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0 answers  ·  posted 12mo ago by celtschk‭  ·  edited 12mo ago by r~~‭

#3: Post edited by r~~‭ · 2021-10-10T19:54:07Z (12 months ago)
Typo
• Does this constructtion always give a topological vector space?
• 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.

#2: Post edited by celtschk‭ · 2021-10-10T11:58:09Z (12 months ago)
• 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 $\mathcal T_k$.
• Now 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.
• 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: Initial revision by celtschk‭ · 2021-10-10T08:42:10Z (12 months ago)
Does this constructtion 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 $\mathcal T_k$.

Now 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.