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August 27

A weird thing

Consider the set of all continuous functions f:[0,1]->R in the sup metric, i.e.

d(f,g)=sup{|f(x)-g(x)|}

Show that this space has a countable dense subset and therefore has a countable basis.

The proof is very easy the book gives a hint (which I won't say for those of you who want to do it yourselves), but immediately I noticed something WEIRD.

Since one point sets in hausdorff spaces are closed, there are at least as many open sets as points in the space (x->({x} complement)). Now the above space has a countable base so there are as many continuous function as real numbers. Is this true??? It's totally outrageous!!! Standard Oil (talk) 05:45, 27 August 2009 (UTC)[reply]

Why is it outrageous? It seems perfectly reasonable. I feel like there should be a simple elementary proof but I'm spacing on it right now. 67.122.211.205 (talk) 06:18, 27 August 2009 (UTC)[reply]

A continuous function on [0,1] is determined by its values at the rational points, so the cardinality of C([0,1],R) is c . There are plenty of relevant countable dense sets of this space, e.g. polynomial functions with rational coefficients, or even simpler, piecewise affine functions with rational entries. --pma (talk) 06:52, 27 August 2009 (UTC)[reply]

Oh yea polynomial functions with rational coefficients you mean Stone-Weierstrass theorem right? Sorry I'm not a math Ph.D and I find these new things very interesting. And the proof is indeed the piecewise affine functions with rational entries, if you mean straight lines joining points with both entries being rational Standard Oil (talk) 17:42, 27 August 2009 (UTC)[reply]

Uncountability of R

Resolved

I want to find an explicit bijection between $\mathbb {R}$ and $2^{\mathbb {N} }$ so that I can conclude that $\mathbb {R}$ is uncountable. Also what is the procedure to convert an arbitrary real number from its decimal representation to its binary representation. Thanks--Shahab (talk) 09:21, 27 August 2009 (UTC)[reply]

For the first question, see #The real numbers and subsets of the natural numbers above. —JAO • T • C 10:07, 27 August 2009 (UTC)[reply]

And for the other question check Binary numeral system#Conversion to and from other numeral systems. So let

\scriptstyle x=\sum _{k}2^{k}\epsilon _{k}

, where

\scriptstyle \epsilon _{k}\in \{0,1\}

, and

\scriptstyle \epsilon _{k}=0

for all positive k but finitely many, and for infinitely many negative k; then,

\scriptstyle \epsilon _{k}=\lfloor 2^{-k}x\rfloor \mod 2

.--pma (talk) 10:29, 27 August 2009 (UTC)[reply]

Thanks.--Shahab (talk) 11:05, 27 August 2009 (UTC)[reply]

Is .99999999999 repeating=1?

I say it gets infinitely close, but never reaches 1, but then again 9/9=1...so maybe it is=1.Accdude92 (talk) (sign) 13:13, 27 August 2009 (UTC)[reply]

What does "9/9=1" have to do with this? 17/17=1, but that doesn't make 0.17171717171717171....17=1? DRosenbach ^{(Talk | Contribs)} 15:39, 28 August 2009 (UTC)[reply]

I'd assume the point was .999... = 9 * .111... = 9 * 1/9 = 9/9 = 1. —JAO • T • C 15:46, 28 August 2009 (UTC)[reply]

I believe someone asked this question several months ago. The proof I've seen is: X = 0.999..., 10X = 9.999..., 9X = 10X - X = 9.999...-0.999... = 9, X = 1. Wikiant (talk) 13:18, 27 August 2009 (UTC)[reply]

Oh, i check the archives, and I found noting.Accdude92 (talk) (sign) 13:20, 27 August 2009 (UTC)[reply]

See 0.999... (a Featured Article). -- Coneslayer (talk) 13:29, 27 August 2009 (UTC)[reply]

