“But this is equivalent to having your set be well-ordered, which is incompatible with the property “closed under division and subtraction by finite integers”″ - Why is this incompatible?
An ordered set is well-ordered iff every subset has a unique least element. If your set is closed under subtraction, you get infinite descending sequences such as 1>0>−1>−2>⋯ . If your sequence is closed under division, you get infinite descending sequences that are furthermore bounded such as 1>12>14>⋯>0. It should be clear that the two linear orders I described are not well-orders.
A small order theory fact that is not totally on-topic but may help you gather intuition:
Every countable ordinal embeds into the reals but no uncountable ordinal does.
Okay, I now understand why closure under those operations is incompatible with being well-ordered. And I’m guessing you believe that well-ordering is necessary for a coherent notion of passing through tomorrow infinitely many times because it’s a requirement for transfinite induction?
I’m not so sure that this is important. After all, we can imagine getting from 1 to 2 via passing through an infinite number of infinitesimally small steps even though [1,2] isn’t well-ordered on <. Indeed, this is the central point of Zeno’s paradox.
Yes, there are good ways to index sets other than well orders. A net where the index set is the real line and the function f:I→X is continuous is usually called a path, and these are ubiquitous e.g. in the foundations of algebraic topology.
I guess you could say that I think well-orders are important to the picture at hand “because of transfinite induction” but a simpler way to state the same objection is that “tomorrow” = “the unique least element of the set of days not yet visited”. If tomorrow always exists / is uniquely defined, then we’ve got a well-order. So something about the story has to change if we’re not fitting into the ordinal box.
A well-order has a least element in all non-empty subsets, and 1 > 1⁄2 > 1⁄4 > … > 0 has a non-empty subset without a least element, so it’s not a well-order.
In general, every suborder of a well-order is well-ordered. In a word, the property of “being a well-order” is hereditary. (compare: every subset of a finite set is finite)
“But this is equivalent to having your set be well-ordered, which is incompatible with the property “closed under division and subtraction by finite integers”″ - Why is this incompatible?
An ordered set is well-ordered iff every subset has a unique least element. If your set is closed under subtraction, you get infinite descending sequences such as 1>0>−1>−2>⋯ . If your sequence is closed under division, you get infinite descending sequences that are furthermore bounded such as 1>12>14>⋯>0. It should be clear that the two linear orders I described are not well-orders.
A small order theory fact that is not totally on-topic but may help you gather intuition:
Every countable ordinal embeds into the reals but no uncountable ordinal does.
Okay, I now understand why closure under those operations is incompatible with being well-ordered. And I’m guessing you believe that well-ordering is necessary for a coherent notion of passing through tomorrow infinitely many times because it’s a requirement for transfinite induction?
I’m not so sure that this is important. After all, we can imagine getting from 1 to 2 via passing through an infinite number of infinitesimally small steps even though [1,2] isn’t well-ordered on <. Indeed, this is the central point of Zeno’s paradox.
Yes, there are good ways to index sets other than well orders. A net where the index set is the real line and the function f:I→X is continuous is usually called a path, and these are ubiquitous e.g. in the foundations of algebraic topology.
I guess you could say that I think well-orders are important to the picture at hand “because of transfinite induction” but a simpler way to state the same objection is that “tomorrow” = “the unique least element of the set of days not yet visited”. If tomorrow always exists / is uniquely defined, then we’ve got a well-order. So something about the story has to change if we’re not fitting into the ordinal box.
Your second example, 1 > 1⁄2 > 1⁄4 > … > 0, is a well-order. To make it non-well-ordered, leave out the 0.
A well-order has a least element in all non-empty subsets, and 1 > 1⁄2 > 1⁄4 > … > 0 has a non-empty subset without a least element, so it’s not a well-order.
Yes, you’re right.
Adding to Vladimir_Nesov’s comment:
In general, every suborder of a well-order is well-ordered. In a word, the property of “being a well-order” is hereditary. (compare: every subset of a finite set is finite)