Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

"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?

share|cite|improve this question
It would be great if you also posted the origin of this quote. – Asaf Karagila Apr 10 '13 at 7:58
@Asaf Our instructor. – Xena 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
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.

share|cite|improve this answer

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 $\mathbb N=\omega$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.