Wikipedia:Reference desk/Archives/Mathematics/2009 August 25

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August 25

Topology

Is there a simple English equivalent meaning for the word "topology" (even into a simple sentence)? Because I'd like to make a related meaning in Arabic beside the Arabic use of Latin name. If not it is still Ok. Thanks, --Email4mobile (talk) 08:31, 25 August 2009 (UTC)

To be precise, the word topology is from Greek, not Latin. The Latin expression would be analysis situs (yet with half Greek origin). --79.38.22.37 (talk) 08:37, 25 August 2009 (UTC)

Thanks for the correction, but can you give me an alternative meaning in English?--Email4mobile (talk) 09:02, 25 August 2009 (UTC)

See -ology. "topos" means place, and "logos" means "word". By analogy,

• biology is the study/theory/science of life (more literally, the word or talk about life)
• sociology is the study of society/social structure

so topology is the theory of "place" or "space". As -ology explains, X-ology can also a subject itself, not only its study. See network topology for example.

But please remember that Wikipedia is not a place where words should be invented.

--Aleph4 (talk) 09:37, 25 August 2009 (UTC)

Inner model theorists are topologists, because they study mice.

(It's sort of a pun.) --Trovatore (talk) 09:46, 25 August 2009 (UTC)

In Tuscany there is also a coarser pun, linked to the jargon word "topa" --pma (talk) 10:02, 25 August 2009 (UTC)

(ec) Do you mean, an other meaning of the word "topology"? As far as I know, the word "topology" is used in maths only, with these meanings: a) the branch of maths devoted to the study of topological spaces; b) the family of all open sets of a topological space: "the Euclidean topology of the real line is generated by all open intervals"; c1) sligthly more generally, the complex of all topological properties of a space: "a horned sphere has a weird topology"; c2) sometimes with a point of view of algebraic topology: "a disk has no topology"; "the projective space has a rich topology"; "a non-orientable manifold has no topology at the top level".--pma (talk) 09:52, 25 August 2009 (UTC)

Thank you all for explanation. Though, topology is also used outside Maths fields; for example: network topology. I'm not sure if the term "the study of space" or "the theory of space" would be the nearest meaning). How about "preserved space"?--Email4mobile (talk) 10:43, 25 August 2009 (UTC)

People probably assumed you were asking about the mathematical usage of the term, as this is the Mathematics Reference Desk. For other meanings see Topology (disambiguation). Don't see how you get to "preserved space", but for general translations (rather than mathematical meanings) you could also ask at the Language Reference Desk. Gandalf61 (talk) 11:07, 25 August 2009 (UTC)

Indeed network topology is still in the mathematical sense, even if it is more used in applied maths. "Study of space" is quite ok as informal explanation of "topology". Preserved space apparently has no clear sense. Still it is not clear if you want other meanings of the same word "topology", or other words, sinonyms or phrases with the same meaning. The original question seems to ask for "an equivalent meaning for the word topology" (the answer then would be: "topology" I guess). --pma (talk) 11:12, 25 August 2009 (UTC)

A network topology is more like a graph (in the graph-theory sense) than a topology. I suppose you could code it as a topology, but it takes a bit of work, especially to make sure you can distinguish nodes of degree less than 3. --Trovatore (talk) 19:39, 25 August 2009 (UTC)

Agreeded. Indeed I just meant that network topology is still a mathematical term, but you are very right that graph-theory is a useful category in topology. --pma (talk) 11:14, 27 August 2009 (UTC)

For a lot of uses "connectivity" or "study of connectivity" has a very similar meaning. You could clarify(?) by calling it the "abstracted connectivity"...maybe...83.100.250.79 (talk) 11:40, 25 August 2009 (UTC)

...uhm maybe not, "connectivity" seems a bit obscure as a term of current language, and not quite appropriate here as a mathematical term. --pma (talk) 11:14, 27 August 2009 (UTC)

I used this online translator and got طوبولوجيا. It seems okay, when I translated it back from Arabic to English it came up with topology. It even translated طوبولوجيا from Arabic into Spanish as topología which is correct. ~~ Dr Dec (Talk) ~~ 13:54, 25 August 2009 (UTC)

