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Is every non empty subset of the integers well ordered and does this mean that every subset contains a least element?

Are the positive rationals well ordered? i believe not.

Is this because of the fact that this set has no minimal element? Does every subset strictly smaller than the set of positive rationals have a least element? i believe it does but not sure

thanks

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    $\begingroup$ The well ordering principal is a theorem about natural numbers... $\endgroup$ Nov 14 '18 at 19:35
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    $\begingroup$ @CarlosBacca In general, an well-order is a total order such that every nonempty subset has a least element. The well-ordering principle says that the natural numbers are well-ordered. The argument is correct- since the positive rational numbers do not have a least element, then it is a nonempty subset of itself without a least element. $\endgroup$
    – Kevin Long
    Nov 14 '18 at 19:41
  • $\begingroup$ Oh, thank you i see $\endgroup$ Nov 14 '18 at 19:42
  • $\begingroup$ im not quite sure the argument is correct actually he said something a bit different you could have the set of all positive rational numbers 1/q such that q is greater 5 This is a subset of the set of positive rationals without a least element $\endgroup$ Nov 14 '18 at 19:51
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When a linear order is non-well-ordered, lots of its proper subsets are also non-well-ordered. For example, consider the set of negative even integers or the set of reciprocals of integer powers of $2$: these are proper subsets of the integers and the positive rationals respectively, and neither has a least element.

The definition of well-orderedness - "Every subset has a least element" - can feel weird at first. It's helpful to think instead in terms of descending sequences: a linear order $(L,<)$ is well-ordered iff it has no infinite descending sequence $a_1>a_2>a_3>...$. Thinking this way it should be clear e.g. that the positive rationals aren't well-ordered, because we can "count down" $${1\over 1}>{1\over 2}>{1\over 3}>{1\over 4}>...$$ An important thing to keep in mind is that a counterexample to well-orderedness - that is, an infinite descending sequence in the linear order in question - will not be unique: we can't speak of "the" descending sequence, but rather "a" descending sequence.

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  • $\begingroup$ A basically-irrelevant comment, to be ignored until the answer is understood: when we move into more advanced set theory - specifically, when we start to consider the axiom of choice - the two notions of well-orderedness above (least elements vs. no descending sequences) turn out to not be quite the same in some contexts. But that's something to ignore right now - I mention it only because if you look around this site for further related questions, you might see this issue mentioned, and I want to get out ahead of any possible confusion. $\endgroup$ Nov 14 '18 at 19:52

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