Topology from seminorms

Question

Let $X$ be a vector space, and $\{p_i\}_{i\in I}$ be seminorms on $X$.

The usual definition of the induced topology to say that it is the coarsest topology for which the maps $p_i$ are continuous. (example : here, definition 4.2.8)

My question is : is there an implicit that the considered topologies are linear (i.e. compatible with the vector space structure of $X$) ?

In other words, the definition should be :

The coarsest topology among the linear topologies for which the maps $p_i$ are continuous.

The only place where I saw it explicitely stated is here, item 7.

EDIT. Clarification.

When one says "the coarsest topology on $X$ such that the $p_i$ are continuous", does one mean :

(1). take all the topologies on the set $X$ and select the coarsest which makes the $p_i$ continuous.

(2). take all the topological vector space structures on $X$ and select the coarsest which makes the $p_i$ continuous.

If (1) is the correct meaning, we should then prove that this topology happens to be compatible with the vector space structure.

Answer 1

Let $X$ be a vector space over $K = \mathbb{R,C}$ with a seminorm $p$. Let $\tau$ be the initial topology on $X$ w.r.t. $p$. Then $(X,\tau)$ is in general NOT a topological vector space, since addition is not continuous or more precisely $\tau$ is not translation-invariant.

An easy example would be $\mathbb{R}$ and $p = |\cdot|$. Then a neighbourhood basis of $x_0 \in \mathbb{R}$ w.r.t. the initial topology of $p$ is $\{x \in \mathbb{R} \ | \ ||x|-|x_0|| < \varepsilon, \ \varepsilon > 0\}$. Therefore the constant sequence $1$ converges to $-1$ and if $+$ would be continuous, $1+(-1) = 0 \to 0 \ (n \to \infty)$ but also $\to -1-1 = -2$ but $-2 \neq 0$ in $\mathbb{R}$, a contradiction. Thus $+$ is NOT continuous w.r.t. the initial topology.

So, to repair this, one must take the initial topology w.r.t. $p\circ\tau_{x_0}$ for all $x_0 \in X$, where $\tau_{x_0}(x) = x + x_0, \ x \in X$. Then $\tau$ is translation-invariant.

Then addition $+: X \times X \to X$ is continuous, since $\tau$ is translation-invariant and $+$ is continuous at $(0,0)$ because $p^{-1}((-1/2n,1/2n))+p^{-1}((-1/2n,1/2n)) \subseteq p^{-1}((-1/n,1/n))$ for all $n \in \mathbb{N}$. Multiplication $m: K \times X \to X$ is continuous since $p\circ m = \mu \circ (|\cdot|\times p)$, where $\mu: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is just multiplication.

See also for example Schaefer, Helmut H.; Wolff, Manfred P. Topological Vector Spaces. Chapter 1 for more such things.