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Dec 17, 2020 at 6:50 vote accept erz
Dec 17, 2020 at 1:52 answer added Mikhail Ostrovskii timeline score: 11
Dec 16, 2020 at 21:47 comment added erz @LiviuNicolaescu as far as i know, both of these statements are true in Banach spaces as well, but this just shows that you cannot approximate a small space with a big one.
Dec 16, 2020 at 13:38 comment added Liviu Nicolaescu In Hilbert spaces if $F\subsetneq G$, then $g(G,F)=1$. Also if $\dim F<\dim G<\infty$ then $g(G,F)=1$.
Dec 16, 2020 at 12:01 comment added erz @LiviuNicolaescu but not only $H$ has to approximate $F$ (which it does at least as well as it approximates $G$), but another way around as well. And if $F$ is too small it does not approximate $H$. In particular, take your example: $g(G,G)=0$, and yet $g(F,G)=1$.
Dec 16, 2020 at 11:58 comment added Liviu Nicolaescu I meant the gap $g(F,H)\leq g(G,H)$. If $ F\subset G$ I believe $g(G,F)=1$.
Dec 16, 2020 at 10:31 comment added erz @LiviuNicolaescu seems unlikely: what if $F$ is too small? Note that the distance is symmetrical.
Dec 16, 2020 at 10:11 comment added Liviu Nicolaescu It looks to me that $F\subset G\implies d(F,H)\leq d(G,H)$ so you could take $\delta=\varepsilon$ and $J=H$.
Dec 16, 2020 at 3:58 history edited LSpice CC BY-SA 4.0
closed to -> close to
Dec 16, 2020 at 3:48 history asked erz CC BY-SA 4.0