The simplest example of this induction step is showing that T^2 embeds in R^3, assuming that T^1 embeds in R^2. Let T := T^1 be embedded in R^2 as the circle of radius 2 about (0,0). At time s, for -1 <= s <= 1, let P_s := { p in R^2 | dist(p, T) = sqrt(1-s^2) } P_s is one circle for s = +-1. Otherwise P_s is the union of two circles, of radii 2 +- sqrt(1-s^2). Now consider the subset X of R^2 x R^1 = R^3 defined by X := {(p,s) in R^2 x R^1 | -1 <= s <= 1 and p in P_s}. It's easy to visualize exactly what X looks like since it's in R^3. Clearly, X is a 2-torus embedded in R^3. That's because what we have here is a T^2 x C, where C is the circle obtained by identifying two intervals by glueing their left endpoints together, and also glueing their right endpoints together. (As a movie, C is just a single point that immediately separates into two points that reach their maximum separation, then retrace their steps by approaching each other until they coincide.) Noting that a T^1 is just (S^1)^1 = S^1: But one may also imagine the new dimension as time, in which case X is a movie of a varying point-set in R^2. This set at time s is just X_s := P_s x {s}. For s = -1 this is one T^1, which immediately separates into two T^1's that move away from each other, reach a maximum separation at s = 0, and then the two T^1's approach each other until coinciding at s = 1. The last paragraph generalizes to showing that if T^n embeds in R^(n+1), then T^(n+1) embeds in T^(n+2). --Dan Robert Munafo wrote: << I'm not sure I get it. I tried to use that description to show that the familiar torus T^2 can exist in R^3, but the resulting surface intersects itself. Placing the initial T^1 (which is a circle) on the X-Y plane, and treating the Z axis as the "time" dimension, and using the X axis as the direction in which the two copies get pushed while time is moving forward, I end up with the surface intersecting itself at various points in the Y-Z plane. I could use a 4th dimension to prevent this self-intersection, but then it's a bit hard to explain how you can twist half of the result around to embed the surface into R^3 in a non-self-intersecting way, or even to explain whether or not it is a Klein bottle. On Wed, Dec 1, 2010 at 18:40, Dan Asimov <dasimov@earthlink.net> wrote: << This is a less elegant but maybe more transparent way to see that T^n embeds in R^(n+1): ---------- The induction step is typified by showing T^3 embeds in R^4, assuming T^2 embeds in R^3. Start from a T^2 = T_0 embedded in R^3. Letting the next dimension be "time", we make a short movie that starts with the T^2 embedded in R^3, which immediately separates into two T^2's that move apart to a maximum separation. (Just push T_0 off itself in R^3 in both directions.) This is the first half of the film; the rest of it is just the first half in reverse time order. ---------- The result can be thought of as two copies of T^2 x [0,1], with both copies of T^2 x {0} identified with each other, and also both copies of T^2 x {1} identified with each other. This is easily seen to be T^2 x S^1 topologically.

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