1

I am having really hard time in my Functional Analysis II class. I never had this problem with Functional Analysis I class, which mostly focused on metrics spaces, normed spaces, banach space and hilbert space. In this class, we are mostly focused on topological vector space. I just find this space so confusing and hard to understand. I have some questions which I think will help me understand what is going on

Suppose that $X$ is a topological vector space (T.V.S) over field $F$

  1. Does this mean that If $x,y\in X$, then $x+y\in X$ (linear space) and If $A,B\subseteq X$, then $A\cup B\subseteq X$ (topology)?
  2. Why is it mentioned in some textbooks that the topology over $F$ is euclidean topology? is this the only topology we can have?
  3. Does it mean that $F$ can only have euclidean topology while $X$ can have it is own separate and different topology?
  4. Does T.V.S have to be Hausdorff? If no, then why is Hausdorff used in most theorems in our lecture notes? What is different about a non-Hausdorff T.V.S?
  5. Does T.V.S imply that addition and multiplication operators are continuous or is it a necessary condition?
  6. Why is a linear space with a discrete topology is not a topological linear space?

Sorry if these questions look trivial, but they are just not clear to me.

gbd
  • 1,963
  • All these question are solved just by looking at a definition of topological vector space, except perhaps for Hausdorffness. So, do you know a definition of topological vector space? What are you studying on? – arnett Feb 27 '22 at 19:58
  • and 5. can be answered by looking at a definition indeed. For 2., "euclidean topology" only makes sense when $F$ is a field between $\mathbb{Q}$ and $\mathbb{C}$, but even then, there are other interesting topologies on those fields as well, so no, it's not the only one. 3. see 2. 4. Depends on convention. Some sources exclude non-Hausdorff TVSs, because the only such ones come with the indiscrete topology, which is not very interesting. 6. Your assertion is not true in general.
  • – Thorgott Feb 27 '22 at 20:01
  • The base field $F$ is either $\Bbb R$ or $\Bbb C$ and always the usual Euclidean topology is considered there. For 1. and the other dots, please cite here the definition of a topological vector space that is being used. (E.g. most of the times Hausdorffness is included in the definition, but it seems not in your class.)
  • – Berci Feb 27 '22 at 20:03
  • 5 is just part of the definition of what a TVS is. So necessary. If $V$ is a TVS over $F$ you know that scalar multiplication, addition and inverse ($x \to -x$) are continuous. – Henno Brandsma Feb 27 '22 at 22:21
  • 4 is for convenience. If we ask for $V$ to be even $T_0$ it will also be Hausdorff. And all TVS's in practice are. It makes proofs easier. – Henno Brandsma Feb 27 '22 at 22:24
  • The only requirement on the topology on $F$ is that we have a topological field. So all field operations are continuous. As in all of maths we give $\Bbb R$ and $\Bbb C$ their standard topologies, in which they are topological fields. – Henno Brandsma Feb 27 '22 at 22:27
  • There are very good reasons not to require Hausdorffness: Spaces $\mathscr L^p$ of $p$-integrable functions are very natural examples but much more importantly, the category of Hausdorff TVS is very different from the category of all TVS. For example, the epimorphims in the former are the continuous linear maps with dense range and only in the latter they are surjective. This is crucial for many applications (i.e., solvability of a pde) where functional analysis is only a tool to solve "algebraic" problems. – Jochen Feb 28 '22 at 06:51