# Topology of the Hamel basis in a TVS

Let $V$ be a complex topological vector space, and let $I$ be a Hamel basis of it. Then as a subset $I\subset V$ acquires an induced topology, becoming a topological space. For a topological space $X$ consider the space $C_\#(X,\mathbb{C})$ of finitely supported complex functions on $X$. Then $C_\#(I,\mathbb{C})$ is algebraically isomorphic to $V$ by definition of a Hamel basis, $$\forall v\in V,\quad v=\sum_{e\in I}\lambda_e^v e,\quad I\ni e\mapsto\lambda_e^v\in\mathbb{C}.$$

Question: Does there exist a topology on $C_\#(X,\mathbb{C})$, such that $C_\#(I,\mathbb{C})$ is isomorphic to $V$ as topological vector spaces?

In finite dimensions $I$ is finite and hence discrete, so that all reasonable topologies on $C_\#(I,\mathbb{C})=\mathbb{C}^I$ coincide and give a positive answer. The question therefore pertains to infinite dimensions.

I am mostly interested in separable Hausdorff $V$. Thank you.

• So if I understand correctly, you are looking for a general construction which, given a topological space $X$, outputs a TVS topology $\tau_X$ on $C_\#(X, \mathbb{C})$. And it should have the property that whenever $V$ is a (separable, Hausdorff) TVS and $I \subset V$ is a Hamel basis with the subspace topology, then $V \cong (C_\#(I, \mathbb{C}), \tau_I)$ as TVSes (i.e. there is a linear homeomorphism between them). Have I got that right? – Nate Eldredge Jun 2 '17 at 14:16
• So your claim would be disproved if we could find a pair of TVSes $V_1, V_2$, not isomorphic as TVSes, with respective Hamel bases $I_1, I_2$ which are homeomorphic in their subspace topologies. Right? – Nate Eldredge Jun 2 '17 at 14:17
• Exactly. Both comments are correct. – Bedovlat Jun 2 '17 at 16:18
• Then just take a linearly independent Cantor set $C$ in $\ell_2$ whose linear hull $lin(C)$ is dense in $\ell_2$ and observe that $lin(C)$ with the subspace topology inherited from $\ell_2$ is not topologically isomorphic to $lin(C)$ endowed with the topology inherited from the Banach space $c_0$. If these two TVS would be isomorphic, then their completions (i.e., $\ell_2$ and $c_0$) would be isomorphic too, but this is not the case. – Taras Banakh Jun 6 '17 at 15:11
• But are the two induced topologies on $C$ equivalent? – Bedovlat Jun 6 '17 at 16:57