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A topological vector space $X$ is defined a vector space over a topological field $\mathbb{F}$ that is endowed with a topology such that vector addition $X\times X \rightarrow X$ and scalar multiplication $\mathbb{F}\times X \rightarrow X$ are continuous functions (where the domains of these functions are endowed with product topology).

I am a bit confused on how exactly the topology is defined/constructed, so how is the basis or subbasis defined, what is the starting point given a vector space $X$? So just to draw a parallel, if $(Y_i,\mathscr{T}_i)$ are topological spaces, then we define the topology on $\prod\limits_{i\in \mathcal{I}}Y_i$, to be the topology generated by the subbasis $\mathscr{B}:=\{\pi_i^{-1}(U_i)|i\in\mathcal{I},\space\space U_i\in\mathscr{T}_i\}$, where $\pi_i$ is the $i^{th}$ projection. So how is it done for a tvs?

Or is there already a topology defined on $X$, and we just augment the topology to be coarsest topology that makes the vector operations continuous? But then how will this work as if the topology is given, the product topology is then already defined so if we augment the topology this will affect the product topology.

Thanks in advance I really will appreciate any feedback to clear up my confusion.

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The topology on a topological vector space is not something you "construct", it is something given as part of the data. That is, a topological field $(V,\tau_V)$ over a topological field $(\mathbb F,\tau_{\mathbb F})$ is a vector space $V$ over $\mathbb F$, equipped with a topology $\tau_V$, such that scalar multiplication $\mathbb F \times V \to V$ is continuous (using the product topology on $\mathbb F \times V$) and addition $V \times V \to V$ is continuous (using the product topology on $V \times V$).

In fact, for a given vector space $V$ over a topological field $\mathbb F$ there might be multiple topologies on $V$ that turn it into a topological vector space.

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  • $\begingroup$ For the last statement: Consider the norm topology vs the weak topology on some normed space. In both cases it is a topological vector space but in the infinite dimensional case, the topologies differ. $\endgroup$ – Tim B. Jan 18 '17 at 12:45
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    $\begingroup$ Or, for a trivial example: if $\mathbb F$ is $\mathbb R$ with its usual topology, and $V = \mathbb R^n$, then $V$ becomes a topological vector space both with the indiscrete topology and with the Euclidean topology. $\endgroup$ – Mees de Vries Jan 18 '17 at 12:52

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