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I'm teaching myself a course on functional analysis but having trouble understanding the notes I've been using. I was hoping I could write out a section of the content and you might be able to help me deduce how these conclusions are drawn since I can't; I have a few areas of confusion, I hope that's ok. The content from the notes is italicized below.

With regards to topological vector spaces: 'Metrizable' means the topology is given by a metric. We seldom use this directly; instead, we use the fact that there is a decreasing base $(N_j)_{j=1}^{\infty}$ of closed, absolutely convex neighbourhoods of $0$. If $p_j$ is the gauge (Minkowski functional) of $N_j$ then the topology is determined by the increasing sequence $(p_j)_{j=1}^{\infty}$ of semi-norms. We could then take for example the metric $\rho(x,y) = \sum_{j=1}^{\infty} \frac{p_j(x-y)}{2^j(p_j(x-y)+1)}$.

Note that $x_n \to x$ as $n \to \infty $ iff $p_j(x_n-x) \to 0 $ as $n \to \infty $, for each $j \in \mathbb{N} $. Here are 2 easy examples: first, let $ \omega$ be the space of all complex sequences. Set $p_j(x) = j \sum_{i=1}^j |x_i| $; then the topology on $\omega$ is simply the product topology. Secondly, let $C(\mathbb{R}^d)$ be the space of all continuous functions on $\mathbb{R}^d$. Set $p_j(f) = j\, \sup_{|x| \leq j} |f(x)| $. The resulting topology is the topology of local uniform convergence.

So, my main confusion is how we conclude in the 2 examples that the topology is precisely the product/local uniform convergence topology? Are we defining seminorms $p_j$ (with the factor of $j$ at the start to ensure they're increasing), and then saying 'these must be the Minkowski functionals for some decreasing base of neighbourhoods of $0$', then using the fact that convergence under the suggested metric occurs only if convergence occurs for each $p_j$ to see what a limit is under in this topology to deduce what the whole topology is? So, say in the fist case we deduce that for convergence to $0$, every finite sum $ \sum_{i=1}^j |x^{(n)}_i| $ of terms of the sequence $(x^{(n)})_{n=1}^\infty$ tends to $0$, then perhaps there's some way I can't see that this type of limit behaviour implies the product topology.

For another thing, while it's obvious that given some set we may induce a Minkowski functional by its very definition, it's not obvious to me that conversely by defining a seminorm $p_j$, we induce some unique neighbourhood for which it is the Minkowski functional: this is a smaller issue however. Despite the fact these are meant to be 'easy' examples, I simply can't see why this collection of seminorms induces precisely the topology stated, in either case. Could anyone explain it to me? I would ask my lecturer but since I wasn't able to attend the course which is now over (it clashed with another) it may be difficult to clear up my confusion. Thank you in advance.

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Here's a related thread citing some sources and you may want to have a look at this one, too –  t.b. Dec 12 '11 at 1:58
    
Many thanks for the suggestions - I have had a quick read through the suggested threads and some of the sources (apologies for not tracking down relevant threads myself, I tried unsuccessfully), but unless I'm overlooking something, I can't see anywhere which actually explains how to deduce what the topology is, given the construction of the space using seminorms as in the wikipedia article on Frechet spaces: I can only see that there is -some- topology induced by these seminorms, not that it is precisely the product/LUC topology. Am I missing something as obvious as the notes seem to suggest? –  Spyam Dec 12 '11 at 3:05
    
Oh, I didn't intend to suggest that those threads immediately answer your question, I meant to say that they might provide for some interesting reading while you waiting for a full-fledged answer. What's seemingly missing in what you quote is that these sets are a fundamental system of neighborhoods of the identity. That is, a set is a neighborhood of zero if and only if it contains some $N_j$. This means a sequence converges to $0$ if and only if it is eventually in $N_j$ for all $j$. If you think about what the metric does, you'll see that in order for $d(0,x)$ to be small, the first –  t.b. Dec 12 '11 at 3:17
    
terms must vanish. In other words, $x$ must lie in the first few $N_j$. –  t.b. Dec 12 '11 at 3:18
    
Ah, I see. I assumed that was what 'base' referred to ("...a decreasing base of closed, absolutely convex neighbourhoods...") but perhaps you're right, this is all quite new to me. I guess I feel like there should be a 'nice'/'obvious' justification for those topologies in some sense, one which you can grasp fairly easily. I can sort of see how the first metric corresponds to pointwise convergence and the second is related to uniform convergence in some nbhd of the origin, but my ideas are not at all concrete. Anyway, I'll wait and see if anyone can clear up my confusions, thanks for the help. –  Spyam Dec 12 '11 at 3:30
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