Suppose $X$ and $Y$ are topological vector spaces, $dim(Y) < \infty$, $\Lambda : X \rightarrow Y$ is linear, and $\Lambda (X) = Y$. Prove that $\Lambda$ is an open mapping.
Thanks in advance.
Edit:
Definition of TVS: Here, a Topological Vector Space is a vector space such that vector addition and scalar multiplication are continuous and also singletons are closed.
I guess I should explicit my comment here since it is a bit big. I insist that there are details missing in the question, as the current statement is false. (The question you referred to has an answer concerning normed vector spaces. We are talking about topological vector spaces here, this is different.) Think for instance of a field $K$ and the map $K_1 \to K_2$ where both $K_i$ are $K$ as vector spaces, $K_1$ is a topological vector space with the discrete topology and $K_2$ with the trivial topology (the axioms of topological vector spaces are trivially checked since the product of two discrete/trivial topologies are discrete/trivial). This is not an open map.
This is what I had in mind when I said "pick two topologies where one is finer than the other". Since I don't need them to be induced by norms, they can be quite horrible. But this can also happen in practice (of functional analysis) when using semi-norms.
EDIT : Now that you say that the exercise is in Rudin, I started worrying about the details.
Here they are. Put the discrete topology on $K$. Let $K_1 \overset{def}= K^n$ with the discrete topology, and let $K_2 \overset{def}= K^n$ with the trivial topology. The product topologies on $K \times K_1$ and $K_1 \times K_1$ are both discrete, so any map with them as the domain is continuous ; it follows that addition and scalar multiplication on $K_1$ is continuous, making it into a topological vector space.
On $K_2$, since the topology is trivial, any map with $K_2$ as a codomain is continuous. Therefore addition $+ : K_2 \times K_2 \to K_2$ and scalar multiplication $K \times K_2 \to K_2$ are both continuous.
The identity map is definitely linear and of finite rank (it is an isomorphism of finite-dimensional vector spaces.
If you don't like this generic example, here is the argument if you take semi-norms. For example, turn $\mathbb R^n$ into an $\mathbb R$-topological vector space with its usual Euclidean norm and let $\mathbb R^n_0$ be $\mathbb R^n$ with the semi-norm $\|x\| = 0$ for all $x \in \mathbb R^n$ (this gives $\mathbb R^n$ the trivial topology since any open ball centered at $p \in \mathbb R^n$ will consist of all $\mathbb R^n$). The identity map $\mathbb R^n \to \mathbb R^n_0$ is linear, but not open (since $\mathbb R^n_0$ has only two open sets).
EDIT 2 : The discussion with OP in the comments clarified that the topological spaces we deal with should be $T_1$ ; it is a general theorem in topological groups that they are $T_0$ if and only if they are $T_1$ if and only if they are $T_2$ ; so we might as well assume they are Hausdorff. With this assumption, the result is true.
Any finite-dimensional Hausdorff topological vector space $V$ over the topological field $K$ is isomorphic (i.e. an isomorphism which is also an homeomorphism) to $K^n$ where $n = \dim_K V$ (exercise ; I don't remember how to do this and I don't feel like digging in details).
Therefore, the map $\Lambda : X \to Y$ factors through a linear projection map $\pi : X \to X/\ker \Lambda$ followed by $\Lambda' : X/\ker \Lambda \to Y$. It is a general result of topological groups that $\pi$ is an open map ; this is because $X/\ker \Lambda$ is equipped with the quotient topology and $U \subseteq X/\ker \Lambda$ is open if and only if $\pi^{-1}(U)$ is open in $X$ ; but for $V \subseteq X$, $$ \pi^{-1}(\pi(V)) = \bigcup_{v \in \ker \Lambda} v + V, $$ which is continuous since translation by $v$ is an homeomorphism in $X$. Therefore it suffices to show that $\Lambda' : X/\ker \Lambda \to Y$ is an open map ; it is an isomorphism (because we assumed it surjective and we factored out its kernel), so this is trivial since both spaces are Hausdorff (all there is to show here is that linear isomorphisms of $K^n$ are also homeomorphisms).
Hope that helps,
