Hyperplane and dimension I'm trying to prove the following:
V is a Vectorspace
$dim V = \infty \Leftrightarrow \exists$ a linear Bijection from $V$ to a Hyperplane of $V$.
Have no idea to start this.

By definition $\dim V=\infty$ means that there exists an infinite basis $\mathcal{B}=\{e_i\}_{i\in I}$ of $V$. As $I$ is an infinite set, if we choose a finite subset $J\subset I$ we still have $|I|=|I\setminus J|$ and thus there exists a bijection $\phi:I\to I\setminus J$. Now define $f:V\to V$ by the formula
$$
f\left(\sum_{k=1}^n\lambda_{i_k}e_{i_k}\right):=\sum_{k=1}^n\lambda_{i_k}e_{\phi(i_k)}.
$$
As every element $v\in V$ has a unique representation in terms of $\mathcal{B}$, this is a well defined function. Furthermore it is clearly linear, injective (since $\phi$ is a bijection), and with range
$$\text{Span}\{e_i\}_{i\in I\setminus J}\subsetneq V.$$
We have thus constructed a linear bijection from $V$ to a proper vector subspace of $V$ (the concept of hyperplane doesn't have so much sense in the infinite-dimensional case).