Indeed if you switch languages from English to Arabic (العربية) at the en.wikipedia page topology, you immediately get طوبولوجيا at the ar.wikipedia page. But I fear this is the information that the OP needs to know the less... --pma (talk) 15:06, 25 August 2009 (UTC)

Sorry, but what's an OP? ~~ Dr Dec (Talk) ~~ 15:14, 25 August 2009 (UTC)

OP=Original post/Original poster. It was clear that طوبولوجيا is a transliteration (and that of course the OP (=Email4mobile) knows it very well) --pma (talk) 15:34, 25 August 2009 (UTC)

The word "طوبولوجيا" already exists in the article but what you aren't aware of is that it is the literal translation of "Topology". "طوبولوجيا" is not an original Arabic word and not even containing a similar word. We use this kind of translation when we can't get any Arabic alternative. For this reason I said in the beginning it would be Ok if there is no alternative. —Preceding unsigned comment added by Email4mobile (talk • contribs) 15:13, 25 August 2009 (UTC)

I see. Well then I doubt very much that there will be an Arabic word for it. In English and all other Eurpoean languages (using the Roman alphabet) the word is a loan word from Greek: Topology in English, Topología in Spanish, Topologie in French, Topologia in Italian, etc. Even in other languages, like Russian, it's a loan word: Топология. They are all more or less just transcriptions of one another. ~~ Dr Dec (Talk) ~~ 15:20, 25 August 2009 (UTC)

You mean transliteration ;) --pma (talk) 15:38, 25 August 2009 (UTC)

I wrote transliterations in the first place, but then after reading the article on transliteration decided that transcription might be better. Oh well... ~~ Dr Dec (Talk) ~~ 15:42, 25 August 2009 (UTC)

"Topology" describes what elements of a set or system are connected to each other, for whatever meaning of "connected" you want to apply to that system. For example, you could describe a topology on the system of Wikipedia articles, in which two articles are connected if any individual editor has edited both of them. Why not look up the Arabic word for topology (I'm sure there is one) and use it? 67.122.211.205 (talk) 05:30, 26 August 2009 (UTC)

"The study of shape" or "the study of shapes" is a good description of topology (as a field studied by mathematicians, rather than as a technical mathematical object). The Oxford English Dictionary has "The branch of mathematics concerned with those properties of figures and surfaces which are independent of size and shape and are unchanged by any deformation that is continuous, neither creating new points nor fusing existing ones". It is related to geometry (هندسة رياضية؟) --- the difference is that in geometry, we worry about exact distances and angles, while in topology, we're allowed to bend things and wiggle them around.

By the way, I took your original question to be asking "I would like to include a short explanation in Arabic of the meaning of the English word 'topology'. If someone can provide me with a short explanation in English, I will translate it into Arabic and be happy." There seems to have been some confusion about this, but I found your request quite clear. Tesseran (talk) 08:55, 27 August 2009 (UTC)

The real numbers and subsets of the natural numbers

The natural numbers are countably infinite and are the smallest infinite set, i.e. they have cardinality ℵ₀. The set of all subsets of the natural numbers has cardinality 2^ℵ₀: assign to each natural number a 0 or a 1 depending on whether the element is in or out of a given subset. Now, we can also show that the cardinality of the real numbers, i.e. the cardinality of the continuum is exactly 2^ℵ₀. The proof I have seen is a proof by contrandiction. Does anyone know of an explicit proof which finds a bijection between the real numbers (or an open subset thereof) and the set of all subsets of the natural numbers? ~~ Dr Dec (Talk) ~~ 11:21, 25 August 2009 (UTC)

I've seen a "proof" that puts a number into binary. Then for each string of 0s and 1s we get a subset of the natural numbers. For example 101 ~ {0,2} and 111101 ~ {0,1,2,3,5}. But this seems to be giving a bijection between the natural numbers and the set of all subsets of the natural numbers... that can't be right! Any ideas?! And besides, what about numbers numbers in the interval [0,1]? Won't we get some overlap? Is it even a bijection? I should stick to my own field and stop playing with fire. My head hurts! ~~ Dr Dec (Talk) ~~ 11:30, 25 August 2009 (UTC)

