Timeline for Is there a stable structure on $[0,1]$ that approximates every continuous function?
Current License: CC BY-SA 4.0
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| Nov 10, 2020 at 14:33 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
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| Nov 7, 2020 at 8:38 | comment | added | user44143 | You are correct. I'll leave up the comment so others can see the problem with the idea. | |
| Nov 7, 2020 at 8:35 | comment | added | Emil Jeřábek | @MattF. How would that be enough? I can in fact give you all continuous bijections $[0,1]\to[0,1]$, the resulting structure is still superstable. But even in this language, I don’t see how to define a single continuous function $[0,1]^2\to[0,1]$ which takes four different values at the corners. | |
| Nov 7, 2020 at 8:28 | comment | added | user44143 | How about adding one unary function for each pair of rationals, $f_{q,r}(x) = \max(0, \min(q + r x, 1))$? That would be enough to define continuous functions that interpolate between the boxes. | |
| Nov 4, 2020 at 13:24 | comment | added | tomasz | @EmilJeřábek: Me neither, but it does not seem implausible. | |
| Nov 2, 2020 at 16:23 | comment | added | Emil Jeřábek | @tomasz They might be easier to control, but I don’t see how to do continuous approximation of continuous binary functions using just unary affine functions. | |
| Nov 2, 2020 at 16:06 | comment | added | tomasz | @EmilJeřábek: That is true, but $+$ is binary. Unary functions are easier to control. | |
| Nov 2, 2020 at 11:36 | comment | added | Emil Jeřábek | If you expand the structure with as simple an affine function as $+$ (restricted to $[0,1]$), it is no longer stable, as it can define the usual order on $[0,1]$. | |
| Nov 2, 2020 at 11:32 | comment | added | Johannes Hahn | Why can't one require $g$ to be continuous? Shouldn't piecewise affine functions work just as well as piecewise constant functions? | |
| Nov 2, 2020 at 10:40 | history | answered | Emil Jeřábek | CC BY-SA 4.0 |