Mathematics desk
< March 10	<< Feb \| March \| Apr >>	March 12 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 11

Topology

I was curious about the definition of a topology. Going by the article, it says after the definition that "All sets in T are called open; note that not all subsets of X are in T".

Considering this:

$X=\lbrace 1,2,3\rbrace \,$

$T=\lbrace \emptyset ,\lbrace 1\rbrace ,\lbrace 2\rbrace ,\lbrace 3\rbrace ,\lbrace 1,2\rbrace ,\lbrace 1,3\rbrace ,\lbrace 2,3\rbrace ,\lbrace 1,2,3\rbrace \rbrace$

1) Isn't T equivalent to the powerset of X? Or is this just the case for simple examples like this?

2) Isn't "not all subsets of X are in T" incorrect in the above example? Damien Karras (talk) 09:24, 11 March 2008 (UTC)[reply]

It means 'not all subsets of X need be in T'. If they all are, we call T the discrete topology on X. Algebraist 10:31, 11 March 2008 (UTC)[reply]

Ok, I think I get you. Apologies for the informality, I'm trying to establish a "visual" ideal of what a topology is. Starting with a set of points (ordered pairs (m,n)) on a plane. If we have a set of points, T within X that are bounded within say, a circle, we can have a smaller subset of points not in T - i.e. a 2 dimensional projection of a torus (or a topologically equivalent shape)? Damien Karras (talk) 12:28, 11 March 2008 (UTC)[reply]

T=\lbrace \emptyset ,\mathbb {R} ^{2},\lbrace (x,y)|x^{2}+y^{2}<1\rbrace \rbrace

is a valid topology on the plane. Does that answer your question? Any collection of subsets satisfying the axioms of a topology is a topology. --Tango (talk) 12:52, 11 March 2008 (UTC)[reply]

Also note that that is only one way of defining it; some texts take T as a family of neighborhoods and derive the open sets from there. GromXXVII (talk) 11:35, 11 March 2008 (UTC)[reply]

Well, a topology just describes what open sets look like in your space (call it X). In general, a topology will not include all possible subsets of your space X. The power set of X is one of the possible topologies on X but it is not the only one. So in the power set, all subsets are open but in a general topology, this may not be true. For example, here is another topology on X, {the empty set, X} with only two elements. You can verify the axioms of a topology and prove that this is a perfectly valid topology (called the trivial topology). In fact this is the smallest topology and the power set topology is the largest topology in the sense that any other topology will be contained in the power set topology and contains the trivial topology. Another not obvious example of a topology would be that a set in R is open if it contains the number zero. Obviously, not all sets are open under this topology. Hope this helps!A Real Kaiser (talk) 04:55, 12 March 2008 (UTC)[reply]

It says in the article "all sets in T are open", yet what about

T=\lbrace \emptyset ,\mathbb {R} ^{2},\lbrace (x,y)|x^{2}+y^{2}<=1\rbrace \rbrace

? Is that not a valid topology? I cannot include a point that exceeds a distance of one, even by the tiniest amount, so it's closed? Damien Karras (talk) 08:28, 12 March 2008 (UTC)[reply]

I believe you’re assuming a metric on the real numbers, which induces a different topology from the one you want to look at. GromXXVII (talk) 11:07, 12 March 2008 (UTC)[reply]

Don't confuse open sets in topology with open sets in real analysis. In real analysis you start with a distance function (a metric) and define an open set to be one in which every point in the set has a neighborhood in the set. Then you can prove various things about open sets: the empty set is open, the whole space is open, a union of open sets is open, a finite intersection of open sets is open, a function is continuous iff the preimage of any open set under the function is open. In topology you turn that on its head: you define the open sets as any set of subsets that contains the empty set and the whole space and is closed under arbitrary union and finite intersection, you define continuity of a function by the open-preimage condition, and so on. The reason is that in topology you're studying properties that are invariant under homeomorphisms, and (as a nontrivial theorem in analysis and a trivial theorem in topology) the open sets are exactly the structure on the space that's preserved by homeomorphisms. Of course, there turn out to be topologies that don't arise from metric spaces. -- BenRG (talk) 11:19, 12 March 2008 (UTC)[reply]

Put simply, all sets in T are open because that's what "open" means. What you generally think of as open is "open with respect to the topology induced by the Euclidean metric", but in topology "open" is just the name we give to members of the set T. --Tango (talk) 14:13, 12 March 2008 (UTC)[reply]

This may be nitpicking, but in the "open if it contains the number zero" example, doesn't the empty set have to be open (as well as closed) in any topology ? Should that be "open if it contains the number zero or is the empty set" ? Gandalf61 (talk) 07:08, 12 March 2008 (UTC)[reply]

