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

Actually I initially thought about how I could map four arbitrary points on a circle to the vertices of a square. I got as far as realising that a pair of adjacent points on the circle will map to one edge of the square. And I figured that the closer together the points on the circle are, the smaller the square (that can be constructed) is. Because of the symmetry of squares, one edge should suffice to give the required homeomorphism (am I right?).

But you can see that the square somewhere inside the circle plus a point somewhere inside the square will mean rays from that point will pass through each vertex and intersect the circle somewhere. Here you fix the location of the square and the interior point.

So the converse must be true; given four (or a pair) of 'fixed' points on the circle, a square can be constructed in the circle's interior such that four (or two) rays intersect each vertex (or a pair of adjacent vertices) of the square and each ray intersects at a point in the interior of the square.

One more thing I've gleaned from reading stuff: the open set thing is really another way to define continuity, but of functions that do the mapping (continuously!).
So . . . nya nya.

 

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I found this, from a course at Cornell (if it isn't a forgery), which I hope settles the confusion about "shape", and "geometric object".

--http://www.math.cornell.edu/~hatcher/Top/TopNotes.pdf

That link is also, I think, quite a good introduction to the subject. It's a pretty big subject.

 

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I should have been hauled up for this geometric error.

Which is that the projection from a point in the interior of a square if you want it to cover an edge of the square at a time (i.e. pairs of adjacent vertices) will always be bounded by the other edges, or by the geometry. Suppose the point of projection is near one vertex, then it has a neighbourhood which is close to either edge meeting at that vertex, and if the point moves toward either, the line of projection through the same vertex will move through nearly 90 deg of arc.

The vertex opposite won't span much of an arc though, and the other vertices will have lines that stay fairly fixed, at almost 90 deg apart. If the point of projection is near the centre (at the centre it's the point the square's diagonals intersect) the lines of projection won't be far from 90 deg apart).

But, of course, the square can be outside the circle, in which case rays that pass through each vertex are constrained to also pass through the circle and intersect at point inside the circle; the circle can be anywhere (and as small as necessary), in the interior of the square, but the rays on the circle must lie in different quadrants in every case (not just anywhere), because of the shape of squares.

'Meh'

 

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Ok, I'm prepared to admit that the word shape is vague, at least in a mathematical sense. It's more a heuristic, the method of mapping, continuously, every point on a circle to every point on a square using a projection from a chosen point is just one way to show they are homeomorphic. I hope I've shown that geometry is involved, and geometry and shape are kind of strongly analogous. The circle is rotation invariant, a square has a fixed set of rotations that leave its orientation and position the same; this has nothing to do with the topological equivalence.

Suppose you choose any four points on a circle, then you have four sections (arcs) of the circle. If you deform these into straight lines you have a quadrilateral, and no square geometry is "in the way". Make two sides of this quadrilateral the same length and parallel, the remaining sides will also be parallel and the same length. If you have a rectangle, contract a pair of sides until you have a square.

So if you want to deform a circle into an n-gon, regular or not, you need to choose n points on the circle with this "method". You could also choose n points in the circle's interior such that they define a convex polygon. I would say there are many more ways to find a homeomorphism.

 

I'm just recapitulating here, kind of thinking out loud.

That a circle divides the plane into two regions, and that this is true for any closed simple curve seems obvious, but how do you prove it? This is probably the kind of question a student of topology would encounter.

That the circle and any closed simple convex curve are homeomorphic is easy to sketch a proof of, using the projection method. A square inside a circle plus a point inside the square gives a function that maps points on the square's boundary to points on the circle, i.e. a one-to-one map. The function can be envisaged as the set of rays over each pair of points along which the boundary point of the square (resp. the circle), can be moved.

You can say the function maps the square onto the circle and its inverse maps the circle onto the square. Since the point of projection is any point in the interior (of either the circle or square, which one depending on which is smaller) there is an infinite set of functions, each 'acting on' an infinite set of projected rays.

Hence, the centre of the circle is just a special point from which projected rays will all be the same 'length'. This isn't relevant to a function that takes points on the circle to points on an exterior (or interior) geometric figure. The notion of a metric, or even scale, isn't relevant since you only need set relations (the square's boundary is a subset of the set of points in the circle's interior, etc). Oh yeah, a circle and a square, or any polygon, forms a topological space.

 

No, this is quite incorrect. A topology is defined on a SET and is a set of its subsets. The simplest topology on a set is not a point, it is \(\{S, \emptyset\}\) where \(S\) is the set in question

 

This is so garbled, I don't quite know how to reply. Yes, a metric (you are implying the Euclidean metric) can be used to define a topology, But this does not imply that the point \(x \in R^n\) "has" a topology, as you asserted.

