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I'm asked to prove that $\mathbb{R}$ and $\mathbb{R}^2$ are not homeomorphic. So far, I've been able to prove that $\mathbb{R}\backslash\{a\}$ and $\mathbb{R}^2\backslash\{b\}$ are not homeomorphic, for $a\in \mathbb{R}$ and $b\in \mathbb{R}^2$. But I don't know how to go on from here. Can anyone give me a hint?

Jakobian
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RBS
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5 Answers5

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Hint. If $f \colon \def\R{\mathbf R}\R \to \R^2$ were a homeomorphism, what does this imply for the restriction $f\colon \R \setminus \{a\} \to \R^2 \setminus\{f(a)\}$?

martini
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  • Well, if $f(a)=b$, we have that both sets need to be connected, which isn't true. But how does this work for the case where we don't remove points from the sets? – RBS Apr 07 '15 at 13:56
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    Suppose that $\def\R{\mathbf R}\R$ and $\R^2$ were homeomorphic. Take any homeomorphism between $f \colon \R \to \R^2$. Now (not in the beginning), remove any point $a$ from $\R$. If we remove the image point $f(a)$ also, the leftover map $f \colon \R -{a} \to \R^2-{f(a)}$ is still a homeomorphism. But such a thing can - as you already proved - not exist. – martini Apr 07 '15 at 13:59
  • Wow, I can't believe I did not see that before. Thanks! – RBS Apr 07 '15 at 14:00
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Hint: Any point in $\mathbb R$ is a cut-point. While if you remove a point in $\mathbb R^2$, it remains connected because is homeomorphic to $S^1 \times \mathbb R$.

Aaron Maroja
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  • Right, and since a homeomophism would preserve connectedness, both $\mathbb{R}\backslash{a}$ and $\mathbb{R}^2\backslash{b}$ would be connected, which is a contradiction. I just don't see how I can expand this to the case where we remove no points from the sets. – RBS Apr 07 '15 at 13:55
  • You may either use @martini 's answer or notice that homeomorphisms take cut-points into cut-points. – Aaron Maroja Apr 07 '15 at 13:56
  • I never learned anything about cut-points (at least they aren't named that way), so I didn't know that. – RBS Apr 07 '15 at 13:59
  • I didn't know either until some days ago. You'll see it's a very useful tool. – Aaron Maroja Apr 07 '15 at 14:19
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For those who don't want to use the idea of cut points, but can rely on the intermediate value theorem, and believe that the range of a continuous 1-1 function $g:\mathbb{R}\to \mathbb{R}$ is an interval (more precisely, an open interval).

Let $f:\mathbb{R}^2\to \mathbb{R}$ be a homeomorphism. For each $x\in \mathbb{R}$, define $f_x:\mathbb{R}\to\mathbb{R}$ as $f_x(y)=f(x,y)$. We have the following easy observations.

(1) $f_x$ is a continuous 1-1 map,

(2) range of $f_x$ is a non-trivial (open) interval,

(3) if $x_1\neq x_2$, then the range of $f_{x_1}$ and the range of $f_{x_2}$ are disjoint.

Thus we have got an uncountable number (corresponding to each $x\in \mathbb{R}$) of pairwise disjoint open intervals (range of each function $f_x$) in $\mathbb{R}$. This contradicts separability of $\mathbb{R}$, since $\mathbb{R}$ has a countable dense set, for example $\mathbb{Q}$.

Fact: In a separable topological space, there does not exist an uncountable number of pairwise disjoint non-empty open sets. If such a collection exists, then any dense set will have to meet each member of the collection and would demand the cardinality of a dense set to be uncountable (by pigeon-hole principle).

Janson
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Let $f\colon X\to Y$ be a homeomorphism, $U\subseteq X$ any subspace of $U$. Equip $U$ and $f(U)$ with the subspace topologies. Show that $U \overset{f}\longrightarrow f(U)$ is a homeomorphism again. Now consider $X=\mathbb R$, $Y=\mathbb R^2$ and $U=X\setminus\{a\}$.

Christoph
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This relies on removing a point which preserves continuity and when you remove a point from $\Bbb{R}$ its disconnected while removing a part from $\Bbb{R}^2$ is connected, thus you cannot have a homeomorphism.

$\textbf{Note:}$ if $f: X \to Y$ is a homeomorphism, then so is

$$f: X \setminus \{x_0\} \to Y \setminus \{f(x_0)\}.$$

This requires proof as well as the plane minus a point being connected. It is trivial to see the reals minus a point are disconnected.

homosapien
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