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Apr
19
revised Set of Linear equation has no solution or unique solution or infinite solution?
cleanups
Apr
9
comment Confusion regarding one formulation of the Axiom of Choice.
I didn't think that the Order Extension Principle was equivalent to BPI, but I wasn't certain. Thanks for the clarification!
Apr
8
comment Confusion regarding one formulation of the Axiom of Choice.
+1: If I recall correctly, Boolean Prime Ideal Theorem is sufficient for every set to be linearly orderable.
Apr
1
comment Confusion regarding one formulation of the Axiom of Choice.
I think Russell put it rather well: AC would be needed to select one sock from each of infinitely-many pairs of socks (assumed indistinguishable), but not from infinitely-many pairs of shoes (as we could just pick the right shoe from each pair, for example).
Apr
1
comment Confusion regarding one formulation of the Axiom of Choice.
However, it is likely that you'll only have a partial ordering, in which case we can use the Order Extension Principle (which is a weak form of AC) to do the job. Now, if there is only one way to order a set, then that set has at most one element--easily proven--but even if that weren't true, then we wouldn't have to choose the order, nor would picking any single element from that set require AC. We need AC when we must make an infinite quantity of arbitrary choices and have no mechanism for specifying how they should all be made (such as "pick the least element each time").
Apr
1
comment Confusion regarding one formulation of the Axiom of Choice.
Digressing from set theoretical context and discussing axioms of set theory is ill-advised. Outside of the context of set theory, there may not be precise analogues, or well-defined concepts. If you can make rigorous the idea of ordering different ways to reach a destination (which could be troublesome, as you may run into different ways of equal length, comfort, etc.), then there's actually no need to use any Choice at all to order that set.
Apr
1
comment Confusion regarding one formulation of the Axiom of Choice.
@algebraically_speaking: That's not quite true. If we need to make any finite number of arbitrary choices, we can, with no need to use AC. So, if there are multiple ways of accomplishing a task, then the set of ways of accomplishing said task is non-empty, and we simply pick an arbitrary way from that set of all ways, with no need for AC. For example, if we know (somehow) that a given set can be ordered, we can just pick any ordering of that set to do the job. However, if we don't know enough about the set to conclude that it is orderable, then we can let Choice do it for us.
Mar
31
answered Confusion regarding one formulation of the Axiom of Choice.
Mar
31
comment Confusion regarding one formulation of the Axiom of Choice.
The Axiom of Choice can be used to show that every set is well-orderable, and from the reasoning you mentioned, the converse is readily proved. That's what Don is saying--he's not saying that without AC, no set is well-orderable (and there are many cartesian products that are non-empty without AC, too).
Mar
31
comment Confusion regarding one formulation of the Axiom of Choice.
@algebraically_speaking: For general ordered sets, there needn't be a lowest element. Consider $\Bbb R$ in the usual order for an example of an ordered set without a lowest element. However, a Cartesian product of ordered sets with least element (for example, a Cartesian product of non-empty well-ordered sets) will certainly be non-empty, for the reasons you describe.
Mar
30
awarded  Yearling
Mar
24
revised Rewriting quadratic form using coordinate transformation
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Mar
23
revised Can we pull $f$ from this equation?
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Mar
22
revised Can we pull $f$ from this equation?
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Mar
22
revised Can we pull $f$ from this equation?
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Mar
22
comment Can we pull $f$ from this equation?
And for $X\ne k,$ $\delta_k(X)=0?$ I think you'll also want to multiply by $c$ to make this work. Still, I like this approach.
Mar
22
answered Can we pull $f$ from this equation?
Mar
21
comment Does a monotone function on an arbitray subset of $\mathbb R$ always have at most countable number of discontinuity?
@Dave: That's a nice fix, too!
Mar
20
revised Equivalence of different definitions of “monotonically normal space.”
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Mar
20
revised Does a monotone function on an arbitray subset of $\mathbb R$ always have at most countable number of discontinuity?
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