Indeed, it can never be an element in any topology on \(R^n\), since these elements are by definition subsets. Note well that \(x \ne \{x\}\), which latter may be in a topology, but is also closed in all possible topologies i. is always the complement of some other element (open set, recall, by definition) in the topology

 

Not with full mathematical rigor, but If any point is randomly chosen in the plane and all rays from it cross the boundary of the "closed curve" an odd number of times then that point part of the "interior division" of the plane. If not then it is part of the exterior, or with zero measure probability, on the boundary.

I regret to see you still think "shape" has anything to do with topography after my post 10, which was (and is still not refuted, I think, but I have not read all post of this thread.):

BTW: I don't think your post 82 quote from Cornell is a "forgery" That is what they taught me there was topography as an undergraduate more than 6 decades ago.
Even my "human = doughnut" in topology was used there. Also there was where I learned how to define "interior" as done above which work for 3D (or higher D) objects too (but if "multiply connected" then a "grazing ray" the just touches a boundary is two "degenerate" contacts to the same point. Like in the calculation of population of excited states energy levels populations with temperature T (LTE) those that are "degenerate levels" (same energy) get counted as two (or whatever the degeneracy is, usually represented by g as I recall.)

 

*sigh*
The point \(x \in \mathbb{R}\) is a real number. \(\mathbb{R}\) with the so-called standard topology is referred to as (the topologist's) real line and written as \(R^1\). The open sets in \(R^1\) (i.e. those elements in the standard topology) are the open intervals of the form \((a,b)\) with say \(x \in (a,b)\)

What is the "centre of mass" of a real number?

 

Perhaps you could explain to we mere mortals WTF this has to do with topology

 

It's becoming obvious that the word shape should never have been invented.

So let's go with some really obvious things about loops of wire.

1) loops made of wire retain their, um, geometry when you bend them. If you bend a loop of wire (like you have with clothes-hangers) into a rough circular, um, geometry it will "keep" it as long as it isn't bent more.

2) if you cast the shadow of a loop of wire onto a flat, "screenlike" surface, the shadow changes when you rotate the loop.

3) you can abstractly "color" this shadow; you can say the wire is a 1-dimensional curve lying on a plane surface (despite the surface being only approximately so). and that it's the boundary of an interior set of points and an exterior set of points. You color the interior "A", and the exterior "B".

4) if you consider the shadow of the wire loop really is 1-dimensional, in a mathematical sense (or that there is such a mathematical sense, whether or not you bend up a coathanger and project its shadow on a wall), then you can consider the entire plane is a union of disjoint sets, A and B (since, logically a 1-dimensional boundary has zero "width").

5) the loop must be embedded in 3-space, because you can project it (physically and mathematically) onto a plane surface.

 

Wouldn't that work only if the boundary was continuously differentiable? If the curve was a polygon, then rays might intersect an edge rather than a single point.

 

More obvious things about wire loops:

If you twist one half of a loop (hold one "side" and twist the other), the shadow looks like a figure eight (if the loop is held the right way); this is independent of the direction you choose to twist the loop.

There are two directions you can twist the loop (clockwise and anti-clockwise). The loop crosses itself, and the crossing geometry doesn't change under rotations of the loop (locally, one part of the continuous loop lies over itself independently of the orientation of the whole loop). There are two twist directions, hence two figure eights in 3-space, but only one figure eight shadow.

The figure eight shadow doesn't separate the plane into two regions, but three (oh dear).

You can't have a knotted loop with one or two crossings in it, you need at least three crossings and they have to alternate.

There is a minimal surface (exemplified by a soap film) having the wire loop as a boundary; the geometry of this surface depends on distances between local sections of the loop boundary.

The minimal surface can be oriented (both sides can be coloured differently) depending on how the loop is deformed (a Mobius loop doesn't have a two-sided minimal surface).

p.s. I don't know how to make this any more exciting than it is already.
p.p.s. I know there are people out there who don't appreciate sarcasm, for some reason that remains obscure to me.

 

Look, we've been discussing continuous plane curves and continuous functions that deform these curves.
Ok?

