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Q&A

Is this topology basis dependent?

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Consider a topological field $K$ and an algebraic(!) vector space $V$ over $K$, that is, $V$ has not (yet) a topology defined on it. I'm particularly interested in the case where $V$ has infinite dimension.

Now be $V^\ast$ the algebraic dual of $V$. Define a topology on $V^\ast$ through pointwise convergence, or equivalently, consider $V^\ast$ as subset under the subspace topology of $K^V$ under the product topology. This topology obviously makes $V^\ast$ a topological vector space over $K$.

Denote with $V^{[\ast]}$ the subspace of $V^\ast$ which consists of all $\phi\in V^\ast$ whose kernel has finite codimension, that is $V/\operatorname{ker}(\phi)$ has finite dimension. Obviously $V^{[\ast]}$ also is a topological vector space over $K$.

Now consider an arbitrary basis $\{b_i:i\in I\}$ of $V$ (where $I$ is an appropriate index set). Then one can define covectors $\{\omega_i:i\in I\}$ by $\omega_i(b_j)=\delta_{ij}$. Of course different bases ${b_i}$ lead to different covectors $\omega_i$.

However (if I made no error in my thoughts) in all cases $\{\omega_i:i\in I\}$ is a basis of $V^{[\ast]}$. Therefore the linear map $f:V\to V^{[\ast]}$ which maps $b_i$ to $\omega_i$ is a vector space isomorphism. Now we can define a set $U\subseteq V$ to be open if its image under $f$ is open. This topology then also makes $V$ a topological vector space over $K$.

My question now is:

Does this topology on $V$ depend on the choice of basis $\{b_i\}$?

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