In these lecture notes on page 71(Example 5.29 a) it is claimed that for a LCS X the weak topology is coarser than the topology of the LCS, but afaik this does not even hold for normed spaces or am I wrong?
I mean it would imply that any weakly open neighborhood is open in the LCS topology or am I wrong?
But any weakly open neighborhood of zero contains a whole non-trivial subspace and this is definitely not the case, if $X$ is a normed space, so there are open sets that are not open in the norm topology! Or where is my argument wrong?
Best Answer
Let $\tau_1$ and $\tau_2$ be two topologies on a given set $X$. We say that $\tau_1$ is coarser than $\tau_2$ if $\tau_1\subseteq \tau_2$. This means that if $U$ is open with respect to $\tau_1$ then it is open with respect to $\tau_2$. The converse need not be true.
Now, given a locally convex space $X$, $\tau_1$ being the weak topology is coarser than $\tau_2$, the original topology of $X$.
Note that the inclusion in general need not be strict. Indeed, the weak topology of a weak topology of a Banach space is the same as the weak topology of a Banach space itself.