The finest locally convex topology is not metrizable Let $E$ be a vector space.
Recall that the inductive topology in $E$ taken w.r.t to the family of spaces $\{ E_{\alpha} \}$ and mappings $g_{\alpha}: E_{\alpha} \rightarrow E$ is the finest topology such that all the mappings $g_{\alpha}$ are continuous.
Let's go a bit deeper and consider the finest locally convex topology in $E$, i.e. the inductive topology taken w.r.t to the empty family of spaces $\{ E_{\alpha} \}$ and empty family of mappings $\{ g_{\alpha} \}$.
I would like to prove that the described topology is metrizable if and only if $E$ is a finite-dimensional vector space. 
The very first idea is to apply the common metrization critirion as follows: we need to check that $E$ is a regular, Hausdorff space that admits a countable basis of the neighbourhoods of $0$.
Is it possible to apply the criterion directly? 
One can note that the described topology is generated, for example, by the basis, consisting of all balanced absorbing sets (which is uncountable in the case of infinite dimension?) and the topology can be defined by the family of all seminorms.
 A: In a nutshell: As a (for simplicity real) vector space, $E$ is isomorphic to $\mathbb R^{(I)}=\{x:I\to \mathbb R: \{i\in I:x_i\neq 0\}$ finite$\}$ and a subset $U$ is a neighbourhood of $0$ (for the finest locally convex topology) if and only if there is a function $f:I\to (0,\infty)$ such that $U$ contains $U_f=\{x\in \mathbb R^{(I)}: |x_i|< f(i)$ for all $i\in I\}$. If $I$ is infinite, given countably many functions $f_n:I\to (0,\infty)$ you can construct $f:I\to (0,\infty)$ such that for all $n\in\mathbb N$ there is $i\in I$ with $f(i)<f_n(i)$ (if $i_n$ are distint elements of $I$ define $f(i_n)=\min\{f_1(i_n)/2,\ldots f_n(i_n)/2\}$ and $f(i)=1$ for all other $i$). Then $U_f$ does not contain any $U_{f_n}$. This implies that the $0$-neighborhood filter does not have a countable basis.
A: To prove this in a different way, recall that a subset $M \subseteq E$ is bounded with respect to the finest locally convex topology if and only if $M$ is contained in a finite-dimensional subspace and bounded there.
Alternative proof. Assume that $E$ is infinite-dimensional, and let $\{x_n\}_{n=1}^\infty$ be a linearly independent sequence in $E$. Then for every sequence $\{\alpha_n\}_{n=1}^\infty$ of non-zero scalars, the sequence $\{\alpha_n x_n\}_{n=1}^\infty$ is unbounded, as it is not contained in a finite-dimensional subspace. As a consequence, it follows from remark 2 in this answer on MathOverflow that $E$ cannot be metrizable.