Yes, the empty set is an open set, so that would need to be included as well. In the example, to be even more nitpicky, he didn’t say if and only if, and I bet you could derive that the whole space was closed, and so the empty set would be open anyway. I don’t think there are any other open sets though. GromXXVII (talk) 11:07, 12 March 2008 (UTC)[reply]

I did wonder whether specifically saying the empty set was open would be redundant in the "open if it contains the number zero" example, but I don't see how that can be. If you are not told that the empty set is open or (equivalently) that R is closed then I don't see how you can derive either of these propositions because:

You can't construct R from the intersection of closed sets because one of these sets must be R itself, which you don't know is closed.
You can't construct R from the union of closed sets because one of these sets must contain 0, and so will be open.
You can't construct the empty set from the intersection of open sets because all the open sets that are given contain 0.
You can't construct the empty set from the union of open sets because the empty set is only the union of other copies of itself, and you don't know that it is open.

Am I missing something here ? Gandalf61 (talk) 11:46, 12 March 2008 (UTC)[reply]

The best thing to do would be to specify that it is open. I still think the construction is possible, although not as easy as I thought it was. The empty set and the whole set are always open and closed for a reason – it’s not arbitrary. So something should break down by not having the empty set open: and the construction could come from that. GromXXVII (talk) 12:45, 12 March 2008 (UTC)[reply]

They are always open because the first axiom of a topology says so. If they could be proven to be open, you wouldn't need to take that as an axiom. What would break down is that you wouldn't have a topology so none of the results we have about topological spaces would apply. There is one way of "proving" the first axiom: The empty set is the empty union of open sets and the whole space is the empty intersection of open sets. That's just a matter of defining the empty union and intersection appropriately rather than defining a topology appropriately - it's still a matter of definition. --Tango (talk) 14:09, 12 March 2008 (UTC)[reply]

Well, you guys are right that I didn't specify the empty set being open. But I didn't need to say if and only if because I am defining my open sets to be all subsets of R which contain the number zero. In definitions, we don't "have to" say if and only if, because a definition already includes that. A set is open therefore it contains zero, by definition. A set contains zero therefore it is open by definition. So, in a definition, I don't need to specify if and only if. So here is my formal definition of a topology on R. A set in R is said to be open if it is either empty or if it contains the number zero.A Real Kaiser (talk) 20:02, 12 March 2008 (UTC)[reply]

math

how does a polynomial divided by a binomial be used in real life situations? —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 16:42, 11 March 2008 (UTC)[reply]

Rational functions#Applications has a few uses. I can't immediately think of any direct real world applications, but rational functions (of which a polynomial divided by a binomial is a special case) are very useful in various areas of maths which do have real world applications. --Tango (talk) 16:50, 11 March 2008 (UTC)[reply]

You can use this to help you find the roots of a polynomial. If you have, say, a fifth-degree polynomial, there is no general algebraic way to find its roots. But if you can discover one root somehow (maybe by guessing), say

x=3

, you can divide the fifth-degree polynomial by

(x-3)

to get a fourth-degree polynomial, which you can solve algebraically (though it's messy; see quartic equation). —Bkell (talk) 18:31, 11 March 2008 (UTC)[reply]

Yes, that's the most obvious use of polynomial division. I was thinking of the case where the binomial doesn't divide the polynomial, otherwise you actually just have a polynomial written oddly and not really a polynomial divided by a binomial. (The question say "divided by" not "dividing by", but I may be reading to much into the wording.) --Tango (talk) 18:35, 11 March 2008 (UTC)[reply]

Mmmh bit of a long shot - but for inverse square relationships somtimes there are equations such as

e^-kx/x² or e^-k(x-a)/(x-a)²

which could be of theoretical interest to physicists.. does that count?87.102.74.53 (talk) 19:04, 11 March 2008 (UTC)[reply]

It could be of use in astrophysics where atronomers need to acurately calculate parabolic paths of bodies. I suppose this could also be applied to ballistics of any sort PiTalk - Contribs 20:01, 11 March 2008 (UTC)[reply]

Wouldn't a parabolic path be the solution to a quadratic?87.102.74.53 (talk) 21:11, 11 March 2008 (UTC)[reply]

That would count if an exponential were a polynomial... --Tango (talk) 21:53, 11 March 2008 (UTC)[reply]

It is, sort of - an infinite one..87.102.17.32 (talk) 13:43, 12 March 2008 (UTC)[reply]

It can be expressed as a power series, sure. A polynomial has a finite number of terms, otherwise it isn't a polynomial - polynomials and power series have some very different properties. --Tango (talk) 18:54, 12 March 2008 (UTC)[reply]

You've never heard the term 'infinte polynomial' then or Euler's claim that "what holds for a finite polynomial holds for an infinite polynomial".

