Timeline for Why does Weihrauch reducibility make use of multi-functions?
Current License: CC BY-SA 4.0
6 events
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| May 28 at 8:21 | comment | added | Sam Sanders | @PeterGerdes I am not a specialist either but as far as I know, "$f$ is a multi-function" in Weihrauch reducibility implies that for $x$ in the domain of $f$, $f(x)$ can depend on the representation of $x$. This is contrary to e.g. reverse math/mainstream where one does assume that e.g. equal reals map to equal reals. Grilliot's trick shows that this omission of function extensionality in Weihrauch reducibility is 'essential' in that otherwise one would always get $\exists^2$. | |
| May 27 at 21:57 | comment | added | Peter Gerdes | Sorry, one more question. Why is the path operator discontinuous much less effectively so? Seems like the leftmost infinite path operation is continuous --if T is the limit of T_i each with leftmost path g_i then the limit of g_i is a path through T no? | |
| May 27 at 21:52 | comment | added | Peter Gerdes | Also you're missing the requirement that the function not merely be discontinuous but effectively so -- is even that too restrictive? All this is probably obvious to someone who regularly deals with Weirauch reducibility but please lay it out for me. | |
| May 27 at 21:49 | comment | added | Peter Gerdes | That's an interesting result, but how does it relate to Weirach reducibility? It's a more restrictive notion so you could imagine every Weirach degree having a functional representative despite Grillot's trick. | |
| Apr 4 at 7:16 | history | edited | Sam Sanders | CC BY-SA 4.0 |
fixed typo
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| Apr 3 at 18:42 | history | answered | Sam Sanders | CC BY-SA 4.0 |