To make a long story short, there is a non-standard representation for every terminating decimal number wherein the final digit is reduced by one and then an infinite string of 9s is appended (after a decimal point if initially dealing with an integer). One of the educational problems here is that the notions of converging sequence and limits are not introduced right alongside the introduction of the number line in elementary school. To say that 0.999... "gets infinitely close" to 1 is to fail to distinguish betweeen a sequence of numbers and simply a number that is the limit of such a sequence. We say that the sequence 0.9, 0.99, 0.999, ... goes to 1 in the limit, while 0.999...=1. —Preceding unsigned comment added by Julzes (talk • contribs) 14:00, 27 August 2009 (UTC)[reply]

In formal terms, 0.999... represents a limit. In this case the limit means that as you add another 9 to the decimal expansion you get closer to 1. This is true: 0.99 is closer to 1 than 0.9, 0.999 is closer to 1 than 0.99, 0.9999 is closer to 1 than 0.999, and so on. If a_n is 0.99...9 where there are n 9s after the decimal place then clearly 0 < a_n < a_n+1 < 1 for any positive whole number n. Actually

a_{n}=1-{\frac {1}{10^{n}}}\ .

So we are getting closer and closer to 1 as we add more and more 9s. Notice that every a_n has a finite number of 9s after the decimal place. Now, this is just one thing we need when we talk about limits. The next thing we need is that if we choose a really, really small distance, say ε, then we can always find an n so that a_n lies within that distance from 1. Actually we can solve: we want a_n to be within a distance of ε from 1, so we want a_n > 1 - ε (just draw a picture: 0 < a_n < 1 and we want it to be within a distance of ε of 1, well inside the interval [0,1] the point 1 - ε is a distance ε from 1. We want a_n to be closer to 1 so we need a_n > 1 - ε). Solving this gives

n>\log _{10}\left({\frac {1}{\varepsilon }}\right)=-\log _{10}\varepsilon \ .

So we can see that 0.999...9 gets closer and closer to 1 as we add more and more 9s, and that we can always add a certain number of 9s so that we get as close to 1 as we like (and if we carry on adding 9s we would get even closer!). All the time we have a finite number of 9s.

The statement 0.999... = 1 is true: I've proven it for you above. All it means is that 0.999...9 gets closer and closer to 1, and that we can get as close as we might like. We never claim that we can add, say ten trillion, 9s and finally get to 1. This is the idea of a limit. ~~ Dr Dec (Talk) ~~ 14:24, 27 August 2009 (UTC)[reply]

No, 0.999... is not "a limit", or at least not in the sense you suggest. It's one (infinite) decimal representation of the real number more usually written as 1. Please read the FA linked above. --Stephan Schulz (talk) 14:41, 27 August 2009 (UTC)[reply]

... which says, in the section Infinite_series_and_sequences (cmd-click)">Infinite_series_and_sequences (cmd-click)">0.999...#Infinite_series_and_sequences''Infinite series and sequences'':

0.999\ldots =\lim _{n\to \infty }0.\underbrace {99\ldots 9} _{n}=\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {9}{10^{k}}}=\lim _{n\to \infty }\left(1-{\frac {1}{10^{n}}}\right)=1-\lim _{n\to \infty }{\frac {1}{10^{n}}}=1.\,

I understand there are various ways of interpreting 0.999..., but surely one entirely reasonable and standard viewpoint is to interpet it as a limit. Gandalf61 (talk) 14:56, 27 August 2009 (UTC)[reply]

Exactly! The limit of my a_n is exactly the infinite string of digits of which Stephan speaks. He says that the infinite string of digits is equal to 1, I say that the limit of a_n is 1. We both come to the same final point. The difference is that defining it as a limit gives you something tangable that you can put your finger on. Where's the problem in that? ~~ Dr Dec (Talk) ~~ 15:02, 27 August 2009 (UTC)[reply]

Let me quote an authoritative source:

Sure, it's a limit. It's also a number. Why do you think a limit shouldn't be a number, or a number shouldn't be a limit? .