• When representing natural numbers in binary each natural number is a finite string. So you get a bijection between the natural number and the finite subsets of the natural numbers. Taemyr (talk) 11:37, 25 August 2009 (UTC)
• (ec)You only get finite sets that way. Finite subsets of natural numbers are countable. I suspect what you (miss-)remember is the proof that reals are uncountable (by trying to enumerate binary values from [0,1] and showing via diagonalization that that fails). --Stephan Schulz (talk) 11:45, 25 August 2009 (UTC)

Ah, I think you're right about that one! ~~ Dr Dec (Talk) ~~ 13:36, 25 August 2009 (UTC)

Oh, okay, cool. Do you have any idea of an explicit bijection from the real numbers to the set of all subsets of the natural numbers? ~~ Dr Dec (Talk) ~~ 11:42, 25 August 2009 (UTC)

Well, you can map each real in binary to a set of natural numbers (n is in the set if the nth bit of the binary representation of real number r is 1). Real numbers have infinite representations (the few with finite ones can be extended with zeros to infinity). --Stephan Schulz (talk) 11:59, 25 August 2009 (UTC)

Nice, straightforward idea but sadly not 1-1, as {1} maps to 0.1₂ and {2,3,4...} maps to 0.0111...₂ and these binary expansion represent the same real number, namely 1/2. Gandalf61 (talk) 12:25, 25 August 2009 (UTC)

True, but it is 1-1 if we only consider mapping infinite binary expansions to infinite subsets of natural numbers.

There probably isn't any pretty bijection between an interval of real numbers and the set of all subsets of natural numbers. That doesn't mean it can't be done, though. For instance, let the above bijection be Y=f(X), where X is a real number, and Y is an infinite subset of natural numbers. We define g(X)=f(X) if X cannot be represented as 1/n for some natural number n. For those numbers, letting n range from 1 to infinity, we define g(1/2n)=f(1/n), and g(1/(2n-1)) to be an enumeration of finite subsets of natural numbers. If we define g(0) to be the empty set, then g is a bijection between [0, 1] and the set of all subsets of natural numbers. --COVIZAPIBETEFOKY (talk) 12:50, 25 August 2009 (UTC)

My suggestion above is simple enough. To get around the 0.1*=1 problem, just define a unique representation for each rational number (by forbidding the ones that ends in 1*). Remember, we start with the numbers, not the strings. The problem I found by now, however, is that it does not map R, but only [0;1[ (or equivalent intervals). But it should be possible to get around this somehow by using a bijection from R to [0;1[. I think x->(1/(x+1)+0.5) (if positive or 0), x->(1/(-x+1)) (if negative) should do the job. --Stephan Schulz (talk) 14:14, 25 August 2009 (UTC)

So you represent 1/2 uniquely by 0.1, then map this to {1} - but now you have no real number that maps to {2,3,4,...}. So some subsets of natural numbers do not have a pre-image, and you still don't have a bijection. (Not saying this can't be patched up somehow to make it into a bijection - just saying it's not as simple as you originally said). Gandalf61 (talk) 15:55, 25 August 2009 (UTC)

Point taken. --Stephan Schulz (talk) 15:59, 25 August 2009 (UTC)

Another example: To each infinite subset of the natural numbers, say $0\leq a_{1}<a_{2}<\cdots$ assign the sequence $n_{1}=a_{1},n_{2}=a_{2}-a_{1},n_{3}=a_{3}-a_{2},\ldots$ of positive natural numbers; interpret this sequence as a continued fraction, and you will get a positive irrational number ${n_{1}+{\frac {1}{n_{2}+{\frac {1}{\ddots }}}}}$. Every positive irrational is obtained in this way. This is a (reasonably nice, I think) bijection between the (positive) irrationals and the infinite subsets of the natural numbers.

All that remains is to map the finite subsets of the natural numbers bijectively onto the rational positive numbers. As both these sets are countable, this is easy to do, but it can even be done "nicely", again using continued fractions.

