proof that finite dimensional vector spaces are manifold here i saw some similar question and answers for my question but didn't really satisfied me .the question is
I want the proof of this proposition : every finite dimensional vector space is a manifold.
clearly every finite dimensional vector space has a countable basis.but i cant understand why is it Hausdorff and locally euclidean .i think the finite vector space induced a topology which makes vector space locally euclidean but do not know how to build this topology and didn't know how to proof it is Hausdorff. please someone explain it completely. thank you
 A: Hopefully this will help to answer your question.
A vector space :
Let $V$ be a non empty set on which two operations, addition $\oplus$ and scalar multiplication $\odot$, are defined.
If the following axioms are satisfied for all ${\bf u, v, w} \in V$ , and all scalars $k, l \in \mathbb{R}$, then $V$ is called a vector space, and the elements of $V$ are called vectors.


*

*${\bf u \oplus v} \in V$

*${\bf u \oplus v = v \oplus u}$

*${\bf u \oplus (v \oplus w) = (u \oplus v) \oplus w}$

*There is an element ${\bf 0} \in V$ , called the zero vector for $V$ , such that $\bf 0 \oplus u = u \oplus 0 = u$.

*For each $u \in V$ there is an element $-{\bf u }\in V$ , called the negative of $\bf u$, such that ${\bf u} \oplus −{\bf u = 0}$

*$k\odot {\bf u} \in V$

*$k {\bf \odot(u \oplus v)} = k  \odot {\bf u} \oplus k  \odot {\bf v}$

*$(k \oplus l) \odot {\bf u} = k \odot {\bf u} \oplus l \odot {\bf u}$

*$k  \odot (m  \odot {\bf u}) = (km) \odot {\bf u}$

*$1 \odot {\bf u = u}$


MacEwan University describe this with example theorem ... and you surely know this.
The standard topology is the set of open ball $B_r$. Let $p,q \in U\subset M$ be points on the manifold where $p=(p_1,p_2\dots,p_n)$ and $q=(q_1,q_2\dots,q_n)$. Let $d(p,q)$ be the distance between these point. 
$$ B_r(p) := \{(q_1,q_2\dots,q_n)| d(p,q)<r\}$$
From the standard topology it is straight forward to see that vector space are topological space.
Now to show it is a topological manifold you have to show that for all point $p$ there exist an open set $U$ containing $p$ such that $x(U)\subset \mathbb{R}^n$ where the coordinate map $x$ and $x^{-1}$ of a chart $(U,x)$ are continuous. 
You already have it by the definition of the standard topology:
$$x^i: U\to\mathbb{R}$$
$$x: p\mapsto (p_1,p_2\dots,p_n)\in \mathbb{R}^n$$
Where $p_1,p_2\dots,p_n$ become the component of the vector.
$$x^{-1}: \mathbb{R}^n\to U$$
$$x^{-1}: (p_1,p_2\dots,p_n) \mapsto p \in U$$
From the scalars multiplication on the vector space one can always construct a subset $U\subset M$ and $S\subset M$ such that for tow different element, $p,q$, of the vector space $U \cap S = \varnothing$. Just construct the set $U=B_r(p)$ and $S=B_r(q)$ where $r=d(p,q)/2$. proving that the vector space is a Hausdorff topological manifold. 
For the locally euclidean this will depend on the metrics $d(p,q)$ you chose. The fact that all chart must invertible there exist is a local chart transition $y\circ x^{-1}$ where $y$ is the euclidean one.
