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"A set A is inductive if every chain in A has an upper bound."

Since $\mathbb{N}$ is a chain, apparently it has an upper bound. But how? I don't understant. How can one find a number greater than every natural number?

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7  
It would be great if you also posted the origin of this quote. –  Asaf Karagila Apr 10 '13 at 7:58
    
@Asaf Our instructor. –  Fanni Apr 10 '13 at 8:02
    
I see. It is a good practice to attribute quotes, because it also gives them context. What was the class the instructor was giving? –  Asaf Karagila Apr 10 '13 at 8:04
    
@AsafKaragila: The bit in quotes is a definition. Not a super common one though. –  Jim Apr 10 '13 at 8:05
    
@Jim: I realized that when I first read the question. But I can give you a few definitions which when devoid of proper context will cause you to raise an eyebrow. Proper context such as who said that? where was it said? All these would appear in a proper citation format. –  Asaf Karagila Apr 10 '13 at 8:07

2 Answers 2

up vote 7 down vote accepted

The set $\mathbb N$ with it's natural partial order is not inductive. Indeed, you have identified a chain that does not have an upper bound.

Maybe you have confused "inductive", meaning a partially ordered set in which every chain has an upper bound, with "a set on which you can perform induction". The set $\mathbb N$ is certainly the latter, but not the former.

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In my view, it relies on the context.

  • If we treat $\mathbb N$ as a subset of $\mathbb N$ itself, then it doesn't has any upper bound(w.r.t $\le$) since no natural number in $\mathbb N$ is greater than or equal to every natural numbers in $\mathbb N$.

  • On the other hand, if we treat $\mathbb N$ as a set on the collection of all sets, then it does have upper bounds(w.r.t $\subseteq$). One of them is $\mathbb N$ itself(note that $\mathbb N=\omega$).

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