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2

$begingroup$

I'm currently struggling with the validity of defining a function using what I think is membership of a recursively defined set. I am confused as to whether this construction requires the axiom of choice, and I am unsure as to whether, even with the axiom of choice, what I am doing is valid, (in a set theoretic ZFC sense). I have not studied super strict axiomatic set theory, and I am unsure about how you can rigorously construct a function.

Here is the proof I am unsure about. I am proving a variant of every sequence having a monotone subsequence. There are two constructions that are a bit different and I am unsure about the validity of both of them.

Let $L$ be an infinite chain. Since $L$ is infinite we can define an injection $Phi : Bbb N to L$. If $Phi (n) >Phi (m)$ for all $n>m$, then I will call $n$ a peak.

If there are a finite number of peaks, then there exists an $N$ such that for all $n ge N$ $n$ is not a peak. Let $f(1)=N$. Given $f(n)$, and given $f(n) ge N$, there must exist a $m gt f(n)$ such that $Phi (m) gt Phi (f(n))$. Let $f(n+1)=m$
(I am unsure about this bit )

If there are an infinite number of peaks, then let $f(1)$ be the first peak, and let $f(n+1)= mathrm min m in Bbb N land
m , mathrm is , mathrm a , mathrmpeak , land m>f(n) $

Summing up, I want to know whether what I have done is valid, how to make it valid, whether the axiom of choice matters, and in general how to tell if a function I construct is ok.

functions elementary-set-theory recursion

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asked yesterday

B.MartinB.Martin

1971110

$endgroup$

add a comment | 

2

$begingroup$

I'm currently struggling with the validity of defining a function using what I think is membership of a recursively defined set. I am confused as to whether this construction requires the axiom of choice, and I am unsure as to whether, even with the axiom of choice, what I am doing is valid, (in a set theoretic ZFC sense). I have not studied super strict axiomatic set theory, and I am unsure about how you can rigorously construct a function.

Here is the proof I am unsure about. I am proving a variant of every sequence having a monotone subsequence. There are two constructions that are a bit different and I am unsure about the validity of both of them.

Let $L$ be an infinite chain. Since $L$ is infinite we can define an injection $Phi : Bbb N to L$. If $Phi (n) >Phi (m)$ for all $n>m$, then I will call $n$ a peak.

If there are a finite number of peaks, then there exists an $N$ such that for all $n ge N$ $n$ is not a peak. Let $f(1)=N$. Given $f(n)$, and given $f(n) ge N$, there must exist a $m gt f(n)$ such that $Phi (m) gt Phi (f(n))$. Let $f(n+1)=m$
(I am unsure about this bit )

If there are an infinite number of peaks, then let $f(1)$ be the first peak, and let $f(n+1)= mathrm min m in Bbb N land
m , mathrm is , mathrm a , mathrmpeak , land m>f(n) $

Summing up, I want to know whether what I have done is valid, how to make it valid, whether the axiom of choice matters, and in general how to tell if a function I construct is ok.

functions elementary-set-theory recursion

share|cite|improve this question

asked yesterday

B.MartinB.Martin

1971110

$endgroup$

add a comment | 

$begingroup$

I'm currently struggling with the validity of defining a function using what I think is membership of a recursively defined set. I am confused as to whether this construction requires the axiom of choice, and I am unsure as to whether, even with the axiom of choice, what I am doing is valid, (in a set theoretic ZFC sense). I have not studied super strict axiomatic set theory, and I am unsure about how you can rigorously construct a function.

Here is the proof I am unsure about. I am proving a variant of every sequence having a monotone subsequence. There are two constructions that are a bit different and I am unsure about the validity of both of them.

Let $L$ be an infinite chain. Since $L$ is infinite we can define an injection $Phi : Bbb N to L$. If $Phi (n) >Phi (m)$ for all $n>m$, then I will call $n$ a peak.

If there are a finite number of peaks, then there exists an $N$ such that for all $n ge N$ $n$ is not a peak. Let $f(1)=N$. Given $f(n)$, and given $f(n) ge N$, there must exist a $m gt f(n)$ such that $Phi (m) gt Phi (f(n))$. Let $f(n+1)=m$
(I am unsure about this bit )

If there are an infinite number of peaks, then let $f(1)$ be the first peak, and let $f(n+1)= mathrm min m in Bbb N land
m , mathrm is , mathrm a , mathrmpeak , land m>f(n) $

Summing up, I want to know whether what I have done is valid, how to make it valid, whether the axiom of choice matters, and in general how to tell if a function I construct is ok.

functions elementary-set-theory recursion

share|cite|improve this question

asked yesterday

B.MartinB.Martin

1971110

$endgroup$

share|cite|improve this question

asked yesterday

B.MartinB.Martin

1971110

share|cite|improve this question

asked yesterday

B.MartinB.Martin

1971110

asked yesterday

B.MartinB.Martin

1971110

asked yesterday

B.MartinB.Martin

1971110