Show $\mathbb{R}^n\setminus \{0\}$ is not homeomorphic to $\mathbb{R}^n $ Is there any simple proof that these two are not homeomorphic?
 A: For $n=1$, you can use the simple argument that $\mathbb{R}$ is connected while $\mathbb{R}\setminus \{0\}$ is not connected.
For $n = 2$, you need to be smarter. Here is an answer using fundamental groups.
We have:
$$\pi_1(\mathbb{R}^2) \cong \{0\}$$
since $\mathbb{R}^2$ is convex (and we can consider linear homotopies). On the other hand,
$$\mathbb{R}^2 \setminus \{0\}\cong S^{1}\times ]0, \infty[$$ 
via the homeomorphism $\mathbb{R}^2 \setminus \{0\}\to S^{1}\times ]0, \infty[: x \mapsto (x/\Vert x \Vert, \Vert x\Vert)$.
So $$\pi_1( \mathbb{R}^2 \setminus \{0\}) \cong \pi_1(S^{1} \times ]0, \infty[) \cong \pi_1(S^{1}) \times \pi_1(]0,\infty [) \cong \mathbb{Z}$$
and the fundamental groups are different. 
For higher dimensions, you can use the same arguments but with higher homology groups instead.
A: It is easy to show that $\mathbb R^n$ is contractible. In fact, the map:
$$H: \mathbb R^n\times [0, 1]\longrightarrow \mathbb R^n$$ given by $$H(x, t):=(1-t)x$$ gives us a homotopy between the identity map of $\mathbb R^n$ and the constant map $c(x):=0$ for every $x\in\mathbb R^n$.
It is also easy to see that if $X$ is homeomorphic to a contractible space $Y$ then $X$ is itself contractible. In fact, let $$f: X\longrightarrow Y$$ be an homeomorphim and let $$H: Y\times [0, 1]\longrightarrow Y$$ be an homotopy between the identity map $\textrm{id}_Y$ of $Y$ and a constant map $c: Y\longrightarrow Y$. Define $$\overline{H}: X\times [0, 1]\longrightarrow X$$ setting $$\overline{H}(x, t):= f^{-1}(H(f(x), t)).$$ Then $\overline{H}$ is continuous and $$\overline{H}(x, 0)=f^{-1}(H(f(x), 0))=f^{-1}(f(x))=x$$ and $$\overline{H}(x, 1)=f^{-1}(H(f(x), 1))=f^{-1}(c(f(x)),$$ which is constant for every $x$. Therefore $\overline{H}$ is a homotopy between the identity map $\textrm{id}_X$ of $X$ and the constant map $$c_f(x):=f^{-1}(c(f(x)),$$ for every $x\in X$. Thus $X$ is also contractible.
Pluggin this into your problem, if $\mathbb R^n-\{0\}$ were homeomorphic to $\mathbb R^n$ then $\mathbb R^n-\{0\}$ would be contractible, which is not the case. 
A: If you want to avoid any use of homotopy or homology, there's this (for $n \geq 2$): Remove any compact subset from $\mathbb R^n$ and the remainder has only one component whose closure in the whole space is not compact. With $\Bbb{R}^n \setminus \{0\}$ you can get two.  Or (equivalently): $\Bbb{R}^n \setminus \{0\}$ has a two-point compactification but $\mathbb R^n$ does not.
A: 
I do not know if this proof is simple.

No they are not.
For $n>2$
The space $\Bbb{R}^n \setminus \{0\}$ is NOT $\textbf{contractible}$ but $\Bbb{R}^n$ is
Contractibility is a topological invariant.
For $n=1$ again they are not since $\Bbb{R}\setminus \{0\}$ is not connected.
