Timeline for Simple constructive proof for the hyperplane separating theorem (HST)?
Current License: CC BY-SA 4.0
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Dec 11, 2022 at 12:26 | history | edited | dodo | CC BY-SA 4.0 |
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Apr 19, 2022 at 22:40 | history | edited | dodo | CC BY-SA 4.0 |
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Apr 15, 2022 at 11:51 | comment | added | Benedict Eastaugh | I would imagine that one could prove this result from an appropriate constructive version of the Hahn–Banach theorem, e.g. the one in §9.3 of Bishop's 1967 book Foundations of Constructive Analysis. | |
Apr 15, 2022 at 10:28 | comment | added | Franka Waaldijk | oh wait you are right sorry, i forgot that these are convex sets in which case the $a_n$ converge even if the other sequences don't... | |
Apr 15, 2022 at 10:24 | comment | added | Franka Waaldijk | by easily fixed i mean not that such a limit can always be constructed... (it can't) but that depending on $d(C,D)$ and the respective diameters of $C,D$ we can find a sufficiently close approximating pair $c_n, d_n$ to construct an $a$ that works. | |
Apr 15, 2022 at 10:17 | comment | added | Franka Waaldijk | @MattF. that idea works if you do it right... still your phrasing just happens to allow more than one limit point... easily fixed. There is an even more simple way to phrase this idea, but i'm quite busy with other stuff, so perhaps you can turn your remark into the desired answer...? thanx :-) | |
Apr 15, 2022 at 6:07 | history | edited | user44143 | CC BY-SA 4.0 |
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Apr 15, 2022 at 5:57 | comment | added | user44143 | I think this is enough: Let $C_n$ and $D_n$ be the intersections of $C$ and $D$ with the ball of radius $n$ about the origin. Choose points $c_n$ and $d_n$ such that $$d(c_n,d_n)<\frac1n+d(C_n,D_n)$$ Let $a_n$ be a unit vector perpendicular to $d_n-c_n$, and let $a$ be the limit of the $a_n$. (One would still have to prove that the limit exists, but I think the construction works.) | |
Apr 15, 2022 at 0:46 | history | edited | dodo | CC BY-SA 4.0 |
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Apr 15, 2022 at 0:41 | comment | added | user44143 | The standard constructive definition is that a set in $\mathbb{R}^n$ is compact iff it is closed and for every $\epsilon$ it can be covered by finitely many balls of radius $\epsilon$. See ncatlab.org/nlab/show/Bishop-compact+space | |
Apr 15, 2022 at 0:29 | comment | added | dodo | @darijgrinberg I updated the question hope it is now clearer. I have to admit that I need help in defining compactness constructively. | |
Apr 15, 2022 at 0:22 | history | edited | dodo | CC BY-SA 4.0 |
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Apr 14, 2022 at 23:46 | comment | added | darij grinberg | How are the convex sets given? How is compactness defined constructively? | |
Apr 14, 2022 at 23:41 | history | asked | dodo | CC BY-SA 4.0 |