What is Topology, and what's the big deal?

Discussion in 'Physics & Math' started by arfa brane, Nov 21, 2014.

  1. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I'm not certain what your "abstract diagram" is of OR if by "pair of boxes" you mean two squares in a plain or two 3D cubes. Any way I don't want to "think about it" I have told you of your errors already - tell if you accept the corrections or not.

    You said earlier that "shape" was a term that should not have been "invented." Not true. You need some and simple way* to note how a 2D square differs from a 2D rectangle and circle (or dozens of other 2D closed curves) but not in topography where they are the all same.

    * Just noting pair wise differences like "circle has no corners", "rectangle has sides of different length", etc. to infinity. "They differ in shape." applies to all.
     
    Last edited by a moderator: Dec 14, 2014
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  3. arfa brane call me arf Valued Senior Member

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    I think no.
    Countable means you start somewhere, say at "1", then "2", "3", and so on. If you have an infinite set to count you have a problem, which is, where do you start? at the smallest value? then what is that value?
    I don't really follow any of this, sorry.

    Projected lines (rays), perpendicular to a given line will either intersect the boundary of a closed curve at single points, or they will intersect an edge (unless the boundary is everywhere smooth, such that there is a single tangent at each point), and if they do intersect an edge, they do not intersect any points in the interior or the exterior of the figure.

    That is, if a ray is "tangent" to an edge you can't determine if the end of that ray is in the interior or not. Your suggestion of degenerate sets of points seems artificial.
     
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  5. arfa brane call me arf Valued Senior Member

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    I mean two boxes, and the diagram is taken from a book by an expert on the subject, one Louis H. Kauffman.
    By all means don't think about it, and please continue to believe you've provided corrections.

    Oddly, you now maintain that shape is a useful concept after all; I challenge anyone at all to present a topological argument which never refers (directly or indirectly) to the shape of objects. That is, the word geometry isn't allowed either, since geometry determines shape.

    Actually that's only a "problem" in geometric topology, which knot theory is a part of.
    Wait, graphs don't have a geometry (there are a possibly infinite number of ways to construct a graph in the plane), and yet planar graphs also "fit" on the surface of a sphere.

    Do the plane and the sphere have different shapes, and does this have any topological relevance?
     
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  7. QuarkHead Remedial Math Student Valued Senior Member

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    Post deleted by me - it rendered as a link in its entirety.
     
    Last edited: Dec 14, 2014
  8. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    You need to tell more specifically how I am confused. That is precisely why I said all the points in a 2D plain are countable. i.e. Consider a point in the plain with Cartesian (x,y) = (abcde... , ABCDE...) Where the lower case letter are the digits of the x and the upper case letter are the digits of the y. Just to be completely clear I give example of ONE point AND where on the number line it map to:
    (254, 3678) then a=2, b=5, c=4 & A=3, B=6, C=7 & D =8 etc. This point and only this point correspond 1 to 1 with the point on the 1D number line with 0 as the origin which is: aAbBcC0D or for the illustration case that 2D point is in 1 t 1 correspondence to x = 23564708.

    Now that the 2D plain is mapped 1 to 1 onto the number line the entire line can be mapped 1 to 1 into the sub range I stated, 0 < x < 2 by taking the recipicals of all (any negative number -1) times -1 and mapping them on the range between 0 and 1 and add 1 to all positive numbers, take the recipicals of that sum and add +1 to it. That maps them onto the range 1 to 2 but the ends of the range are not need.

    For example a very slightly negative number like -0.00003 when you add a -1 to it becomes -1.00003 and its recipical is approximately -0.9999 which when multiplied by -1 becomes 0.9999 or nearly 1, I. e. is in the range 0< x <1. likewise a very large negative number maps into a slightly greater than zero number.

    Any positive number that the 2D plain mapped on to after adding 1 to it has a recipical between 0 and 1, but as we have stuffed all the negative numbers into that range we add another 1 to that recipical to move the total into the range to between 1 and 2.

    For example a positive number slightly greater than 0, like 0.00004 becomes 1.00004 after adding the first 1. Then the recipical of that sum is slightly less than 1 but when we add the second 1 it is slightly less than 2.Likewise a huge positive number is even slightly greater when the first 1 is added so its recipical is slightly grater than 0 but when we add the second 1 it is slightly more than 1.
    I. e. all the positive numbers are thus mapped into number between 1 and 2.

