Post History
#3: Post edited
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
- 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
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.