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March 10

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Extending proof to uncountable subsets of R

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Show that the following theorem is equivalent to the completeness axiom (i.e. that every bounded subset of the real numbers has a least upper bound):

Any nondecreasing sequence has a limit.

Going in the direction "limit => completeness axiom", any proof that I can think of will show that the set has a least upper bound, and therefore that any countable subset of the real numbers has a least upper bound. How do you extend this to uncountable subsets? Widener (talk) 00:34, 10 March 2012 (UTC)[reply]

Hint: a set bounded above but without a least upper bound will generate a nondecreasing sequence. Rschwieb (talk) 02:13, 10 March 2012 (UTC)[reply]
Use the density of the rationals.--121.74.121.82 (talk) 05:09, 10 March 2012 (UTC)[reply]
Am I missing something? a nondecreasing sequence like an=n does not have a limit. Shouldn't that be Any bounded nondecreasing sequence has a limit? 84.197.178.75 (talk) 18:21, 10 March 2012 (UTC)[reply]
Yes, that is what I meant. (talk) 10:47, 11 March 2012 (UTC)[reply]
I think it's true that any uncountable bounded subset of can be formed by taking a countable subset of with least upper bound and adding an uncountable number of reals to that subset. As long as every real number added is less than or equal to , will again be the least upper bound. Widener (talk) 11:51, 11 March 2012 (UTC)[reply]
Hmm interesting question, had me thinking for a while. Like one of the above posters said, the separability of the real numbers (i.e. density of rationals) is essentially here. Let A be any set with upper bound. If it had a greatest element it would have an sup, so suppose it has no greatest element. Take the set K of rationals in A; obviously for each x in A there's some y in K such that x is less than y. Now enumerate K as x1,x2..., and define a new sequence y1,y2... as follows. Let y1 be any point of K, and inductively let yn be any point of K such that yn is bigger than y(n-1) and x1,...,x(n-1) (I'll let you verify why such a yn always exist). The sequence y1,y2,... is non decreasing, bounded, and for every x in A there's some yn that's bigger than x. The limit of this sequence is the sup of A.
Just wondering, can anyone give a totally ordered set such that it doesn't have the least upper bound property but every non decreasing sequence converges? Money is tight (talk) 14:37, 11 March 2012 (UTC)[reply]
.--130.195.2.100 (talk) 03:01, 12 March 2012 (UTC)[reply]
This isn't necessarily obvious though. Is there a proof of this? Widener (talk) 05:16, 12 March 2012 (UTC)[reply]

What is the eccentricity of the following conic section? (666*x)^2-(666*y)^2+(x+y+666^2)=2012 123.24.112.17 (talk) 16:56, 10 March 2012 (UTC)[reply]

From the cartesian coordinates representation in Conic section article, I get A=666^2, B=0 and C=-666^2; so B*B-4*A*C > 0 making it a hyperbola, and since A+C=0 its a rectangular one. And Hyperbola mentions that for a rectangular hyperbola, the eccentricity is . But I'm not familiar with the subject so I can't tell if this is correct... 84.197.178.75 (talk) 18:49, 10 March 2012 (UTC)[reply]

Big or Little

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When I'm writing out a power series in a single variable, I use the big O notation for the truncated terms, don't I? For example:

Does the point of evaluation need to be mentioned/taken into account for non-entire functions? Fly by Night (talk) 19:00, 10 March 2012 (UTC)[reply]

Yes, you'd use O here. The function is O(x3) for x → 0, but not o(x3).
The notation O(x3) is only meaningful if "for x → 0" or something similar is specified or understood from context. Of course, if you were dealing with xa for some nonzero a, then O(x3) would mean the same as O(1), so there would be no point using O(x3), and it's unlikely you'd be referring to a nonzero a here. These remarks are applicable to any function with an asymptotic expansion, entire or not. 96.46.204.126 (talk) 19:49, 10 March 2012 (UTC)[reply]
It's unlikely to use for with , but it's very likely to use it for . -- Meni Rosenfeld (talk) 21:23, 10 March 2012 (UTC)[reply]
Thank you. I made a mistake in not considering the case a=±∞. 96.46.204.126 (talk) 19:38, 12 March 2012 (UTC)[reply]
See Big O notation section Usage: Infinitesimal asymptotics for this exact example. 84.197.178.75 (talk) 19:56, 10 March 2012 (UTC)[reply]