    SUMMARY: All the points in the plain were first mapped 1 to 1 onto the number line then the negative numbers of the number line were compress into the and 0 < x < 1 and all the positive numbers of the line were mapped 1 to1 into the range 1<x< 2. Hence there are no more points in the 2D plain that in the line segment between 0 & 2. - Strange, but true, these infinities are the same order, (equally large.) and of course "countable."

    DO YOU STILL THINK I AM "CONFUSED"? or was I correct to say the points in the 2D plain are "countable" and can map, 1 to 1 into the range 0 < x < 2.

    BTW the point in the plain at the origin (0.... , 0....) maps into the 0 of the number line. No mathematical mapping needed and certainly not done by taking recipicals.
    I should also note that my mapping is not the most simple one. I avoided it to help make the point that there are an infinite number of mappings, 1 to 1. For example, if you like you can map the entire 2D plane onto only even integers! Telling exactly how, is a little more complex.
     
    Last edited by a moderator: Dec 15, 2014
  9. arfa brane call me arf Valued Senior Member

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    Billy T: what I was trying to point out, and I should have distinguished the natural numbers from the real numbers, is that you can't count an infinite set of points if that set is a subset of the real number line, as any edge of a polygon is.

    You can't count an infinite set of real numbers because, as I also tried to say (not very well), you can't find the smallest number so you can even start counting.
    Or more exactly, you can't find a one-to-one correspondence between points on the real line and the natural numbers.

    I don't know what you're trying to do in your last post, but counting does not involve negative numbers for a start, nor do you start at zero (you start at 1, the minimum number of "things that can be counted"; 1 is the smallest, and the first, natural number).
     
  10. QuarkHead Remedial Math Student Valued Senior Member

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    1,740
    Yes, for the reason arfa (sort of) gave - the fact that there is 1-1mapping from the 2-plane to the real numbers misses the point. You need to show a 1-1 correspondence with the natural numbers which are, by definition, countable (but infinite)

    Since the 2-plane \(R^2 \equiv R^1 \times R^1\) it will suffice to show that the real numbers are not countable. Here is a version of Cantor's argument in compressed form.

    Suppose the real numbers are countable. This means that they can be enumerated i.e. every real number in its decimal expansion can be given an index from the natural numbers. It will suffice to work only on those elements in the expansion the the "right" of the decimal point.

    Like
    \(x_1 = .a_{11}a_{12}a_{13}.......\)
    \(x_2 = .a_{21}a_{22}a_{23}.......\)
    ............................................................
    \(x_n=.a_{n1}a_{n2}a_{n3}...........\)

    Now construct another real number from the diagonal of this array i.e.\(d=.a_{11}a_{22}a_{33}.........\) which is clearly real number. And now construct another number \(b\) by the rule that \(b_{mm}=a_{mm}-1\) whenever \(a_{mm}\ne 0\) and \(b_{mm} =1\) whenever \(a_{mm}=0\)

    Clearly, by construction this new number, which is clearly real, can be equal to no \(x_k\), thus contradicting the assumption that a complete enumeration of the real numbers is possible, hence the reals are uncountable

    Nifty, eh? (I take no credit for the general idea, which was Cantor's)
     
  11. someguy1 Registered Senior Member

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    727
    Yet the positive rationals are countable, and you can't find the smallest one of those either. Right?
     
  12. QuarkHead Remedial Math Student Valued Senior Member

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    someguy1, I suspect he meant the "strictly positive integers"

    Many people use this term in place of "the natural numbers" to ensure that nobody should so foolish as to include zero in the latter
     
  13. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I agree. That is why I said at end of my post that mapping plane's points onto just even integers was possible but much more complex than my mapping onto the number line. The Point is that the number of points in the plane is the same order of infinity as the integers. (Or even just prime numbers according to "proofs" that there is no largest prime.) Currently the largest known prime is only the 48th of the "Mersenne Primes." 3 is the first of that form 2^n -1 with n=2 and 7 is the 2nd, 31 is the 3d. The new one and the largest known of all primes is 2^n -1 with n = 57,885,161. In based 10 notations, it has 17,425,170 digits! After that they rapidly "thin out" Perhaps there is a "largest Mersenne prime" but I think not. For every known Mersenne prime, the must be at least a million non Mersenne Primes.

    Finding the 49th Mersenne Primes will not be easy - The 48 was found with 360,000 CPUs peaking at 150 trillion calculations per second. It and a few earlier ones were produced by the 17th-year GIMPS - the longest continuously-running global "grassroots supercomputing" project in Internet history. (Much larger than "SETI -at home.") Compared to finding a larger non- Mersenner prime, finding the 49th may be the lesser task as you know want to guess and test.

