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This question may be trivial, or overly optimistic. I do not know (but I guess the latter...). I am a group theorist by trade, and the set-up I describe cropped up in something I want to prove. So this question is out of my comfort zone, but I am happy to clarify anything if needed.

I have a countable set $S$ equipped with a partial order $<$ and a minimal element $0$ (so $0<x$ for all $x\in S\setminus\{0\}$). I want to perform induction on chains which contain $0$, so $0<x<\dotsb<y<z$, in this order. As in: if property $P$ holds for $0, x, \dotsc, y$ then $P$ holds for $z$. Obviously I can perform induction on a finite chain. My question is:

What are my options if I want to to perform induction on infinite chains?

I believe one option would be transfinite induction, and in order to apply this I would need to prove that every chain not containing $0$ has a minimal element. But this condition on chains is unlikely to hold in my setting. So I am wondering: do I have any other options?

[An example to keep in mind is the chain with elements from $\{2^n\mid n\leq m\}\cup\{0\}$ for some fixed integer $m$, with the natural ordering inherited from $\mathbb{Q}$. So the chain $0<\dotsb<2^{m-1}< 2^m$. This example makes me think the question is not trivial - standard induction will not work.]

This question may be trivial, or overly optimistic. I do not know (but I guess the latter...). I am a group theorist by trade, and the set-up I describe cropped up in something I want to prove. So this question is out of my comfort zone, but I am happy to clarify anything if needed.

I have a countable set $S$ equipped with a partial order $<$ and a minimum element $0$ (so $0<x$ for all $x\in S\setminus\{0\}$). I want to perform induction on chains which contain $0$, so $0<x<\dotsb<y<z$, in this order. As in: if property $P$ holds for $0, x, \dotsc, y$ then $P$ holds for $z$. Obviously I can perform induction on a finite chain. My question is:

What are my options if I want to to perform induction on infinite chains?

I believe one option would be transfinite induction, and in order to apply this I would need to prove that every chain not containing $0$ has a minimum element. But this condition on chains is unlikely to hold in my setting. So I am wondering: do I have any other options?

[An example to keep in mind is the chain with elements from $\{2^n\mid n\leq m\}\cup\{0\}$ for some fixed integer $m$, with the natural ordering inherited from $\mathbb{Q}$. So the chain $0<\dotsb<2^{m-1}< 2^m$. This example makes me think the question is not trivial - standard induction will not work.]

This question may be trivial, or overly optimistic. I do not know (but I guess the latter...). I am a group theorist by trade, and the set-up I describe cropped up in something I want to prove. So this question is out of my comfort zone, but I am happy to clarify anything if needed.

I have a countable set $S$ equipped with a partial order $<$ and a minimal element $0$ (so $0<x$ for all $x\in S\setminus\{0\}$). I want to perform induction on chains which contain $0$, so $0<x<...<y<z$, in this order. As in: if property $P$ holds for $0, x, \ldots, y$ then $P$ holds for $z$. Obviously I can perform induction on a finite chain. My question is:

What are my options if I want to to perform induction on infinite chains?

I believe one option would be transfinite induction, and in order to apply this I would need to prove that every chain not containing $0$ has a minimal element. But this condition on chains is unlikely to hold in my setting. So I am wondering: do I have any other options?

[An example to keep in mind is the chain with elements from $\{2^n\mid n\leq m\}\cup\{0\}$ for some fixed integer $m$, with the natural ordering inherited from $\mathbb{Q}$. So the chain $0<\ldots<2^{m-1}< 2^m$. This example makes me think the question is not trivial - standard induction will not work.]

This question may be trivial, or overly optimistic. I do not know (but I guess the latter...). I am a group theorist by trade, and the set-up I describe cropped up in something I want to prove. So this question is out of my comfort zone, but I am happy to clarify anything if needed.

I have a countable set $S$ equipped with a partial order $<$ and a minimal element $0$ (so $0<x$ for all $x\in S\setminus\{0\}$). I want to perform induction on chains which contain $0$, so $0<x<\dotsb<y<z$, in this order. As in: if property $P$ holds for $0, x, \dotsc, y$ then $P$ holds for $z$. Obviously I can perform induction on a finite chain. My question is:

What are my options if I want to to perform induction on infinite chains?

I believe one option would be transfinite induction, and in order to apply this I would need to prove that every chain not containing $0$ has a minimal element. But this condition on chains is unlikely to hold in my setting. So I am wondering: do I have any other options?

[An example to keep in mind is the chain with elements from $\{2^n\mid n\leq m\}\cup\{0\}$ for some fixed integer $m$, with the natural ordering inherited from $\mathbb{Q}$. So the chain $0<\dotsb<2^{m-1}< 2^m$. This example makes me think the question is not trivial - standard induction will not work.]

This question may be trivial, or overly optimistic. I do not know (but I guess the latter...). I am a group theorist by trade, and the set-up I describe cropped up in something I want to prove. So this question is out of my comfort zone, but I am happy to clarify anything if needed.

I have a countable set $S$ equipped with a partial order $<$ and a minimal element $0$ (so $0<x$ for all $x\in S\setminus\{0\}$). I want to perform induction on chains which contain $0$, so $0<x<...<y<z$, in this order. As in: if property $P$ holds for $0, x, \ldots, y$ then $P$ holds for $z$. Obviously I can perform induction on a finite chain. My question is:

What are my options if I want to to perform induction on infinite chains?

I believe one option would be transfinite induction, and in order to apply this I would need to prove that every chain not containing $0$ has a minimal element. But this condition on chains is unlikely to hold in my setting. So I am wondering: do I have any other options?

[An example to keep in mind is the chain with elements from $\{2^n\mid n\leq m, n~\&~m\in\mathbb{Z}\}\cup\{0\}$, with the natural ordering inherited from $\mathbb{Q}$. So the chain $0<\ldots<2^{m-1}< 2^m$. This example makes me think the question is not trivial - standard induction will not work.]

This question may be trivial, or overly optimistic. I do not know (but I guess the latter...). I am a group theorist by trade, and the set-up I describe cropped up in something I want to prove. So this question is out of my comfort zone, but I am happy to clarify anything if needed.

I have a countable set $S$ equipped with a partial order $<$ and a minimal element $0$ (so $0<x$ for all $x\in S\setminus\{0\}$). I want to perform induction on chains which contain $0$, so $0<x<...<y<z$, in this order. As in: if property $P$ holds for $0, x, \ldots, y$ then $P$ holds for $z$. Obviously I can perform induction on a finite chain. My question is:

What are my options if I want to to perform induction on infinite chains?

I believe one option would be transfinite induction, and in order to apply this I would need to prove that every chain not containing $0$ has a minimal element. But this condition on chains is unlikely to hold in my setting. So I am wondering: do I have any other options?

[An example to keep in mind is the chain with elements from $\{2^n\mid n\leq m\}\cup\{0\}$ for some fixed integer $m$, with the natural ordering inherited from $\mathbb{Q}$. So the chain $0<\ldots<2^{m-1}< 2^m$. This example makes me think the question is not trivial - standard induction will not work.]