is it possible to construct a continuous bijective map from $\mathbb{R}$ to $\mathbb{R^{2}}$ is it possible to construct a continuous bijective map from $\mathbb{R}$ to $\mathbb{R^{2}}$. if it is, please give an example.If not, how to prove?  
 A: If $f:\mathbb{R}\to \mathbb{R}^2$ is continuous and injective, then $f$ cannot be surjective. The reason is as follows. For each $n\in \mathbb{N}$, since $[-n,n]$ is compact, it is homeomorphic to $f([-n,n])$. Then due to invariance of domain, $f([-n,n])$ is a nowhere dense closed subset of $\mathbb{R}^2$. Therefore, by Baire category theorem, $f(\mathbb{R})=\cup_{n=1}^\infty f([-n,n])$ is also nowhere dense in $\mathbb{R}^2$.
Remark: With an almost identical argument, the result can be generalized as follows. If $m\ne n$, then there is no continuous bijection from $\mathbb{R}^m$ to $\mathbb{R}^n$.
A: The answer is no.
Suppose that $f:\Bbb R\to\Bbb R^2$ is a continuous bijection. For $n\in\Bbb Z^+$ let $I_n=[-n,n]$, and let $K_n=f[I_n]$. Then $\Bbb R^2=\bigcup_{n\in\Bbb Z^+}K_n$, so by the Baire category theorem some $K_m$ has non-empty interior. Choose $x$ and $r$ so that $B(x,r)\subseteq K_m$, and let $g=f\upharpoonright I_m$. Then $g$ is a continuous bijection from the compact set $I_m$ onto $K_m$, so $g$ is a homeomorphism. (In case this isn’t familiar, note that $g$ is closed: every closed subset of $I_m$ is compact, $g$ preserves compactness, and every compact subset of $K_m$ is closed, since $K_m$ is compact. Finally, a closed, continuous bijection is clearly a homeomorphism.)
Let $J=g^{-1}[B(x,r)]$; $B(x,r)$ is connected and open, so $J$ is an open interval in $I_m$ homeomorphic to $B(x,r)$. But this is impossible: $J$, being an interval, has cut points, and $B(x,r)$ has none.
