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Mar 23, 2020 at 13:08 vote accept ABIM
Mar 23, 2020 at 12:44 vote accept ABIM
Mar 23, 2020 at 13:08
Mar 21, 2020 at 13:15 answer added Henno Brandsma timeline score: 2
Mar 19, 2020 at 17:51 comment added ABIM True, I guess there should be some type of topological regulairty like contractability?
Mar 19, 2020 at 10:59 comment added KP Hart How does that work for the space of rationals where you take each $K_n$ a singleton (using some enumeration of $\mathbb{Q}$)? Don't you get the topolog of pointwise convergence that way?
Mar 19, 2020 at 6:28 comment added user131781 This gives the metrisability condition in the second one. Here it is the completeness which is tricky. For this you need some version of the Kelley condition, i.e., that a function is continuous whenever its restriction to compacta is. I am not a point set topologist but flicking through my home library suggests that your conditions might not suffice.
Mar 19, 2020 at 6:23 comment added user131781 There are two points here, one very elementary, one rather subtle. The first one involves the metrisability. This is more transparent in the following version. If a uniformity is defined by a sequence $(d_n)$ of pseudometrics, then it can be be specifies by a single one. The standard ploy is to use $\sum \frac 1{2^n}\frac {d_n}{1+d_ n}$. Separability plays no role.
Mar 18, 2020 at 23:15 comment added Pietro Majer Also note that X can be isometrically embeded in a separable Banach space E, so that C(X,Y) is a subspace of C(Y,E), a Frėchet space.
Mar 18, 2020 at 20:15 comment added abx Might be an overkill, but look at Bourbaki's General Topology X, §3.1, Corollary.
Mar 18, 2020 at 18:55 comment added ABIM Haha, ya same here (not the biggest fan from working from home). Thanks for the tip, I made the modification :)
Mar 18, 2020 at 18:54 history edited ABIM CC BY-SA 4.0
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Mar 18, 2020 at 18:28 comment added Jochen Wengenroth Corona prevents me from checking the bookshelf in my office. Just note that the formula is not completely correct, if the metric $d_X$ is not bounded the series may diverge. You should replace $d_X$ by $\min\{d_X,1\}$.
Mar 18, 2020 at 18:02 history edited YCor CC BY-SA 4.0
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Mar 18, 2020 at 17:24 history asked ABIM CC BY-SA 4.0