    BTW: I have known Cantor's clever demonstration that one can not list all the numbers for 55 years, but I had forgotten his name. Thanks for that. It is easier for most to understand just in the following words than math notation.
    Assume you could make a list of all the numbers (most with very long or even infinite decimal representations.) Here is a number not in the list: Change the first digit of the first number, the second digit of the second number etc. That is a new number, not identical with any in the list. Ergo you can not list all the numbers.

    Again: The critical point is that the number of unlistable* points in the plain is an infinity no larger than the infinity of integers.

    Or putting that into words: For any chosen point in the plain there is a unique integer it, and only it, can be mapped onto. (You will never "run out of" integers.)

    There are higher order infinities. If I recall correctly one can even be defined, which is still in the plain! I. e. thru every point in the plain an infinity of lines can be drawn, each of which has an infinite number of points. - I .e. "a one point in the plain to an infinity of points" can be defined for each and every point in the plain. Infinities get very strange. They are like people: some are bigger than them selves!

    * Cantor's proof shows the points on the line (or in the plain as they can be in 1 to 1 mapped onto the line) are unlistable, but not that they are uncountable, I think.
     
    Last edited by a moderator: Dec 15, 2014
  14. QuarkHead Remedial Math Student Valued Senior Member

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    No, the integers are countable, the 2-plane (as a set) is not

    Again, the integers are countable
     
  15. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    We agree the integers are countable (that is the definition of countable) and that the real numbers (irrational included) can not be listed.

    As I think we agree the points in a plain can be mapped 1 to1 uniquely onto the real numbers, I think we agree they are the same order or "size" of infinity.

    Where we disagree is you think the fact that the irrationals (etc.) can not be listed proves that they can not be counted. That may be true, but you (and Cantor) have not proved they can not be counted - you and he assert that "can not be listed" = "can not be counted."

    Doing that will be tough, if not imposable, as generally it is hard to prove equality above. To do that you must prove a negative assertion. For example, PROVE that the irrationals cannot* be placed in 1 to 1 correspondence with the integers.

    *Proving a negative assertion is always tough. For example, try to prove unicorns do not exist (any where in this vast universe). Proof of negative assertions in math are possible as it is a closed tautology.
     
    Last edited by a moderator: Dec 16, 2014
  16. someguy1 Registered Senior Member

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    They're synonymous, by definition. A set is listable, or countable, if it can be bijected to the natural numbers. It's a definition. So if a set is listable it's countable and vice versa, simply because those two words are technical terms that mean the exact same thing.
     
  17. arfa brane call me arf Valued Senior Member

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    Here is a reasonable proof of the Jordan Curve Theorem for polygons. http://www-cgrl.cs.mcgill.ca/~godfried/teaching/cg-projects/97/Octavian/compgeom.html
    A more general proof must include any kind of closed curve or figure.
    In Kauffman's book, the outline of a proof for smooth simple curves is given, using much the same method (a chosen direction and a line with rays perpendicular to it).

    Note I used the word "figure"; Kauffman uses the word "symbol" for closed curves, and then uses them as notational symbols in "polynomials".

    So, um, can we get back to what this theorem is, and what it means? Why does Kauffman say:
    "a fundamental theorem of mathematics is the underpinning of a notation for that same mathematics"?

    I mean, all the boundary of any closed figure does, is separate the plane into two disjoint regions (sets of points)--so what?
     
  18. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    No the definition of "countable" does not mention lists. (or even the procedures for "counting" which is wise as there are procedures other than making a list.*)

    *Here is an example of a non-list counting procedure that assigns a unique positive integer to each counted item (even if they are infinite in number):

    My grocery store sell eggs, 12 in a carton or "box" here after. Typically the boxes sit on a table about 10 side by side in about 5 rows deep or 50 boxes in each layer. (I'll assume these numbers.) Thus, without making any list, I know there are 50x12 = 600 eggs in each layer. To the eggs in bottom or first n=1 layer I assign the integers 1 thru 600. To the eggs in the second n=2 layer I assign the integers 601 to 1200. Etc. for any value of n you like. If there are an infinite number of layers, that is no problem (except would break a lot of eggs) as each of the infinite number of eggs has a unique integer assigned to it.

    Some here want to say that if you can not make a list, you can not count. I am asking for some proof that list making is the only way to count and giving an alternative method of counting an infinite number of eggs, just to illustrate there can be more than one way to count, even an infinite number of items. The burden of proof is on those who assert that making a list is the ONLY way to count. I.e. they must PROVE, that making a list is the only way to count the points in 2D plain. I don't need to give an alternative procedure as I do not assert there is one. I have even agreed that their assertion MAY be true - just asking for some proof that it is.

