How to Prove that a finite-dimensional space can not be isomorphic to an infinite-dimensional one? I know, I know that all people will say that because of their differences and that's why it's not possible, but if someone could help me with an example or with the prove, I would really appreciate it.
The differences are:
(1). There are endomorphisms $T$ with $\ker(T)=\{0\}$ which are not surjective.
(2). Not in every case a linear form $\phi$ is representable by a vector $v$ in presence of a scalar product, i.e., there doesn't exist a vector $v$ that $\phi(.)=\langle v,.\rangle$.
(3). Not all linear mappings are continuos.
(4). You can equip a vector space with two different norms such that the unit ball in respect to the first norm in unbounded in respect to the second.
And sorry for my English I'm not very good.
 A: Here is a way to prove it.
Proof: Let $V$ and $W$ be infinite dimensional and finite dimensional vectors spaces respectively. Let $\dim W=n$. Suppose that $V$ is isomorphic to $W$. Then there exists a linear map $T : V\to W$ which is injective and surjective. Let $v_1,\dots,v_{n+1}$ be a linearly independent list in $V$. Because an injective linear map  takes a linearly independent list to a linearly independent list, we should have that the list $Tv_1,\dots,Tv_{n+1}$ is linearly independent. But this contradicts the theorem that any list of length $n+1$ cannot be linearly independent in an $n$ dimensional space. Thus, $T$ is not injective and $V$ is not isomorphic to $W$.
Edit: Additionally it also shows that no linear map from an infinite dimensional vector space to a finite dimensional vector space is injective.
A: That's because, if $T:U\to V$ is an isomorphism, and $B$ a basis of $U$, then $T(B)$ is a basis of $V$. In particular, two isomorphic vector spaces must have the same dimension.