 

I don't think differentiable or not has anything to do with it - The rays either cross the boundary or grazes (shares point or points} with the boundary and I told how to count this low probability case in post 91. I. e. I said:

"... a "grazing ray" that just touches a boundary is two "degenerate" contacts to the same point. Like in the calculation of population of excited states energy levels populations with temperature T (LTE) those that are "degenerate levels" (same energy) get counted as two (or whatever the degeneracy is, usually represented by g as I recall.) "

When writing that I had in mind a ray tangents to the circumference of a circle starting from a point outside the circle and did not want their common point to be counted as 1. (1 is odd number and in my definition of the interior that would make the starting point of that ray falsely inside the circle.)

If that ray did cross and then exit at a nearby point, which in the limit become the same point, that point is two points that have degenerated into one point. A point can have two identities - like the mid point of X. It is part of the line segment with positive slope and also part of the line segment with negative slope.

In the case of an external ray tangent to the circle, there is the point where the ray enters the the circumference and the point where it exits the circumference boundary - these differently defined points just happen to be the same point.

 

Billy T: I'm not sure you've followed what I was pointing out.

Suppose you have a triangle, then you don't want to choose a point such that a ray from it is collinear with an edge of the triangle. The reason you don't want that situation is obvious (I thought); the rays from the chosen point won't intersect the boundary of the triangle at single points but rather a set of points which is an infinite set.

Really you want the intersections to be sets containing (elements which are) single points on the boundary, rather than a whole edge. The reason you want that is so you can count how many times (find each set's cardinality) a ray passes through the triangle (or whatever geometric figure) since, as you say, an odd number of times means the ray ends in the interior, and an even number means otherwise.

There is at least one solution: choose a line somewhere instead of a point, and project perpendicular lines from it into the triangle. Then the problem reduces to finding a direction for a set of parallel lines (some of which pass through more than one edge) such that no line is collinear with any edge of the figure, whatever it looks like.

 

Alright, let's have another look at a loop of wire which has been twisted so it has a single crossing, so that when you project the shadow of this crossed-over loop you see a figure eight.

The crossing in 3-space can have some arbitrary distance between the local sections, or arcs, and this doesn't affect the shadow. So, you can consider the crossing as a feature that divides the loop into two parts; generally, any crossing in any loop (knotted or otherwise) divides the loop this way, and links are divided too.

You consider the 'local structure' of crossings; Kauffmann uses bracket polynomials which use notation that is based on the Jordan Curve Theorem: any closed simple curve divides the plane into two disjoint regions (as discussed here and probably lots of other places); when a bracket contains a crossing 'symbol', we are meant to interpret it as part of the larger structure (as outlined). Kauffmann states that this is a remarkable fact about knots: the notation used in his bracket polynomial formalism is based on a theorem.

 

I don't want to discusses whether or not an infinite number of points is "countable" (I think yes) but be that as it may be (highly dependent on what countable means, as no they can not actually be counted even at 10 per second rate).
Hell even all the points in a 2D plain is countable and no larger infinity than the number of points on the line segment given by 0<x < 2.

I told that when the ray enters a (or many) boundary points, but does not leave the boundary by crossing (even made "cross" bold in post first telling the method), you account it again as the ray leaves that point. I.e. each going thru without crossing of a boundary point is degenerate pair of points so even if the ray goes thru a section of a straight line boundary the number of points counted is an even infinite number.

I also noted that there can be more than point identified differently point at a single point, like the crossing of two line segments in X, the crossing point belongs to two different line segments. Take one away from the other and that point is still there in both line segments. There can be as many points at the origin of a Cartesian coordinate system as you can define, which is infinite as the number of DIFFERENT lines passing thru the origin given by the equation y = mx where m is the slope of the line is infinite.

If a ray does cross into a circle and then exit only an nano meter of its length later obviously the ray crosses the circle boundary twice. As the length it is inside the circle goes to zero, it is only natural to say that the one tangent point it passes thru is to be counted as two points "degenerated into one point." This concept of degenerate (identical) energy levels in physics is not only commonly done, but they MUST each be counted separately when telling their LTE populations. I.e. the number of electrons in such a degenerate state is essentially twice the number in a level only very slightly differ in energy - that is the way nature is.

BTW, that is what "Zeeman Splitting" is all about - a weak magnetic field with split apart the degenerated energy levels making each have a slightly different energy.

 

No it divides the SHADDOW into two loops that touch - not the twisted wire loop - it is still unified.

 

Ok, think about it like this: you've twisted an otherwise looped piece of wire so its projection looks like a figure eight.

Draw the following abstract diagram: a pair of boxes which you connect with a pair of lines. The lines can be crossed over each other, or not.
What does this diagram say about the loop of wire? It says that if you uncross the lines, the rest of the loop is the same.