We should make an authomatic link to the article on 0.999.., from all questions containing the string "999..". --pma (talk) 15:13, 27 August 2009 (UTC)[reply]

In a certain setting they are interchangable. Given a sequence of real numbers (x_n) the limit, if it is defined, will be a real number. And any real number x is the limit of a sequence, namely the sequence (x_n) where x_n = x for all n. ~~ Dr Dec (Talk) ~~ 15:20, 27 August 2009 (UTC)[reply]

Hehe! Correct. But it's quite a bit funny writing here proofs that 1=0,999.. every two weeks or so ;-) (indeed, repetita juvant)--pma (talk) 15:30, 27 August 2009 (UTC)[reply]

The statement is true for ordinary real numbers but it's worth reading the 0.999... article for cases where it isn't true. For instance there's a game where each position can be given a value, but the game's value equivalent of 0.9999... is less than that for 1. Dmcq (talk) 14:41, 27 August 2009 (UTC)[reply]

It's quite simple to accept that numbers are limits and vice versa. After all, what do people think is going on when a limit is evaluated? —Anonymous Dissident^Talk 15:30, 27 August 2009 (UTC)[reply]

Hmmm, that's not exactly correct. Not every well-defined limit is a number; some limits are functions. Let f : R → R be a smooth function then

\lim _{\varepsilon \to 0}{\frac {f(x+\varepsilon )-f(x)}{\varepsilon }}

isn't a number; it's a function. ~~ Dr Dec (Talk) ~~ 15:42, 27 August 2009 (UTC)[reply]

c'mon... of course everybody here's talking of real numbers ;) --pma (talk) 15:56, 27 August 2009 (UTC) [reply]

Okay, okay... Sorry! :oP ~~ Dr Dec (Talk) ~~ 15:58, 27 August 2009 (UTC)[reply]

I never meant to imply all limits are numbers; in saying "that numbers are limits and vice versa" I meant it in the way that one might say "men drink on a Friday night", or "teenagers vandalise the car park" – some do, but not all... if that makes sense. —Anonymous Dissident^Talk 16:02, 27 August 2009 (UTC)[reply]

I know what you wanted to say, I was just being silly. But the statment "number are limits and visa versa" means that both numbers are limits and limits are numbers. ~~ Dr Dec (Talk) ~~ 16:08, 27 August 2009 (UTC)[reply]

Indeed, you're correct. I hope you don't mind, but I've appended a </small> tag to the end of your comment so the rest of the refdesk is not out of format. —Anonymous Dissident^Talk 16:14, 27 August 2009 (UTC)[reply]

Line bundles and cohomology

I've been working on trying to classify line bundles over hypersurfaces. The classification that I'm looking for is more refined than that of the Stiefel–Whitney class. Let M be a real n-dimensional smooth hypersurface embedded in Rⁿ⁺¹. For my purposes, the space of line bundles over M coincides with the quotient space $\Omega ^{1}(M)/\Lambda ^{1}(M)$ , where $\Omega ^{1}(M)$ is the space of all differential one-forms on M and $\Lambda ^{1}(M)$ is the space of all exact differential one-forms on M. Now, I want to understand this space a little better.

Let $U\subset V\subset W$ be nested vector spaces, then we construct the formal isomorphism $W/U\cong (W/V)\times (V/U).$ Now, notice that $\Lambda ^{1}(M)\subset \Gamma ^{1}(M)\subset \Omega ^{1}(M)$ where $\Gamma ^{1}(M)$ is the space of closed differential one-forms on M. We get that

\Omega ^{1}(M)/\Lambda ^{1}(M)\cong (\Omega ^{1}(M)/\Gamma ^{1}(M))\times (\Gamma ^{1}(M)/\Lambda ^{1}(M))=(\Omega ^{1}(M)/\Gamma ^{1}(M))\times H_{dR}^{1}(M)\ ,

where $H_{dR}^{1}(M)$ denotes the first de Rham cohomology class. We can make sense of the space $\Omega ^{1}(M)/\Gamma ^{1}(M)$ by noting that the exterior derivative d is such that $d:\Omega ^{1}(M)\to \Omega ^{2}(M)$ , the kernel of d is given by $\Gamma ^{1}(M)$ and the image is given by $\Lambda ^{2}(M)$ , i.e. the space of exact differential two-forms on M. It follows that $\Omega ^{1}(M)/\Gamma ^{1}(M)\cong \Lambda ^{2}(M).$ Finally, our original space $\Omega ^{1}(M)/\Lambda ^{1}(M)$ can be shown to have the following property:

