# Well-ordering on natural numbers

Let $$\omega=\{0,1,2,3,\ldots\}$$. We say that $$\omega$$ is a well-ordered set. But I can't understand why. By the definition of well-ordering, there should be no infinite descending chain, but if I start from infinity, how can I reach 0 in finite descents? Or is this not allowed? Is this nonsensical to choose infinity? Then what about $$\omega+1$$? Is this set well-ordered? How can I reach to $$0$$ from $$\omega$$?

• This wikipedia article might help you: en.wikipedia.org/wiki/Well-order – Omar S Aug 6 at 7:32
• @Omar I have read it. But couldn't get my doubt cleared from it. – user180446 Aug 6 at 8:21

Every descending chain in $$\omega$$ has to start somewhere - specifically, it must start with a finite value $$n$$. One cannot "start from infinity" within $$\omega$$ since every element of $$\omega$$ is finite.
$$\omega + 1$$ is well-ordered in the obvious way. Suppose we have a descending chain beginning with $$\omega$$. Then the next element in the chain must be some $$n < \omega$$: that is, some finite $$n$$. After that point, it's clear there can only be finitely many steps until reaching zero.
• So we can not land into the floating infinities in between? So $\omega$ is like some island in the infinite sea surrounding us? And we can do our usual math only on these islands, not in the sea? – user180446 Aug 6 at 8:20
• There are no infinities in between. Although there are infinitely many natural numbers, no specific natural number is infinite. So, a specific number before $\omega$ will be finite. – badjohn Aug 6 at 8:25
• @user180446 $\omega+1$ is not "a larger infinity", it's just a countable set (so same size as $\omega$!) with a slightly different well-order. – Henno Brandsma Aug 7 at 8:25