Timeline for What is the cardinality of Dana Scott's $D_{\infty}$?
Current License: CC BY-SA 4.0
31 events
| when toggle format | what | by | license | comment | |
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| Feb 1, 2021 at 16:18 | vote | accept | YKY | ||
| Jan 29, 2021 at 16:22 | comment | added | Alex Nelson | @ToddTrimble Thanks for the clarification. I'm afraid I've got sets on the brain ;) | |
| Jan 29, 2021 at 13:06 | comment | added | YKY | @YemonChoi Fine, I'm trying to find a simple way to explain that without losing the intuition... | |
| Jan 28, 2021 at 19:47 | comment | added | Yemon Choi | I'm afraid I do not have the time to go over other people's projects. Could you please at least update your original question to take account of the errors pointed out by @ToddTrimble? | |
| Jan 28, 2021 at 19:17 | comment | added | YKY | @YemonChoi If you're interested you may take a look: drive.google.com/file/d/1AhQS3fp4WMFIDEhn_q4vNs-YJaq4Z-Fr/… . I'd be very grateful if you can give some comments or spot some mistakes therein. My email is there, thanks :) | |
| Jan 28, 2021 at 14:52 | comment | added | Yemon Choi | @YKY I am instinctively sceptical: how are you getting a Hilbert space of continuous functions? It can't be an $L^2$-space... | |
| Jan 28, 2021 at 12:51 | comment | added | YKY | @YemonChoi I think I have found a model of untyped $\lambda$-calculus in the Hilbert space of continuous functions, Dana Scott's work being the inspiration. The structure of $D_{\infty}$ somewhat suggested Hilbert space to me, so I asked this to know more background :) | |
| Jan 27, 2021 at 19:45 | comment | added | Yemon Choi | @AndrejBauer regading your second comment/example, I guess I was thinking carelessly of colimits of finite sets / finite-dimensional vector spaces / etc. But as you point out, these are limits here, not colimits. | |
| Jan 27, 2021 at 19:43 | comment | added | Yemon Choi | @AndrejBauer Thank you for correcting my earlier misunderstanding/stupidity. I am still not sure that the cardinality of $D_{\infty}$ itself is of interest, but of couse the asymptotics of the cardinalities of the stages in the construction could be interesting. I was just a bit puzzled by the OP seeking some expression for the cardinality of $D_\infty$ in terms of the cardinality of $D_0$ | |
| Jan 27, 2021 at 15:54 | comment | added | Andrej Bauer | @YemonChoi: That is incorrect, $D_\infty$ is either trivial or uncountable, depending on $D_0$. And it is not at all the case that a limit of finite object need be finite. Already the cartesian product of countably many sets of size two is uncountable. | |
| Jan 27, 2021 at 15:42 | comment | added | LSpice |
TeX note: please use $\|D\|$ \|D\| rather than $||D||$ ||D|| for better spacing. I edited accordingly. (I also think that $D_n^{D_n}$ D_n^{D_n} is much more common than $D_n{}^{D_n}$ D_n{}^{D_n}, but the latter was obviously intentional, so I didn't change it.)
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| Jan 27, 2021 at 15:41 | history | edited | LSpice | CC BY-SA 4.0 |
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| Jan 27, 2021 at 15:16 | history | edited | Paul Taylor |
deleted "cardinal" tags because they're misleading
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| Jan 27, 2021 at 13:56 | answer | added | Paul Taylor | timeline score: 11 | |
| Jan 27, 2021 at 13:19 | comment | added | YKY | @AlexNelson If its ordinal is like $\epsilon_0$ then wouldn't its cardinal number be $\aleph_0$? | |
| Jan 26, 2021 at 23:20 | comment | added | Noah Schweber | @AlexNelson And in fact $\epsilon_0$ is countable. | |
| Jan 26, 2021 at 19:44 | comment | added | Todd Trimble | @AlexNelson Ordinal exponentiation is about something else altogether. | |
| Jan 26, 2021 at 18:34 | comment | added | Alex Nelson | When $|D_{0}|=\omega$, you get $|D_{\infty}|\cong\varepsilon_{0}$, don't you? | |
| Jan 26, 2021 at 16:46 | comment | added | Yemon Choi | In light of @ToddTrimble's comment, I am not sure if your question remains well-founded, since his comment seems to indicate that some of your assertions about cardinalities is incorrect. Moreover, I don't quite understand what you expect the cardinality to be; if each $D_n$ is finite then the cardinality of $D_\infty$ is going to either be finite or countably infinite. Todd's comment also seems to indicate that cardinality is not really the right notion of "size" for $D_\infty$ anyway | |
| S Jan 26, 2021 at 14:27 | history | suggested | gmvh |
Added top-level tag
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| Jan 26, 2021 at 12:29 | review | Suggested edits | |||
| S Jan 26, 2021 at 14:27 | |||||
| Jan 26, 2021 at 10:59 | comment | added | Todd Trimble | Scott's models are directed-complete partial orders (DCPOs), which form a cartesian closed category, and the exponential $A^B$ of two DCPOs does not consist of all functions from $B$ to $A$, but only those which preserve directed joins. Therefore it is not correct to say that $|A^B| = |A|^{|B|}$ (the cardinal exponential on the right counts all functions). I may come back to say more about the question, but there is nothing shocking going on. | |
| Jan 26, 2021 at 10:10 | comment | added | YKY | @Wojowu This is achievable by a suitably defined notion of continuous functions, known as Scott-continuous. | |
| Jan 26, 2021 at 10:07 | comment | added | YKY | PS: Scott was trying to find a model for the untyped $\lambda$-calculus, in which functions can apply to themselves. | |
| Jan 26, 2021 at 10:04 | comment | added | Wojowu | I don't know anything about domain theory, but let me reiterate - there is no set $A$ with more than one element which is equipotent to the set of functions $A\to A$. | |
| Jan 26, 2021 at 9:59 | history | edited | YKY | CC BY-SA 4.0 |
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| Jan 26, 2021 at 9:56 | comment | added | YKY | Thanks for asking this, maybe I messed up the description. What you asked is precisely Dana Scott's contribution, and it started the field known as domain theory. The correct notation is $D_{\infty} \cong D_{\infty} \rightarrow D_{\infty}$. This seemingly paradoxical statement is achieved by a slightly different notion of function application (always going from $D_{n}$ to $D_{n-1}$). Independent of this issue, I guess my question is still valid, right? | |
| Jan 26, 2021 at 9:44 | comment | added | Wojowu | I'm afraid I might be missing something. Assuming $A^A$ means the set of $A$-indexed sequences from $A$, and $\cong$ is something that implies bijection, then it is impossible to have a set with more than one element satisfying $A^A\cong A$. So either I'm missing what $A^A$ means or what $\cong$ means. | |
| Jan 26, 2021 at 9:30 | history | edited | YKY | CC BY-SA 4.0 |
added 54 characters in body
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| Jan 26, 2021 at 9:24 | history | edited | YKY | CC BY-SA 4.0 |
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| Jan 26, 2021 at 9:19 | history | asked | YKY | CC BY-SA 4.0 |