 $\Omega ^{1}(M)/\Lambda ^{1}(M)\cong \Lambda ^{2}(M)\times H_{dR}^{1}(M)\ .$

My question falls into two parts:

Can anyone construct the isomorphism explicitly, or at least give some clues as to its nature?
Can anyone say anything about the topology or geometry of these two spaces?

~~ Dr Dec (Talk) ~~ 13:53, 27 August 2009 (UTC)[reply]

1. Since for plain vector spaces there is no natural isomorphism for

\textstyle W/U\cong W/V\times V/U

I'd say you need an additional structure to define an explicit one, like a scalar product. In your case M is naturally a Riemannian manifold, so I guess you'll use a bit of Hodge theory (which implies changing a bit the functional setting).

2.

\Omega ^{k}(M)

is a Fréchet space wrto the uniform convergence of derivatives of all orders; you are taking a quotient over a closed subspace, hence you still have a Fréchet space; the same for the RHS.

--pma (talk) 16:31, 27 August 2009 (UTC)[reply]

Thanks pma; I'll try to find some details on Fréchet spaces. I had heard of them before, but I've never studied them. On another point: I'm not assuming that M is a Riemannian manifold. There's really no need. My construction only requires that M has a torsion-free connexion. In fact, my work started in the affine setting (where we have a volume form on each tangent space instead of a metric), but that turned out to be too strict. There's really no need for any structure. (Although maybe the isomorphism will, in the end, depend upon the choice of connexion). ~~ Dr Dec (Talk) ~~ 16:50, 27 August 2009 (UTC)[reply]

OK, but since you took M embedded in Rⁿ⁺¹, if you want you already have a Riemann structure; and in any case for a finite dimensional manifold M you can always pick one. But I believe you, that you need less! --pma (talk) 17:21, 27 August 2009 (UTC)[reply]

What is an axiom?

And yes, I have looked at the relevant Wikipedia article, so I'll be more specific with my confusion...

What is the distinction between an axiom and a definition. For example, if I define a new function $f(x)=x^{2}+x$ , have I introduced an axiom? Or, perhaps a slightly different example, do the complex numbers depend on any additional axioms to the reals? Or even, when defining a particular group (as opposed to defining what groups are), am I introducing axioms? The reason I ask is that I heard someone saying that it is a mistake to say that the complex numbers depend on any additional axioms [to the reals] which led me to reconsider what I understood the word to mean.--Leon (talk) 16:06, 27 August 2009 (UTC)[reply]

The way I understand it, the key difference between an axiom and a definition is that an axiom is a proposition or base from which other abstractions and definitions stem. Such an axiomatic proposition is seen to be self-proving or -evident; unable to be disproved. I guess axioms are the logical principles that underscore what makes good mathematical sense – what makes 1+1=2. Don't take my word for it, though. —Anonymous Dissident^Talk 16:12, 27 August 2009 (UTC)[reply]

That makes sense—to some extent. But it's difficult to see from what a definition does "stem out". And how can definitions be disproved?--Leon (talk) 16:18, 27 August 2009 (UTC)[reply]

As my dictionary says: An axiom is "a proposition regarded as self-evidently true". A definition is "a statement of the exact meaning of a word or the nature or scope of something." An axiom gives you (something assumed to be) a fact; a definition gives you meaning. Clearly ~~some~~ all definitions can be axioms, but not all axioms can be definitions. ~~ Dr Dec (Talk) ~~ 16:21, 27 August 2009 (UTC)[reply]

