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March 10

Extending proof to uncountable subsets of R

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 $\{a_n\}$ has a limit.

Going in the direction "limit => completeness axiom", any proof that I can think of will show that the set $\{x\in ℝ:\mathrm{\exists }n\in \mathbb{N}:x=a_n\}$ 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)

Hint: a set bounded above but without a least upper bound will generate a nondecreasing sequence. Rschwieb (talk) 02:13, 10 March 2012 (UTC)

Use the density of the rationals.--121.74.121.82 (talk) 05:09, 10 March 2012 (UTC)

Am I missing something? a nondecreasing sequence like a_n=n does not have a limit. Shouldn't that be Any bounded nondecreasing sequence $\{a_n\}$ has a limit? 84.197.178.75 (talk) 18:21, 10 March 2012 (UTC)

Yes, that is what I meant. (talk) 10:47, 11 March 2012 (UTC)

I think it's true that any uncountable bounded subset of $ℝ$ can be formed by taking a countable subset of $ℝ$ with least upper bound $L$ and adding an uncountable number of reals to that subset. As long as every real number added is less than or equal to $L$, $L$ will again be the least upper bound. Widener (talk) 11:51, 11 March 2012 (UTC)

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)

${\omega }_1+{\omega }^{*}$.--130.195.2.100 (talk) 03:01, 12 March 2012 (UTC)

This isn't necessarily obvious though. Is there a proof of this? Widener (talk) 05:16, 12 March 2012 (UTC)

Eccentricity of a conic section

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)

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 $\sqrt{2}$. 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)

Big or Little

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:

$e^x=1+x+\frac{x^2}{2}+O(x^3).$

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)

Yes, you'd use O here. The function $e^x-1-x-\frac{x^2}{2}$ 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 x → a 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)

It's unlikely to use $O(x^3)$ for $x\to a$ with $0<a<\mathrm{\infty }$, but it's very likely to use it for $x\to \mathrm{\infty }$. -- Meni Rosenfeld (talk) 21:23, 10 March 2012 (UTC)

Thank you. I made a mistake in not considering the case a=±∞. 96.46.204.126 (talk) 19:38, 12 March 2012 (UTC)

See Big O notation section Usage: Infinitesimal asymptotics for this exact example. 84.197.178.75 (talk) 19:56, 10 March 2012 (UTC)