Also try searching for "finite polynomial" - it's a common phrase So just stop posting wrong stuff please.87.102.17.32 (talk) 19:41, 12 March 2008 (UTC)[reply]

"A polynomial has a finite number of terms" is not wrong. All due respect to Euler, but I reckon the vocabulary was quite limited in his time and he thus resorted to using this odd language (and of course, his statement is dead wrong). "Finite polynomial" is definitely not common, based both on my experience and a quick google test:

"Polynomial" - 5000000 ghits.
"Finite polynomial" - 5000 ghits, most seem to be taken out of context.
"Infinite polynomial" - 2000 ghits.

When people speak of polynomials they mean something with certain properties, most of which unsatisfied by a general power series. -- Meni Rosenfeld (talk) 20:17, 12 March 2008 (UTC)[reply]

I don't see much difference in properties between an infinite and finite polynomial, (or to change the semantics on its head - "infinte and finite power series"). Are there really any major differences I should be aware of, excluding the number of 'nomials' in each? 87.102.17.32 (talk) 20:31, 12 March 2008 (UTC)[reply]

For a start, with power series you have to worry about convergence, which you obviously don't for polynomials. --Tango (talk) 21:00, 12 March 2008 (UTC)[reply]

Right, but as far as you're concerned a the exponential function is not a polynomial.87.102.17.32 (talk) 21:02, 12 March 2008 (UTC)[reply]

By the standard definition of polynomial (which is pretty much universal), an exponential is not a polynomial. --Tango (talk) 21:04, 12 March 2008 (UTC)[reply]

"Polynomial of infinite degree not a polynomial."?? where is this standard definition that excludes the infinite case87.102.17.32 (talk) 21:26, 12 March 2008 (UTC)[reply]

[outdent]See polynomial. One trademark characteristic of polynomials is that if you differentiate one enough times, you end up with zero. Another is that a polynomial can only grow so fast asymptotically (at a rate which is called, unsurprisingly, "polynomial"). Another is that it (for a positive degree) always has a complex root, in fact, as many as the degree of the polynomial if you count multiplicities. Polynomials exist over any ring and do not depend on a topology (being composed of just additions and multiplications, not limits). The list goes on, while your "infinite polynomial", a term which is virtually never used, is basically just any analytic function. -- Meni Rosenfeld (talk) 22:16, 12 March 2008 (UTC)[reply]

I get that these are properties of finite polynomials - again if you include polynomials of infinite order you lose that 'trademark characteristic' . or at least you would have to differentiate infinite times (I know that is meaningless)

Don't quite know what you meant by 'for positive degree always has a complex root' - I assume that was a reference to Fundamental theorem of algebra, though it sounded like you were saying polynomials always have complex numbered solutions for polynomial=0?

Look at Polynomial#Elementary_properties_of_polynomials this extends to polynomials of infinite order. I'm just trying to suggest not making a semantic distinction between polynomials and 'power series' - as far as I see it infinite series are a subset of polynomials and the exponential function, sin etc are in that subset, and as such inherit the properties of finite polynomials - being careful to note of course that when an operation on the function is dependent on the degree, that operation will never reach a final state in the case of degree=infinity.

I'm convinced that it's productive to treat both finite and infinite power series as examples of the same set. And yes all members of that set are analytical functions.

87.102.32.239 (talk) 23:43, 12 March 2008 (UTC)[reply]

Yes, no doubt finite power series are a special case of power series. In fact, this case is so special that it was even given a special name - "polynomial". But that's really backwards. Polynomials come first as a composition of the basic operations that exist in any ring - addition and multiplication. You can then discuss what happens when you take limits - a nontrivial feature of reals and complexes, not shared with most rings - of polynomials, and end up with power series. But almost everything which makes polynomials what they are is lost in the limiting process. -- Meni Rosenfeld (talk) 13:22, 13 March 2008 (UTC)[reply]

It could be used to approximate a more complicated function with a simple pole (i.e. something that behaves like it's been divided by (x+a)), though for what real life purpose I'm not sure.

It's not really an answer or anything, but this came up on my watchlist and made me laugh:

Wikipedia:Reference desk/Mathematics‎; 22:17 . . (+192) . . ConMan (Talk | contribs) (→math: possible answer)

No clue about the function. -mattbuck (Talk) 00:10, 12 March 2008 (UTC)[reply]