--Aleph4 (talk) 22:08, 25 August 2009 (UTC)

On a different, but related topic. Is there anyway of giving a cananonical example for a set with cardinality ℵ_n for some natural number n? ~~ Dr Dec (Talk) ~~ 16:43, 25 August 2009 (UTC)

More or less by definition, ℵ_n is the cardinality of the set of ordinals of cardinality at most ℵ_n-1. Algebraist 16:46, 25 August 2009 (UTC)

Could you please explain that in a little more detail, if you don't mind? The ordinal article says that an ordinal are is the order type of a well-ordered set. So they are numbers. So the cardinality of an ordinal will be 1. So all ordinals have cardinality less than ℵ₀. ~~ Dr Dec (Talk) ~~ 17:00, 25 August 2009 (UTC)

(if you didn't mean to make a joke) "the cardinality of the ordinal x", refers to x as an ordered set, e.g. the set of all smaller ordinals, according to the von Neumann representation. --79.38.22.37 (talk) 17:18, 25 August 2009 (UTC)

A standard definition for ordinal numbers is very similar to the one for natural numbers, which defines each natural number as the set of all natural numbers less than it (where 0 is the empty set). That is, an ordinal number is often defined as the set of all ordinal numbers less than it. The finite ordinal numbers are natural numbers. The first infinite ordinal number is the set of all natural numbers. The second infinite ordinal number would be the set containing all natural numbers and the set of all natural numbers. And so on. --COVIZAPIBETEFOKY (talk) 17:12, 25 August 2009 (UTC)

Hold on, forgot to make my point (I guess you got it, but I'll put it explicitly anyways). So, the first infinite ordinal number has cardinality equal to ℵ₀, and so do many many many more after that. However, there will eventually be an ordinal number with cardinality ℵ₁, and ℵ₂, and so on. --COVIZAPIBETEFOKY (talk) 17:15, 25 August 2009 (UTC)

I see! ~~ Dr Dec (Talk) ~~ 17:15, 25 August 2009 (UTC)

Two more examples of sets of size aleph1:

• Example 1: Let P the set of all subsets of the rational numbers. (So P has size continuum.)

On the set P define an equivalence relation E by declaring two sets A and B (both subsets of the rationals) to be in relation E if they are order isomorphic (i.e., there is an order preserving bijection from A onto B), or neither of the two is a well-order. (For example, the sets {0,1,2} and {-1, 1/2, 7} are in relation E.)

The set of E-equivalence classes has cardinality aleph1. This is easy to see as each equivalence class (except for the class of non-well-ordered sets) naturally and bijectively corresponds to a countable ordinal. (So this example is just a thinly disguised version of the first uncountable ordinal, mentioned above.)

• Example 2: Let P be the same set as above. Now define a relation L as follows:

(A,B) is in L iff there is an order preserving embedding from A into B, and there is an order-preserving embedding of B into A.

Laver's theorem says that there are exactly aleph1 equivalence classes. ("Fraisse's conjecture"; highly non-trivial.)

--Aleph4 (talk) 21:19, 25 August 2009 (UTC)

Is this correct?

Resolved

 – Accdude92 already has the correct answer

Find F(x)+g(x) F(x)=X/X+2 G(X)=2X=3

I got 2x(squared)+8X+6\X+2 Accdude92 (talk) (sign) 14:12, 25 August 2009 (UTC)

I'm sorry, but I don't understand the problem. Do you mean find F(x) + G(x) when

$F(x)={\frac {x}{x+2}}\ ,$

$G(x)=2x-3?\,$

In that case it's just addition of fractions and some simplification. I get

$F(x)+G(x)={\frac {2(x^{2}+x-3)}{x+2}}\ .$

If x and X are different and g and G are different then I have no idea. ~~ Dr Dec (Talk) ~~ 14:18, 25 August 2009 (UTC)

Oh, I see, you have G(x) = 2x + 3, and in this case you're right:

${\frac {x}{x+2}}+(2x+3)={\frac {2(x^{2}+4x-3)}{x+2}}\ .$ ~~ Dr Dec (Talk) ~~ 14:20, 25 August 2009 (UTC)

Darn keyboard. Its really Find F(x)+g(x) F(x)=X/X+2 G(X)=2X+3 —Preceding unsigned comment added by Accdude92 (talk • contribs) 14:21, 25 August 2009 (UTC)

fibonacci spiral

for what value of $\omega$ does $|z|=re^{\omega arg(z)}$ plot a fibonacci spiral in the argand plane? —Preceding unsigned comment added by 92.1.60.82 (talk) 23:55, 25 August 2009 (UTC)

See Golden spiral. Black Carrot (talk) 06:09, 26 August 2009 (UTC)