How to proof that $\mathbb{R}$ and $\mathbb{R}^2$ are not homeomorphic? So I'm asked to proof 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?
 A: 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 $S^1 \times \mathbb R$.
A: Hint. If $f \colon \def\R{\mathbf R}\R \to \R^2$ were a homeomorphism, what does this imply for the restriction $f\colon \R \setminus \{a\} \to \R^2 \setminus\{f(a)\}$?
A: 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 $U=X\setminus\{a\}$.
A: 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).
