Is $\mathbb{C}-\{0\}$ homeomorphic to $\mathbb{R}$? I am trying to see if $\mathbb{C}-\{0\}$ homeomorphic to $\mathbb{R}$ with both having metric topology. 
They both are connected, non-compact. what other topological property can I look at? Or is there a new way to approach the problem?
 A: If $x,y\in\mathbb R$ then $t x+(1-t)y\in\mathbb R$ for each $t\in[0,1]$, so $\mathbb R$ is convex. In particular, this implies that $\mathbb R$ is path connected, as the map $p:[0,1]\to\mathbb R$ with $p(t)=tx+(1-t)y$ is continuous. Now let $f,g:[0,1]\to\mathbb R$ be paths with $f(0)=g(0)=x$ and $f(1)=g(1)=y$. Define $H:[0,1]^2\to\mathbb R$ by $$H(s,t) = (1-t)f(s) + tg(s). $$ For any fixed $s$, the map $t\mapsto (1-t)f(s) + tg(s)$ is the line segment connecting $f(s)$ and $g(s)$ (and hence is continuous), so by convexity, $H(s,t)\in\mathbb R$ for all $(s,t)\in[0,1]^2$. Moreover, for each fixed $t$, the map $s\mapsto (1-t)f(s)+tg(s)$ is continuous as the linear combination of continuous functions, so $H$ is continuous. It follows then that $H(s,0)=f(s)$ and $H(s,1)=g(s)$ for all $s\in[0,1]$ and that $H(0,t)=x$ and $H(1,t)=y$ for any $t\in[0,1]$. We conclude that $H$ is a path homotopy
As for $\mathbb C\setminus\{0\}$, consider the path $p(t) = e^{2\pi i t}$ and the integral of $z\mapsto\frac1z$ along this path:
$$\oint_p \frac1 z\ \mathsf dz=\int_0^1 2\pi i e^{2\pi it}e^{-2\pi i t}\ \mathsf d t = 2\pi i\int_0^1\ \mathsf dt = 2\pi i. $$ Since the integral is not zero, by Cauchy's integral theorem, $p$ is not path homotopic to the constant path $c(t)=1$ (as the integral of any function along a trivial path is zero). It follows that $\mathbb C\setminus\{0\}$ has more than one path homotopy class, and so is not simply connected. This implies that $\mathbb R\not\cong\mathbb C\setminus\{0\}$. 
A: Small Inductive Dimensions do not match. For each $x\in \mathcal R$, the set $B_x=\{x-r,x+r :r>0\}$ is a local base at $x$, and for all $b\in b$, the set $\;\bar b\backslash b\;$ has just $2$ members. No $z\in S=\mathcal C\backslash \{0\}$ has a local base with this property. For if $\;D_z\;$ is a local base at $z$ then some $d\in D_z$ is a subset of $\{y:|y-z|<|z|/2,\;$  and for each $t\in [0,2 \pi), $ there is a least real  $r\in (0,|z|/2]$ such that $z+r e^{i t}\not \in d.$  Then $$z+r e^{i t}\in \bar d\backslash d.$$
