Timeline for Can the Lawvere fixed point theorem be used to prove the Brouwer fixed point theorem?
Current License: CC BY-SA 3.0
11 events
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| Apr 23, 2017 at 17:43 | comment | added | Todd Trimble | Sorry for the delay in responding. Actually, the root of the problem is that I badly misidentified the topology on $\mathbf{P}D$ (it's codiscrete, which seems painfully obvious in retrospect), and in fact $D$ is not an internal sup-lattice as defined here. Except for that, I believe the rest was correctly argued. I'm still investigating possible repairs... | |
| Apr 20, 2017 at 20:11 | history | bounty ended | Qiaochu Yuan | ||
| Apr 19, 2017 at 22:45 | history | edited | Todd Trimble | CC BY-SA 3.0 |
added 216 characters in body
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| Apr 19, 2017 at 22:38 | comment | added | Will Sawin | If it helps, I think the map $sup_L^X$ from ${\mathbf P}(L \times X)$ to $L^X$ is not well-defined, because it does not always produce continuous maps. From an arbitrary downward-closed subset of $L \times X$, by taking the supremum, we can produce an arbitrary function from $X$ to $L$ and not just a continuous function. | |
| Apr 19, 2017 at 22:31 | comment | added | Todd Trimble | @WillSawin Thanks for your comments. Yeah, something seems fishy, but I'm not quite sure where the real root of the problem is. Thinking... | |
| Apr 19, 2017 at 21:39 | comment | added | Will Sawin | It seems to me like a retract of $D^f$ does not exist for an arbitrary continuous map $f$. To admit a retract, a map to a Hausdorff space must have closed image, as the image is exactly the locus where the composition of the retract with the original map is equal to the identity. If we take $X$ as the interval with the discrete topology, $Y$ with the interval with the usual topology, and $f$ the obvious map, then $I^X$ is the space of $I$-valued functions on $I$ with the topology of pointwise convergence, which is clearly Hausdorff, but within which the continuous functions are not closed. | |
| Apr 19, 2017 at 21:35 | comment | added | Will Sawin | How can the compact-open topology on $[I^{op}, \Omega]$ be the order topology when $\Omega$ has the indiscrete topology? | |
| Apr 19, 2017 at 12:03 | history | edited | Todd Trimble | CC BY-SA 3.0 |
typo; added a link
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| Apr 19, 2017 at 10:53 | history | edited | Todd Trimble | CC BY-SA 3.0 |
added 18 characters in body
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| Apr 19, 2017 at 10:44 | history | edited | Todd Trimble | CC BY-SA 3.0 |
removed a slight ambiguity
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| Apr 19, 2017 at 5:19 | history | answered | Todd Trimble | CC BY-SA 3.0 |