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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 $\omega$?

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

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  • $\begingroup$ 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? $\endgroup$ – user180446 Aug 6 at 8:20
  • $\begingroup$ 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. $\endgroup$ – badjohn Aug 6 at 8:25
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    $\begingroup$ @badjohn thanks. $\endgroup$ – user180446 Aug 6 at 9:58
  • $\begingroup$ @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. $\endgroup$ – Henno Brandsma Aug 7 at 8:25

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