Convergence in Fréchet spaces and the topology actually i have two questions concerning Fréchet spaces (i am not familiar with these spaces so i need some help). I have a vector space $V$ with a distance $$d(x,y) = \sum_{n\in \mathbb{N}} \frac{|x-y|_k}{2^k(1+|x-y|_k)} $$ where $|.|_k$ is a separating family of semi-norms.
My questions are the following :
1/ I have noticed that a sequence $(x_n)$ converges to a $x$ for $d$ if and only if for all $M\in \mathbb{N}$ $$\lim_{n\to \infty} \max_{k\leq M}|x_n-x|_k =0$$ However i did not find this result in textbook (probably because it is too easy or because it is wrong). Could someone tell me if it really holds of if i made a misstake ?
2/ The topology induced by the familly of semi-norms is the set of subsets of $V$ which are unions of sets of the form  $$\cap_{j\in J}\{ y\in V : | |y|_k - a_j | <\epsilon_j \}$$ where $J\subset \mathbb{N}$ is finite. Is there a very short proof that this topology is the same as the one induced by the metric above ?
 A: 1/ It seems the following. 
Yes, both the necessity and the sufficiency can be easily checked. Let $\{x_n\}$ be a sequence of vectors of the space $V$.
Necessity. Assume that the sequence $\{x_n\}$ converges to a vector $x\in V$, $k\in\Bbb N$ be an arbitrary index, and $\varepsilon>0$ be arbitrary. Choose a number $\delta>0$ such that $\frac{2^k\delta}{1-2^k\delta}<\varepsilon$. Since the sequence $\{x_n\}$ converges to a vector $x$, there exists a number $N$ such that $d(x_n,x)<\delta $ for each index $n>N$. Let $n>N$ be an arbitrary index. Then 
$$\frac{|x_n-x|_k}{2^k(1+|x_n-x|_k)}\le d(x_n,x)<\delta,$$ 
so $$|x_n-x|_k<\frac{2^k\delta}{1-2^k\delta}<\varepsilon. $$ 
Sufficiency. Assume that there exists a vector $x\in V$ such that for each index $k\in\Bbb N$ a sequence $\{|x_n-x|_k\}$ converges to the zero. Let $\varepsilon>0$ be an arbitrary number. Choose a natural number $M$ such that $\frac 1{2^M}<\frac{\varepsilon}{2}$. Since for each index $k\le M$ the sequence $\{|x_n-x|_k\}$ converges to the zero, there exists a number $N$ such that $|x_n-x|_k<\varepsilon/2$ for each index $k\le M$ and each index $n>N$. Let $n>N$ be an arbitrary index. Then 
$$d(x_n,x) = \sum_{k\in \mathbb{N}} \frac{|x_n-x|_k}{2^k(1+|x_n-x|_k)} =$$
$$\sum_{k=1}^M \frac{|x_n-x|_k}{2^k(1+|x_n-x|_k)} +
\sum_{k=M+1}^\infty \frac{|x_n-x|_k}{2^k(1+|x_n-x|_k)}<$$
$$\sum_{k=1}^M \frac{\varepsilon}{2^{k+1}} +
\sum_{k=M+1}^\infty \frac{1}{2^k}<$$
$$\frac{\varepsilon}2+\frac 1{2^M}<\frac{\varepsilon}2+\frac{\varepsilon}2<\varepsilon.$$
