Timeline for What is the concept of being n-admissible used in the proof of the Iteration Theorem?
Current License: CC BY-SA 4.0
12 events
when toggle format | what | by | license | comment | |
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Jul 30, 2020 at 19:04 | vote | accept | Paulo Henrique L. Amorim | ||
Jul 30, 2020 at 10:30 | answer | added | Daniel Fischer | timeline score: 1 | |
Jul 30, 2020 at 2:27 | history | edited | Paulo Henrique L. Amorim | CC BY-SA 4.0 |
edited body
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Jul 30, 2020 at 2:21 | comment | added | Paulo Henrique L. Amorim | Can you submit you first comment as an answer? This explanation give a precise definition of what n-admissible is in the given context. | |
Jul 28, 2020 at 14:46 | comment | added | Paulo Henrique L. Amorim | I just did a big mistake, I was thinking about sucessor of $n$ instead of how $f(n)$ will be defined in case $S(n) \notin I$, but as long $n \neq 1$, we have that $(\exists !q)(n = S(q))$, thus $f(n) = f(S(q)) = g(f(q)$, and we have that $q \in I$, because the way $I$ was defined and because condition (4) too. Sorry for this, Im new to abstract mathematics. From your comment I wonder why the author dont restricted $S(n) \notin A$, can you see a good reason? | |
Jul 28, 2020 at 14:40 | comment | added | Daniel Fischer | I'm not sure what you're confused about. What do you mean by "the image from $f(n)$"? | |
Jul 28, 2020 at 14:36 | comment | added | Paulo Henrique L. Amorim | Im a bit confused, because if we let some $f$ be a restriction of such an $F$ to the initial segment $I_n = \{1,...,n\}$ of $P$, where $S(n) \notin I$, but we have $1 \in I$, $I \subseteq P$, $n \in I$ and we can define $f$ to met the conditions to be $n$-admissible, but I cant see what will be the image from $f(n)$. | |
Jul 28, 2020 at 13:56 | comment | added | Daniel Fischer | The way you wrote it, you're close, but slightly misunderstanding. During the proof, we do not yet have an $F$ that we could restrict, so you'd have to throw in a couple of "hypothetical" there. If we had the additional condition $S(n) \notin A$, then an $n$-admissible function would be a function that could be a restriction of such an $F$ to the initial segment $I_n := \{1, \dotsc, n\}$ of $P$. Since $A$ might be a larger segment, a function is $n$-admissible if its restriction to $I_n$ could be the restriction of such an $F$. The conditions make the "could be" precise. | |
Jul 28, 2020 at 13:34 | history | edited | Paulo Henrique L. Amorim | CC BY-SA 4.0 |
typos; added 31 characters in body
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Jul 28, 2020 at 13:25 | history | edited | Paulo Henrique L. Amorim | CC BY-SA 4.0 |
Adding the rest of the Iteration Theorem proof
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Jul 26, 2020 at 16:05 | history | edited | Paulo Henrique L. Amorim | CC BY-SA 4.0 |
adding some references
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Jul 26, 2020 at 13:48 | history | asked | Paulo Henrique L. Amorim | CC BY-SA 4.0 |