Prove two topologies are the same in a vector space I'm trying to solve a problem that states the following:
Let $E$ be an $n$-dimensional real vector space. Prove that:


*

*Any basis $\{e_i\}_{i=1}^n$ of the real vector space $E$ determines a topology which transforms the map $\phi:E\longrightarrow\mathbb{R}^n$, defined as $\phi(v)=(c_1,c_2,...,c_n)$ 
where $v=\sum_{i=1}^nc_ie_i$, into a homeomorphism.

*Such a topology is independent of the choice of the basis. Then we say that $E$ is a real topological vector space.


I have solved the first section of the problem, but I'm desperately stuck in the second one. I know by intuition that I should use the fact that the following diagram is conmutative and $L$ (the application associated to the change of basis) is linear, but I am unable to find a formal proof. 
Here is the conmutative diagram I'm talking about
How can I prove the double inclusion between the topologies associated to two different basis $B$ and $B'$?I've been trying for quite a while now, can anyone please help?
 A: I suppose that this is what the commutative diagram is intended to express, but I am not sure about the notation. The point is that if $\phi$ and $\psi$ are the homeomorphisms associated to two different choices of bases, there is an invertible $n\times n$-matrix $A$ such that $\psi(v)=A\phi(v)$, and left multiplication $m_A$ by the invertible matrix $A$ defines a homeomorphism $\mathbb R^n\to \mathbb R^n$ (with inverse given by $m_{A^{-1}}$). But this exactly says that $id_E$ can be written as $\psi^{-1}\circ m_A\circ \phi$ so it is a homeomorphism from the topology on $E$ defined by $\phi$ to the topology on $E$ defined by $\psi$. 
A: Let $A=\{e_1,...,e_n\}$ and $B=\{f_1,...,f_n\}$ be bases for $E.$ For $x\in E$ there are (unique) reals $x_{A,1}... x_{A,n}$ and $x_{B,1}...x_{B,n}$ such that $x=\sum_{i=1}^nx_{A,i}e_i=\sum_{j=1}^nx_{B,j}f_j.$
In particular for each $f_j\in B$ we have $f_j=\sum_{i=1}^n (f_j)_{A,i}.$ Note that  for each $j,$ not all of $(f_j)_{A,i}$are $0$ (for $i=1,..., n$).
Define $\|x\|_A=\max_j|x_{A,j}|$ and $\|x\|_B=\max_j|x_{B,j}|.$ The following is very useful:  
(1).....There exist positive $M_A$ and $M_B$ such that for all $x\in E$ we have $$\|x\|_B\leq M_A\|x\|_A \;\text { and }\; \|x\|_A\leq M_B\|x\|_B.$$ Proof: For $x\in E$ and $1\leq i\leq n$ we have $$|x_{A,i}|=|\sum_{j=1}^n(f_j)_{A,i}x_{B,j}|\leq \sum_{j=1}^n |(f_j)_{A,i}|\cdot |x_{B,j}|\leq$$ $$\leq (\sum_{j=1}^n|(f_j)_{A,i}|)\|x\|_B.$$ Let $M_B=\max \{\sum_{j=1}^n|(f_j)_{A,i} : 1\leq i \leq n\}.$ Then $M_B>0$ (see "Note" in second paragraph above) and $|x_{A,i}|\leq M_B\|x\|_B$ for each $i$ from $1$ to $n$.
So by def'n of $\|x\|_A$ we have $\|x\|_A\leq M_B\|x\|_B.$ 
By interchanging $A,B$ in the above argument (with appropriate notational changes) we obtain $M_A>0$ and $\|x\|_B\leq M_A\|x\|_A.$
(2)....Let $T_A$ and $T_B$ be  topologies on   $E.$  Let $C_A$ be a base for $T_A$ and let $C_B$ be a base for $T_B.$ In order that $T_A=T_B$ it suffices that whenever $x\in E$ and whenever $\{i,j\}=\{A,B\}$ and $x\in t_i\in C_i,$ there exists $t_j\in C_j$ with $x\in T_j\subset T_i.$
(3).... Now let $T_A$ and $T_B$ be the topologies on $E$ as defined in your Q . A base for $T_A$ is $\{D_A(x,r): x\in E\land r>0\}$ where $$D_A(x,r)=\{y\in E: |y_{A,i}-x_{A,i}|<r \text { for } i=1,...,n\}=\{y\in E: \|y-x\|_A<r\}.$$  And a base for $T_B$ is $\{D_B(x,s): x\in E\land s>0\}$ where $$D_B(x,s)=\{y\in E:\|y-x\|_B<s\}.$$  Now using (1) show that the sufficient conditions of (2) are met.  
