there an easy way to see this? (Other than intuitively of course!) – Alex Provost Apr 26 '16 at 17:29
  • @AlexProvost: in the sense of Alexandroff. – Jack D'Aurizio Apr 26 '16 at 17:31
  • I don't understand. Wouldn't that mean that removing a point from the closed cylinder should yield a space homeomorphic to the strip? But the closed cylinder minus a point has nontrivial fundamental group. – Alex Provost Apr 26 '16 at 17:40
  • My argument is just an alternative to the other one. If we remove some point from $\mathbb{R}\times [0,1]$, the fundamental group may still be trivial, but that never happens for $\mathbb{R}^2$. – Jack D'Aurizio Apr 26 '16 at 17:49
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    It is incorrect to speak of "the" compactifications of these spaces . What is true is that their one-point (Alexandroff) compactifications are not homeomorphic. The one-point comp'n of $R\times [0,1]$ is homeomorphic to a closed disc in $R^2$ with a pair of its boundary points identified. – DanielWainfleet Apr 26 '16 at 18:24
  • @user254665: thank you, now fixed. – Jack D'Aurizio Apr 26 '16 at 18:46
  • I added the words "one-point" in my edit. I like the simple way you show they are not homeomorphic, without needing homotopy theory.+1. – DanielWainfleet Apr 26 '16 at 18:57
  • I just saw another flaw which I leave to you to fix. I think you didn't say what you meant to say: In $S^2$ every point $p$ has a nbhd base $B_p$ such that $b$ \ ${p}$ is connected for all $b\in B_p$ .But $u$ has no such nbhd base. – DanielWainfleet Apr 26 '16 at 19:03
  • @user254665: I hope my answer is clear now. – Jack D'Aurizio Apr 26 '16 at 19:08
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The property of simple connectivity will distinguish between $\Bbb{R} \times[0,1]$ and $\Bbb{R}^2$. When we remove one point from $\Bbb{R} \times[0,1]$ then it is simply connected but removing one point from $\Bbb{R}^2$ then it is not simply connected.

Yasir Mahsud
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To riff off Jack D'Aurizio's idea to use compactifications (and trying to avoid Algebraic Topology notions...):

The Cech-Stone compactification of $\mathbb{R}^2$ has a connected remainder, while that of $\mathbb{R} \times [0,1]$ has two components in its remainder. The proof is similar to that of the Cech-Stone compactification of $\mathbb{R}$.

The Freudenthal compactification should have the same properties for its remainder as well.

Henno Brandsma
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There are already more elementary answers given, but the Invariance of Domain theorem immediately shows they are not homeomorphic, as your strip is not open in $\mathbb{R}^2$.

Lasse Rempe
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Here’s another variation on the idea of using compactifications to avoid algebraic topology: $\Bbb R\times[0,1]$ has a two-point compactification, but $\Bbb R^2$ does not.

The two-point compactification of $\Bbb R\times[0,1]$ is pretty evident. Suppose that that $\Bbb R^2$ has a compactification $X=\{p,q\}\cup\Bbb R^2$, where $p\ne q$. Let $U$ and $V$ be disjoint open nbhds of $p$ and $q$; $K=\Bbb R^2\setminus(U\cup V)$ is a compact subset of the plane. Let $G=U\cap\Bbb R^2$ and $H=V\cap\Bbb R^2$; $G$ and $H$ are open in $\Bbb R^2$ and partition $\Bbb R^2\setminus K$.

$K$ is compact, so let $r_0=\max\{\|x\|:x\in K\}$. For each $r>r_0$ let $C_r$ be the circle of radius $r$ centred at the origin. $C_r$ is connected, so for each $r\ge r_0$ we must have $C_r\subseteq G$ or $C_r\subseteq H$. Let $I_G=\{r>r_0:C_r\subseteq G\}$ and $I_H=\{r>r_0:C_r\subseteq H\}$. Each $C_r$ is compact, so $I_G$ and $I_H$ are open subsets of the connected set $(r_0,\to)$; without loss of generality $I_G=(r_0,\to)$ (and $I_H=\varnothing$). It follows that $H$ is a bounded open set in $\Bbb R^2$ and hence that $W=X\setminus\operatorname{cl}_{\Bbb R^2}H$ is an open nbhd of $q$ in $X$. But then $V\cap W=\{q\}$ is open in $X$, and $q$ is isolated, which is impossible.

Brian M. Scott
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  • Every remainder of $\mathbb R ^2$ is connected as it is the intersection of continua $\text{cl}_{\gamma \mathbb R ^2}\mathbb R ^2 \setminus B(0,n)$ , whereas clearly $\mathbb R \times [0,1]$ has a non-connected remainder. – Forever Mozart Apr 28 '16 at 22:44
  • @ForeverMozart: That comment essentially makes the connection between my answer and Henno’s. Mine is intended to be as elementary as possible. – Brian M. Scott Apr 28 '16 at 22:49