TITLE: A name for a weak topology
QUESTION [5 upvotes]: Let $V$ be a real vector space and let $V'$ be the algebraic dual of $V$, i.e. the space of all the linear functionals $V\to\mathbb{R}$. Then there exists the weakest topology $\tau$ which makes all the elements of $V'$ continuous, and $\tau$ is locally convex and Hausdorff. For example, if the dimension on $V$ is finite, then $\tau$ is the usual Euclidean topology. 
I have two questions:

Is there a commonly used name for $\tau$? 
Let $\tau'$ be the maximal locally convex topology on $V$, i.e. the weakest topology that makes all the seminorms on $V$ continuous. Of course $\tau'$ is finer than $\tau$, but $\tau'$ and $\tau$ could coincide in some cases (for example, for finite dimensional spaces). What is the exact relationship between $\tau$ and $\tau'$?

REPLY [6 votes]: I got into trouble in following prof. Johnson's suggestion, and now I am quite convinced that the topologies $\tau$ and $\tau'$ coincide only when $V$ is finite-dimensional. 
In fact, let $\{x_i\}_{i\in I}$ be a Hamel basis for $V$, and let us consider the seminorm
$p(\sum_{i\in I} v_i x_i)=\sum_{i\in I} |v_i|$
(this seminorm is well-defined since every element in $V$ admits a unique representation as a finite linear combination of elements of the basis). Then the set $U=p^{-1}(-1,1)$ is open in $\tau'$, and does not contain any nontrivial linear subspace of $V$.
On the other hand, since $\tau$ is the weak topology with respect to a family of linear functionals (in fact, with respect to all the linear functionals), every $\tau$-neighbourhood of $0\in V$ must contain a finite-codimensional linear subspace of $V$. Therefore, if the dimension of $V$ is not finite, then the set $U$ introduced above is open with respect to $\tau'$, and not open with respect to $\tau$.