Then how are the group axioms "self-evidently true", for instance? And in my above examples, do any of the constructions I speak of introduce new axioms?--Leon (talk) 16:29, 27 August 2009 (UTC)[reply]

They would have been four facts about certain mathematical structures which later became known as groups. We've been talking about the problems of axiomatic mathematics above. Your question is a matter of definition. Axioms talk about properties, definitions give definitions: We say that an A is a B if it has the properties {p₁, …, p_n}. (The resulting axiom would be that an A with the properties {p₁, …, p_n} is a B. ~~ Dr Dec (Talk) ~~ 16:37, 27 August 2009 (UTC)[reply]

I think there is little, if any, difference in underlying meaning, and use of one term rather than the other is largely dictated by custom and tradition. For example you could define the properties of evenness and primeness on the natural numbers, then you could define an evenprime number as a natural number that is (a) even and (b) prime. In this context, the properties of evenness and primeness are axioms - they are the evenprime axioms. You could then show that only one natural number, namely 2, satisfies the evenprime axioms. Now you can turn this process on its head and define 2 as the unique evenprime natural number - and you have an axiomatic definition of 2. Gandalf61 (talk) 16:42, 27 August 2009 (UTC)[reply]

(ec) Leon's objection is right. "A proposition regarded as self-evidently true" is not very good as definition of axiom, with the due respect to OED. I happened to write it above; the word axiom is from the Greek word ἀξίωμα, and may be explained as "what makes something worth (ἄξιος) of being given a certain name". E.g. the axioms of groups explain what properties are needed to be worth of being called a group. There is nothing of self-evident, although of course people try to choose axioms as natural and elementary as possible. A definition is also of more local and particolar use (e.g. like in Leon's example, when one sais : "let us define f(x)=x²+x.."). --pma (talk) 17:09, 27 August 2009 (UTC)[reply]

I quoted the OED because I thought that the difference between axiom and definition could be resolved by the use of a dictionary. Axiom, from the Greek axiōma meaning what is thought fitting. The OED translation, IMHO, cuts staight to the heart of the issue. ~~ Dr Dec (Talk) ~~ 17:17, 27 August 2009 (UTC)[reply]

Also, you're not going to write down the group axioms unless your have a good idea of what you're about to call a group is. As such, when you write down the group axioms you're going to write down self-evident truths about these structure. People didn't invent groups; they discovered them. Moreover, they didn't invent them by writing down some random axioms and saying "Right, put the kettle on, let's see where these take us." ~~ Dr Dec (Talk) ~~ 17:23, 27 August 2009 (UTC)[reply]

Sure, so what? well, no matter...--pma (talk) 18:30, 27 August 2009 (UTC)[reply]

Well, if my last comment was so trivial as to require an answer of "so what?", then surely my original point is quite clear. ~~ Dr Dec (Talk) ~~ 19:18, 27 August 2009 (UTC)[reply]

First: thanks guys for your help!

Second, in the cases of my particular examples, can you tell me if those constructions constitute invocations of further axioms? By further, I mean ones on top of the real line and/or the group axioms?--Leon (talk) 17:19, 27 August 2009

The way I was taught was that a definition introduces a new term (without the definition a term would be meaningless), whereas an axiom uses all previously defined terms, but puts them together in a way that is not self-evident (provable) from their definitions. -- 128.104.112.102 (talk) 17:29, 27 August 2009 (UTC)[reply]

In consumer behavior modeling, we make the distinction between "external validity" and "internal validity." Roughly speaking, a statement is internally valid if it is consistent with truths within the confines of the model. A statement is externally valid if it is consistent with truths external to the model. It appears (to me) that the distinction between axiom and definition is similar. Within the confines of a model, I can define 1+1 to equal 3 and then go on to make other statements that would be true *given* the definition (i.e., I establish internal consistency). However, the definition is not consistent with truths external to the model. Hence, I do not have external consistency. In short, it seems to me that it is sufficient that definitions achieve internal consistency, but necessary that axioms achieve external consistency. Wikiant (talk) 17:25, 27 August 2009 (UTC)[reply]

