# Does this construction always give a topological vector space?

Consider an algebraic vector space $V$ over $\mathbb R$ or $\mathbb C$.

Now for each possible basis $B_k = \{e_i^{(k)}|i\in I\}$ of $V$, one can define an inner product $\langle\cdot\vert\cdot\rangle_k$ by $\langle e_i^{(k)}\vert e_j^{(k)}\rangle_k = \delta_{ij}$ (for vector spaces of infinite dimension, this might not cover all possible inner products).

Now each of those inner products of course defines a corresponding topology $\mathcal T_k$ in the usual way, and of course for each $\mathcal T_k$, we have that $(V,\mathcal T_k)$ is a topological vector space.

Now define $\mathcal T = \bigcap_{k} T_k$, that is, $\mathcal T$ contains all the sets that are open in all the topologies $\mathcal T_k$. It is easy to check that $\mathcal T$ again is a topology on $V$.

But since not every topology on $V$ turns $V$ into a topological vector space, my question is:

Is $(V,\mathcal T)$ also a topological vector space, and how can I see that it is or isn't?

Clearly for finite-dimensional vector spaces, all $\mathcal T_k$ are the same, thus $\mathcal T$ trivially makes $V$ a topological vector space. However I don't see how to answer the question for infinite-dimensional vector spaces.

## 1 answer

Let $F$ be either $R$ or $C$ and $U$ be open in $T$, that is, by definition of $T$, $U$ is open in $T_k$ for every $k$. Since, for every $k$, $(V,T_k)$ is a topological vector space, $+:V\times V\rightarrow V$ and $\cdot:F\times V\rightarrow V$ are continuous with respect to $T_k$. So, for every $k$, $+^{-1}(U)\in T_k$ and $\cdot^{-1}(U)\in T_k$, thus $+^{-1}(U)\in T$ and $\cdot^{-1}(U)\in T$. Therefore, $+$ and $\cdot$ are continuous with respect to $T$ which, in turn, implies that $(V,T)$ is a topological vector space.

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