A Can falling factorials be a Schauder basis for formal power series?

lugita15
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We usually talk about ##F[[x]]##, the set of formal power series with coefficients in ##F##, as a topological ring. But we can also view it as a topological vector space over ##F## where ##F## is endowed with the discrete topology. And viewed in this way, ##\{x^n:n\in\mathbb{N}\}## is a Schauder basis for ##F[[x]]##.

Now in contrast, ##\{(x)_n:n\in\mathbb{N}\}##, where ##(x)_n## denotes the falling factorial, is not a Schauder basis for ##F[[x]]##. That’s because if ##\Sigma_na_n(x)_n## never converges in the standard topology on ##F[[x]]## if infinitely many of the ##a_n##’s are nonzero. But my question is, does there exist some alternate topology on ##F[[x]]## which makes ##\{(x)_n:n\in\mathbb{N}\}## a Schauder basis for ##F[[x]]## as a topological vector space over ##F## endowed with the discrete topology?
 
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A schauder basis per the wikipedia article is typically defined based on a norm, which I think means you don't get to pick any particular topology on the field. Do you only care about topologies defined by a norm on the vector space, or would any topology that happens to permit you to compute unique limits suffice?
 
Office_Shredder said:
A schauder basis per the wikipedia article is typically defined based on a norm, which I think means you don't get to pick any particular topology on the field. Do you only care about topologies defined by a norm on the vector space, or would any topology that happens to permit you to compute unique limits suffice?
I’m not interested in normed vector spaces at all. As Wikipedia says “Schauder bases can also be defined analogously in a general topological vector space.”
 
It feels like the answer has to be no, but it's tough. For example can you just define a topology where the sequence ##\sum_{i =0}^n a_i(x)_i## converges to ##\sum_{0}^{\infty} a_i x^i##? I poked around a little bit and can't generate an obvious contradiction where you just assume the normal topology on the polynomials and then pick some open sets around the infinite series that make those limits true.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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