As I understand it, a definition just gives a more succinct label to a concept that we can already express. For example if we define the complex numbers as the quotient space R[x]/(x² + 1), that lets us talk about the concept using the term "complex numbers" instead of having to explain "the quotient space R[x]/(x² + 1)" every time we want to refer to that space. Axioms declare properties of or relations between objects we've already defined that we want to take as true but that don't follow from their definitions. However I don't see why you couldn't lump most axioms about an object into the definition of the object so I'm not really sure how clear the distinction always is. I don't know axiomatic set theory, but from the naive foundations that get you to the real numbers you can get to the complex numbers just with definitions. AFAIK you should be able to get to pretty much all the math we know and love just with definitions once you accept the underlying axioms, which is why those axioms are chosen. I'm not really sure though what would prevent us from "defining" the complex numbers completely from scratch with a definition that includes all the axioms. Rckrone (talk) 19:23, 27 August 2009 (UTC)[reply]

DISAMBIGUATION:

In epistemology, an axiom is a proposition, expressible as a complete sentence, that is self-evidently true, in the sense that you can't understand what it says until you know the fact that it asserts, and that is at the base of knowledge in that other knowledge depends on it. See self-evidence. My statement that I am conscious is a self-evident proposition. Your statement that I am conscious is not a self-evident proposition.
In mathematics, an axiom is something else entirely. Not unrelated, though.

Michael Hardy (talk) 20:33, 27 August 2009 (UTC)[reply]

Part of this is a creep in meaning over time. To Euclid, mathematical (geometric) axioms were indeed self-evident. It is a relatively recent development to call any assumed statement an "axiom". To the Greeks, this would have been absurd. It took at least the development of non-Euclidean geometry in the 19th century before there was general awareness that distinct, reasonable axiom systems could contradict each other. Our current understanding of formal logic is not even that old. — Carl (CBM · talk) 00:06, 28 August 2009 (UTC)[reply]

Diophantine variances

For certain pedagogical purposes, I want to fiddle around with some variances of small-integer-valued random variables without getting distracted by the arithmetic involved in actually computing the variances.

For that reason, it could be convenient if simultaneously

The values of the random variables are small integers;
The expected values are small integers;
The variances are small integers. (Don't worry about the standard deviations. In other words, the varinces need not be squares of integers.)

Is there a way—maybe even an algorithm?—for producing such things? Michael Hardy (talk) 20:38, 27 August 2009 (UTC)[reply]

If I have this right, you're looking for a sequence of integers whose sum is 0 and whose sum of squares is divisible by the number of elements in the sequence? Are there any other constraints? If not it should be pretty easy to massage the values of an arbitrary sequence to make it work. For example if you had any k numbers with a sum of 0 and a sum of squares n > k, you could tack on n-k zeros to your sequence and you'd be done. Then translate the whole thing if you want some other mean besides zero. Rckrone (talk) 21:15, 27 August 2009 (UTC)[reply]

Well, of course there are other "constraints" if you mean things not precisely definable, but recongizable by people with common sense. Just the usual stuff: don't make the whole thing too complicated, don't add coincidences from which naive students may make inductive leaps, don't add things likely to distract attention from where you want it, etc. I'll fiddle with this more and see what I think. Michael Hardy (talk) 02:13, 28 August 2009 (UTC)[reply]

....to clarify further: If one had an algorithm that would churn these out by the truckload, then one would look among the results for cases that looked good from the point of view of the other "constraints". So if you can provide such an algorithm, that would take care of everything except what I would need to do myself because only I know the whole story of the context in which I would use these. Michael Hardy (talk) 02:53, 28 August 2009 (UTC)[reply]

The probability distribution f(x) = (4−x)/10 for x = 0,1,2,3 has mean value 1 and variance 1. Bo Jacoby (talk) 21:40, 27 August 2009 (UTC).[reply]