A proof about a proposition on Fréchet spaces Here is the proposition: assume that $V$ and $W$ are both Fréchet spaces, $u$ is a continuous map from $V$ to $W$. If we suppose that $u$ is injective, and $dim_C$$W/u(V)$ is finite, then $u(V)$ is a closed subspace of $W$.  
I thought out a proof, but did not use the condition that $dim_C$$W/u(V)$ is finite, so I hope anyone could help me to find out the reason. The following is my proof:  
The reference is Functional Analysis written by Walter Rudin. In this book, there is the Theorem $1.27$ on Page $21$, which states that Suppose that $Y$ is a subspace of a topological vector space $X$, and $Y$ is an Fréchet space (in the topology induced from $X$). Then $Y$ is a closed subspace of $X$.
So I tried to prove that the image of $u$ is also a Fréchet space in $W$, then I can use the Theorem $1.27$ to prove the proposition in the beginning.
We notice that there is a generalized open mapping theorem in the same book, it is the Theorem $2.11$ on Page $48$. It supposes that 


*

*$X$ is an Fréchet space.

*$Y$ is a topological vector space.

*$u$ from $X$ to $Y$ is continuous and linear.

*$u(X)$ is of the second category in $Y$.  


Then  


*

*$u(X)=Y$

*$u$ is an open mapping

*$Y$ is an Fréchet space.  


Here "be of the second category" is the same as in the definition of Baire Category.
Now if we replace $X$ by $V$, replace $Y$ by the image of $u$, and view $u$ as a 1-1 map from $V$ to $u(V)$, then according to the open mapping theorem, $u(V)$ is a Fréchet space, so it is closed in $W$.
As I am not familiar with functional analysis,I do not know where I am wrong. Please tell me, thank you very much! 
 A: 
Now if we replace $X$ by $V$, replace $Y$ by the image of $u$, and view $u$ as a 1-1 map from $V$ to $u(V)$, then according to the open mapping theorem, $u(V)$ is an Fréchet space, so it is closed in $W$.

We don't know yet that $u(V)$ is a Fréchet space in the topology induced by $W$. We know that it is a Fréchet space in the topology transported from $V$ via $u$, but that could theoretically be strictly finer than the topology induced by $W$.
However. If we can show that $u$ is an open mapping to $u(V)$ with its subspace topology induced by $W$, then we know that $u$ is a homeomorphism from $V$ to $u(V)$, and then we know that $u(V)$ is a Fréchet space, hence closed.
Let $F \subset W$ be an algebraic complement of $u(V)$. By assumption, $F$ is finite-dimensional, hence $F$ is a closed subspace. Now consider the space $X = W/F$. Since $F$ is a closed subspace of $W$, $X$ is a Fréchet space, and
$$V \xrightarrow{u} W \xrightarrow{\pi} X$$
is a continuous bijection between Fréchet spaces, hence a topological isomorphism, in particular open.
Now, for an open $U \subset V$, the set $u(U) = u(V) \cap \pi^{-1}((\pi\circ u)(U))$ is open in $u(V)$, since it is the intersection of $u(V)$ with the open (in $W$) set $\pi^{-1}((\pi\circ u)(U))$. So $u \colon V \to W$ is an embedding, hence $u(V)$ is a Fréchet space, hence closed.
