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Let $E$ be a vector space.

Recall that the inductive topology in $E$ taken w.r.t to the family of spaces $\{ E_{\alpha} \}$ and mappings $g_{\alpha}: E_{\alpha} \rightarrow E$ is the finest topology such that all the mappings $g_{\alpha}$ are continuous.

Let's go a bit deeper and consider the finest locally convex topology in $E$, i.e. the inductive topology taken w.r.t to the empty family of spaces $\{ E_{\alpha} \}$ and empty family of mappings $\{ g_{\alpha} \}$.

I would like to prove that the described topology is metrizable if and only if $E$ is a finite-dimensional vector space.

The very first idea is to apply the common metrization critirion as follows: we need to check that $E$ is a regular, Hausdorff space that admits a countable basis of the neighbourhoods of $0$.

Is it possible to apply the criterion directly? One can note that the described topology is generated, for example, by the basis, consisting of all balanced absorbing sets (which is uncountable in the case of infinite dimension?) and the topology can be defined by the family of all seminorms.

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  • $\begingroup$ It would be helpful if the people who voted to close this question as "unclear what you're asking" explained what they found unclear. And then the OP can try to improve the question... $\endgroup$ – user1729 Jan 24 '18 at 14:20
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In a nutshell: As a (for simplicity real) vector space, $E$ is isomorphic to $\mathbb R^{(I)}=\{x:I\to \mathbb R: \{i\in I:x_i\neq 0\}$ finite$\}$ and a subset $U$ is a neighbourhood of $0$ (for the finest locally convex topology) if and only if there is a function $f:I\to (0,\infty)$ such that $U$ contains $U_f=\{x\in \mathbb R^{(I)}: |x_i|< f(i)$ for all $i\in I\}$. If $I$ is infinite, given countably many functions $f_n:I\to (0,\infty)$ you can construct $f:I\to (0,\infty)$ such that for all $n\in\mathbb N$ there is $i\in I$ with $f(i)<f_n(i)$ (if $i_n$ are distint elements of $I$ define $f(i_n)=\min\{f_1(i_n)/2,\ldots f_n(i_n)/2\}$ and $f(i)=1$ for all other $i$). Then $U_f$ does not contain any $U_{f_n}$. This implies that the $0$-neighborhood filter does not have a countable basis.

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  • $\begingroup$ Thank you! As for me, it was difficult to design a similar argument by myself, though i find it quite easy to assimiler $\endgroup$ – hyperkahler Jan 26 '18 at 22:20

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