I am reading functional analysis from an applied math textbook and I have to admit some of the exercises confuse me, in the sense I am not sure what is asked of me.
"Show that each of the seminorms inducing a locally convex topology is continuous with respect to this topology"
I tried a naive approach (seminorm p is after all a function from a vector space to R) so $\forall \epsilon >0$, I picked $\delta = \epsilon$ and went ahead with the usual definition of continuity to show that for every $x,y\in V$ $|p(x)-p(y)|< \epsilon$. But I was told this is not correct - I understand that this way I am not using at all that these seminorms are the ones producing a locally convex v.s, but in my mind all seminorms are continuous (maybe I am missing the "with respect to this topology part").
The next approach I tried (which I am not confident of) is by assuming that the sets in my textbook associated with the norms that induce a locally convex space are $B(I_o,\epsilon) = \{v\in V :p_i(u)\leq \epsilon , \forall i\in I\}$. These are neighborhoods of zero and addition and multiplication are continuous operations because of the seminorm homogeneity and triangle inequality properties.
I also tried to use that $f$ is continuous at $x$ iff $f(B_x)\succ B_{f(x)}$ but I failed.
Could you help me realize what I need to do and how I do it to answer this question?
Let $x_0\in X$ and $\epsilon> 0$ be given. Then set $U := \{x\in X : p(x-x_0) < \epsilon\}$. As a subbasis element of the locally convex topology, this set is open in $X$ and for all $x\in U$ you have $|p(x)-p(x_0)|\le p(x-x_0) < \epsilon$.