An uncountable countable set
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From: Han.deBruijn on 20 Sep 2006 15:56 stephen(a)nomail.com wrote: > Is sqrt(2) a number? Yes. Han de Bruijn
From: MoeBlee on 20 Sep 2006 16:01 Tony Orlow wrote: > MoeBlee wrote: > > Tony Orlow wrote: > >> If omega is the successor to the set of all finite naturals, > > > > But it's not the successor of the set of all finite naturals. > > > > MoeBlee > > > > It's not the least set that includes them all? It is. But it is not THE successor of any one of them. You're confused, probably because the English word 'successor' doesn't quite work out here as an informal mathematical term. We standard ordering of the class of ordinals is the membership relation. So, for ordinals x and y, 'x < y' just means xey. (Imagine '<' is in a special font so that it's different from other '<' symbols used for, for example, the standard ordering of the reals that is much different from the standard ordering of ordinals.) Then, for any x (whether x is an ordinal or not), successor(x) = xu{x}. Now, if y = xu{x} for some x, then y has an immediate predecessor, viz. x. But if there is no x such that y=xu{x}, then y does NOT have an immediate predecessor. So we don't get an unqualifed definition predecessor(y), since some y do not have such an immediate predeccessor. We can define 'predecessor(y)' only if for some x, we have y = xu{x}. And w is such a set that there is no x such that w = xu{x}. We can say that all members of w are predecessors of w. But no predecessor of w is an IMMEDIATE predecessor of w. And the successor of w is wu{w}, and, in very very casual terminology we might say that w has many other successors (any ordinal greater than w) but that only wu{w} is "THE successor" or, perhaps it would be helpful to say "the IMMEDIATE successor." Now, that's pretty complicated, due to the fact that we're using English instead of formal symbols. But if you just read a set theory book, especially with a formal context, you'd see that it's all very clear. > For all finite naturals, > they represent the set of all naturals less than them. I'd drop 'represent' and just say they 'are'. > The notion of > limit ordinals is an extension thereof. Omega is the least ordinal which > serves as a superset of the naturals. It is successor to the set with no > end, in that sense. No, it is not the SUCCESSOR of the least unbounded ordinal. It IS the least unbounded ordinal. Look, if you keep insisting on putting your OWN terminology over set theory, we just keep getting more and more confusions. If we're talking about T-theory, then fine, define your own T-terms. But in set theory, w is NOT the successor of w. MoeBlee
From: Tony Orlow on 20 Sep 2006 16:04 MoeBlee wrote: > Tony Orlow wrote: >> Omega is successor to N > > No, it is not not. Why do you keep repeating your error? In the limit ordinal sense, it's what comes after the set of finite naturals. > >>>> and any ordinal being >>>> the set of all preceding naturals, omega is the set of all preceding >>>> naturals. It's larger than all naturals. DO you disagree with that >>>> simple statement? >>> It contains terms whose meaning we do not agree on. >>> >>> What is the test for "largeness"? >> Number of successions, in this case, resulting in the value. A larger >> number of successive increments produces a larger value. > > w is larger than any natural number where 'larger' is taken in the > sense of the standard ordinal ordering. But w is not larger in the > sense you just mentioned, since w is not any number of successions away > from any natural number. No, there is no such number in any theory including omega. You'd need uncountable sequences. > >> Can you really not say that an infinite count is greater than any finite >> count? That seems to be the basis for omega. > > What does "basis for" mean? I mean the justification for the limit ordinals as being the next greater thing than any of the finite successor ordinals, considered a closed set. > > 'w' has a definition. That definition is not that of "having infinite > count greater than any finite count". That's a theorem based on the logic that if there were any finite size to the set, then one could come up with a finite natural not in the set, and so the set size must be greater than any finite size. Yesno? > >> There are a number of ways to do that. "Larger" can mean "farther from >> 0", geometrically. > > So from undefined T-terminology we move on to yet more undefined > T-terminology. > > MoeBlee > (sigh)
From: MoeBlee on 20 Sep 2006 16:09 MoeBlee wrote: > No, it is not the SUCCESSOR of the least unbounded ordinal. It IS the > least unbounded ordinal. Just to be clear, w IS bounded with regard to ordinals, since w is greater as an ordinal than any member of w. But w is not bounded with regard to the standard ordering of naturals, since there is no member of omega that is greater than or equal to all members of w. 'bounded' is relative to some particular ordering. MoeBlee
From: stephen on 20 Sep 2006 16:06
Han.deBruijn(a)dto.tudelft.nl wrote: > stephen(a)nomail.com wrote: >> Is sqrt(2) a number? > Yes. > Han de Bruijn Really?? Below is the definition you have apparently embraced: > *The number* is each of its representations. The number 3 > 1) either is realized by a fundamental set like the following > fundamental set of 3: > {III, Dik, {a,b,c}, {father, mother, child}, {sun, moon, earth}, > ...} > 2) or is completely determined by a series of digits > like 3.000000 or 3.00 or 3 or 3/1 or 6/2. (6/2 can also be interprtede > as an exercise.) So what is a fundamental set of sqrt(2)? What series of digits completely determines sqrt(2)? You are in disagreement with WM, who has clearly stated that sqrt(2), pi, e and all the irrationals are not numbers. Stephen |