From: orjanjo@math.ntnu.no (Orjan Johansen) Newsgroups: sci.math Subject: Re: Quick Question Date: 13 Sep 1998 14:43:44 GMT In article , Jim Nastos wrote: > >I'm wondering if anyone could provide an example of a spacefilling curve >for me ... perhaps, along with it, an algorithm to create a curve thatr >may seem to fill space but actually doesn't. Start with a square, and one edge of the square. Then divide this up into four smaller squares with an edge such that the designated edges of the smaller squares form a curve starting and ending at the same points as the edge of the large one. Once you have this you can repeat the process, and in the limit all the squares you get will contain a bit of the final curve. (View with a fixed width font.) _________ __ __ | | | | | | | | | | | | | | | | | | | | | |____|____| |_________| __| |__ | | | | | | | | | |_____|_____| | | | |____|____| As you see, there will be some overlap between the edges. > This is out of personal curiosity ... if possible, could you also >explain why the space-filling curve is such? Since the squares get arbitrarily small, the final curve will be dense in the original square. But the image of a curve is closed, and so it must be all of the square. >Does there need to exist a bijection between the points on the curve >and the points in the area/space? They are the same points. However, the curve itself is a function from an interval to the square. It is onto, but not one-to-one - the curve intersects itself an infinite number of times. No continuous function from the interval to the square can be a bijection. The simplest way to see this is to use connectedness - removing a point from a square will keep it connected, removing a point from an interval need not. Greetings, Ørjan. -- A pro-spam bill has passed the US Senate, and is now in the House of Representatives. See and . ============================================================================== Newsgroups: sci.math From: "Jeroen Bruijning" Subject: Re: Quick Question Date: Mon, 14 Sep 1998 10:02:13 GMT Orjan Johansen wrote in message <6tgln0$kt6$1@due.unit.no>... >No continuous function from the interval to the square can be a >bijection. The simplest way to see this is to use connectedness - >removing a point from a square will keep it connected, removing a >point from an interval need not. This is true, but the reason quoted is insufficient. You need to use a few elementary topological facts, namely, that [0,1] is compact and that the square is Hausdorff. This implies that a continuous function is closed, therefore, a continuous bijection is a homeomorphism. This in turn makes the connectedness argument valid. To see that compactness is essential, consider wrapping the half-open interval [0,1) around a circle exactly once. This is a continuous bijection, but not a homeomorphism as the connectedness argument as quoted above shows: the circle with a point removed stays connected, while [0,1) with a (mid-)point removed gets disconnected. J.