    I did note that giving such a proof would be hard as you must "prove non-existence." If you have such a proof perhaps it can be adapted to make the atheists rejoice?
     
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  19. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    To arfa brane; I don't have access to Kauffman's book. You spoke of "boxes" in earlier post. I asked if you were referring a cube or a square (or something else). You did not give the simple reply but said the figure in Kauffman's book. That is not an answer for me. Please tell something useful for me. The following is also not clear for the same reason.
    I certainly agree that a closed curve devides the 2D plain into two distinct regions ("interior" and "exterior") and even gave to method to test which region any arbitarly chosen point of the plain that point is in.

    I noted that a set of points, of area measure zero, that are on the boundary is ill defined - neither inside nor outside but suggested they would be "not in the interior." (Rays from interior points, by my rule, cross the boundary curve an odd number of times.)

    BTW: Is the method in Kauffman's book basically the same as mine (rays from points in the interior ALWAYS cross the boundary an odd number of times; but stated more precisely in "mathematical terms."? If not (is something else) can you briefly describe it?
     
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  20. Guest254 Valued Senior Member

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    Billy T, can you please, please stop posting utter nonsense. You have absolutely no clue what you're on about.

    I, and it seems many others, stopped posting on this forum because of the unbearably low scientific standards. You are contributing to the problem -- the fact you're a moderator only makes this offense even worse.

    You do *not* know what you're talking about when it comes to even the most basic of mathematics. And this instance is not a one-off. For example, here:

    http://sciforums.com/threads/what-is-time.143040/page-44#post-3249112

    you make the most basic errors on material that is known to second year undergraduates. Stop trying to make out you know more than you do -- it's incredibly transparant.
     
  21. QuarkHead Remedial Math Student Valued Senior Member

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    1,740
    First, to be almost as rude as Guest, you seem not to understand the difference between ordinal numaabers and cardinal numbers. I can illustrate how these can be brought into register but only in the finite case.

    Although I 100% agree with someguy's last, perhaps it may have helped you had he said "an enumerated list".

    To take your every-day example, and ignoring the fact that, in a sense, packing eggs in boxes of 12 is already an enumeration........

    Suppose I sneak into your store and very carefully take every sigle egg out of its box and pile them up. How many are there? To find out, you "count" them, right? IOW, to each egg you assign a number. Since this procedure would be futile unless you assigned some sort of order (5 follows 4, for example), you have effectively "listed" the eggs in the order to which you assigned each number.

    What if the last number, an ordinal, is not finite? i.e. you exhausted your egg pile and the natural numbers at the same time. No problem, call it \( \omega\).

    Now I ask what is the "size" of this pile of eggs - note NOT what number you assigned to the last egg - you answer would of course be a cardinal number, which we call \(\aleph_0\), the first transfinite cardinal number.

    It was shown by my borrowing from Cantor that there exist other cardinals that are strictly greater that \(\aleph_0\). That which is assigned to the "size" of \(\mathbb{R}\) is called \(\mathfrak{c}\) and it assumed (as far as I am aware, not proved??) that \(\mathfrak{c}= 2^{\aleph_0}\)
     
  22. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    That post of mine notes that Fourier transformer can fail to reproduce EXACTLY even a singled valued function if (1) there are discontinuities of it or (2) there is a finite step at one point even if the direviative from the right and from the left are equal but the function is not continuous. I've been thru this before (in 2008) here is the exchange:
    * Back in 2008, as the “sheriff of nonsense” I pointed out errors in posts of others and took special delight when noticing one of James R's as he had made only 4 in a decade.
    Here in full is Jame's reply (post 71):
    “You're right, of course, Billy T.”

    Perhaps you will argue with James? - Tell him he is also wrong?

    BTW 26 (or more ?) have "liked" my posts.
     
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  23. Guest254 Valued Senior Member

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    I don't care what conversation you had with someone called James in 2008. The post I linked to aptly demonstrates your lack of understanding of the (elementary) subject. Let me elaborate by including the implicit quotes:
    No, the overshoot is not there in the full (infinite) Fourier series. The overshoot exists in each finite Fourier sum because the Fourier series fails to converge uniformly in a neighbourhood of the discontinuity. This is considered elementary real analysis. And I know it is elementary, because I have to teach it in an elementary course on Fourier series to second year undergraduates.

    I don't want to argue about this with you. You're wrong. Stop wikipedia-mining and gracefully accept it.
     

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