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- From: scavo@cie.uoregon.edu (Tom Scavo)
- Subject: Re: Is this thing a fractal?
- Message-ID: <1992Nov5.181057.15032@nntp.uoregon.edu>
- Keywords: dimension.
- Sender: news@nntp.uoregon.edu
- Organization: University of Oregon Campus Information Exchange
- References: <1992Oct28.175819.4696@nntp.uoregon.edu> <1992Oct29.193113.13763@murdoch.acc.Virginia.EDU> <1816@spam.ua.oz>
- Date: Thu, 5 Nov 92 18:10:57 GMT
- Lines: 58
-
- In article <1816@spam.ua.oz> ahanysz@spam.ua.oz (Alexander Hanysz) writes:
- >
- >Now I'm going to ask that eternally difficult question: WHY? The
- >Mandelbrot set is such a complicated object, how do we know that its
- >boundary has topological dimension 1?
-
- A proof was given earlier in this thread by Gerald Edgar.
-
- >More fundamentally, does anyone know a nice _definition_ of topological
- >dimension? It's hard to argue when you don't know what the terms mean.
- >All I've been told about topological dimension is "a line has dimension
- >1, a surface has dimension 2, a volume..." I have yet to see a formal
- >definition.
-
- I suspect that others have the same question so here goes:
-
- A set has topological dimension 0 if every point has arbitrarily
- small neighborhoods whose boundaries do not intersect the set.
-
- A set S has topological dimension k if each point in S has
- arbitrarily small neighborhoods whose boundaries meet S in a
- set of dimension k-1 , and k is the least nonnegative integer
- for which this holds.
-
- See, for example, Section 14.5 in Devaney's _A First Course in Chaotic
- Dynamical Systems_. (Does anyone know of a better reference?)
-
- For instance, the rationals have topological dimension 0 since the
- boundary of a disk with irrational radius fails to intersect the
- rationals (that is, the sum of a rational and an irrational is
- irrational). The irrationals also have topological dimension 0 ,
- I believe. Suppose it were true that (1) if A is a subset of X ,
- then the t-dim A <= t-dim X ; and (2) for any A in R^n , t-dim A
- = n if and only if A contains a nonempty open subset of R^n .
- Then by (1) the irrationals have topological dimension no bigger than
- 1. But by (2) they can not have topological dimension equal to 1
- since they contain no open subset. Hence, they have topological
- dimension 0 . (Can anyone think of a simpler proof?)
-
- Here's another example. Consider the fractal generated by the
- following geometric transformation:
-
- _____
- | |
- | |
- --------------- --> ----- -----
-
- Each leg of this "hat" is one-third the length of the original line
- segment. Now the fractal dimension of the resulting image is
- log 5 / log 3 , but the topological dimension is 1 since any disk
- intersects the curve at a set of discrete points having topological
- dimension 0 . (There's a related exercise in Devaney that I can't
- solve: Can a fractal that is totally disconnected, with topological
- dimension 0 , have a fractal dimension larger than 1 ?)
-
- --
- Tom Scavo
- scavo@cie.uoregon.edu
-