Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

After reading around these various references, I think that there are two slightly different views of the essence of torushood.

If you think of it geometrically, as being the envelope you get when you wipe a circle around a circular orbit, then a sphere does seem to be a degenerate torus, got by shrinking the orbit to a point, as you originally suggested. But if you think of a torus as a topologist does, as a kind of surface, the essence of torushood is the hole (or, if you like, the 'handle'). This is what makes the torus have a different homotopy group, why you can cut along a circle which doesn't separate the surface, etc...

Seen from the topology perspective, that degenerate case isn't a sphere at all, it is a very odd entity: two spheres, A and B, in the 'same place' but with distinct surfaces, which pass through one another at the poles: so if you go to the north or south pole on A and keep going, you move into B, and vice versa. So if you draw a great circle through the poles and cut the surface in half along it, it won't fall apart into two hemispheres, but will contract back 'through' itself into a cylinder: just like, well, a torus. (To see why, consider that to get a torus you need to sweep the circle through 360 around the 'orbit', but you only need to sweep it 180 to get a complete spherical surface when you rotate about its center. So the full toroidal 360 sweep gets you two spheres. Another way is to visualize the spindle torus, see http://mathworld.wolfram.com/Torus.html, and imagine the extreme case where the 'inner' surfaces come to coincide with the 'outer' ones in the limit. Now imagine what happens to a circle on the original 'normal' torus which went 'through' the hole. In the spindle torus it goes from the outside, 'into' the torus, then back out. In the limit, it starts 'outside' on sphere A, say, then goes 'inside' at the pole onto sphere B, then back 'out' at the opposite pole. The 'same' circle starting at the other side of the sphere only intersects this circle at the actual poles: everywhere else it is on different surfaces.)

But, Im sure that this is all rather too, er, hairy for this website, and it might be better to just stick to the geometrical view of things. In which case, my objection to the original graphic is misplaced, and I hereby withdraw it.

Hey, this has been quite interesting. Thanks for prodding me to check out other points of view here ! :-)

The second one is easiest, so I'll do it first. M is the lub of A, N is the lub of B. What does this mean? Take A and M: it means that every x in A is less than M (upper bound) and that there isnt any smaller number that is an upper bound: so if you have any K, say, less than M, then some x in A will not be less than K (thats why it is the LEAST upper bound. If your head is spinning, draw a picture with A as a kind of cloud or blob, and M as a horizontal line just touching the top of the cloud. If you were to lower the line (K) then it would clip off some of the cloud (x) which would be above it, right?) Similarly for B and N.

OK, now, think about A union B. This is like taking both clouds together and treating them as a single cloud. So, is N or M an upper bound of that? Well, one of them has to be (the highest line, right?) That is, max{N,M} is either N or M, depending which is bigger: if N<M then max{N,M}=M, otherwise =N. We will argue by cases. Assume N<M first. If N<M then everything in B is <N which is <M, and everything in A is <M, so everything in (A union B) is <M, so M is an ub (thats all just saying the highest of the two lines is above both clouds); and if there were a smaller one, say K, then K would have to be a ub of (A union B) and hence an ub of A: but it can't be, because M was a lub of A (remember? In other words, if you lower the top line then you lose part of the highest cloud). So M is the lub in this case. If M<N, the argument is exactly the same with B and N instead of A and M, and N is the lub in that case. So: if N<M then the lub is M and max{M,N}=M, and if M<N then the lub is N and max{M,N}=N. Either way, the lub is max{M,N}. (What if M=N? I leave that to you.)

The first problem use that 'leastness' critically. Take some X1 in S - we know there is one because S isn't empty (that condition in the statement of the problem was a clue, in fact) - and ask yourself: is this X1 the lub L of S already? It might be. If it is, we can just define the sequence to be L,L,L,.. on for ever. But if not, then there must be another X2 in S which is closer to L. (Why? Because if there weren't, then X1 would be the upper bound, right? That's just another way of saying the same thing: if there's nothing larger then X1 in S, then X1 is an upper bound of S.) So, if X1 isnt the upper bound, we can find an X2 which is closer to L. In fact, we can find an X2 which is less than half the distance from L that X1 was, i.e. an X2 such that X2>(X1+L)/2. Why? Because if we couldn't, then (X1+L)/2, which is <L, would be an upper bound, so L wouldnt be the LEAST upper bound. So, go on in that way: you have X1 to Xn chosen, and if Xn is L then stick with L from then on, and if not then choose an Xn+1 > (Xn+L)/2, ie which halves the distance between your best hit so far and the lub L. You keep on creeping closer. Then L is the limit of that sequence Xn as n goes to infinity. I'll let you prove that last part. :-) (The point is that it doesnt matter exactly which Xn+1 you pick: they can't be larger than L, and if they are smaller you can always get half-way closer.)

These aren't the shortest or most elegant proofs, and I havnt written them in a nice mathematical style. But if you can see how the necessary thinking goes, you can learn to do that other stuff yourself.

Pages: **1**