ContinuityWe consider a pair of projectively-identical, skew rulers, one red, and one blue, made equivalent by projectivity via an intermediate, green ruler.Consider just one, black, skew line of the set of skews that directly link end-points of intervals on the red ruler to end-points of intervals on the blue ruler, and how – strictly projectively – to move that skew. When we know how to do that for one, we can generalise to the others. A red line in red point M joins a red point, say R, in the red ruler, to a point, say Q, in the green ruler. A blue line in blue point N joins that point Q to a blue point, say B, in the blue ruler. The join of the red point R in the red ruler to the blue point B in the blue ruler is our single, black, skew line. Now all these joins, and the points they join, are coplanar, in, say, plane γ. If we wish to move the skew line, we need only contrive to move the green point Q in its green line, and that might be accomplished by rotating plane γ around the line joining M to N. The question now is whether or not we can projectively move point Q, and rotate plane γ, smoothly and continuously. Incidentally ...We have a perspectivity in Q, meaning that every skew interval bridging the rulers is equivalent to the interpunctual interval MN, and each is equivalent to every other.Also, two or more intervals formed by the same skew, moved, are equivalent. Bear in mind that each skew is just that - skew - so can never touch any other - this means that two, successive dispositions of the same, moved skew cannot be incident with each other in any way at all. |
Projective Motion of a Point in a LineIf we wish to projectively move a point along a line, which is to say by elementary incidence alone, two, successive, ‘instantaneous’ positions of that point must form an interpunctual interval on the line — and then, three, successive such positions must form two, contiguous, interpunctual intervals on the line.If the projective motion of the point is to be constant and uniform, and expressed as some constant number, say v, of intervals per constant, successive interval of time (we might call v the ‘projective velocity’ of the moving point), then the interpunctual intervals must be projectively identical. The positions must therefore be successive end-points of the intervals of some projective ruler, and the projective motion of the point must consist of the translation from one end-point to the next. Given three, successive end-points of two, contiguous intervals, and a “terminator”, the corresponding projective ruler can be found. Please note carefully that the interval end-points are the only positions available to the projectively moving point, and that the moving point and the end-point are one and the same thing. There are, therefore, no intermediate points on the line where the moving point may temporarily, if briefly, lie. A projectively-moving point cannot slide on a line like a bead on a string, because the positions are ordinarily discrete - that is to say, distinct, non-continuous. Incidentally ...Recall that, projectively, a point is a place, a somewhere, and nothing else at all: we therefore here seem to consider how a place may be moved from place to place. This raises fairly obvious questions.Among them -
The places in question are, however, specified by incidence — in this case, of lines, because the incidence of two lines, if it exists, is a place, on a plane. So the question acquires a different emphasis, and becomes, ‘Can these places be projectively merged (made to degenerate), by twos, into an extension (that is to say, a line), such as to make the projective motion non-discrete (that is to say, continuous) – exactly, in fact, like a bead sliding on a string?’ |
The “Fundamental” TheoremThe validity of the famous “Fundamental Theorem” is keyed to continuity, as will be seen. |