We know of a charcterization of spaces homeomorphic to [0,1], as being metric continua with 2 noncut points. We have as well a characterization of spaces homeomorphic to the unit circle. I can't find characterization of spaces homeomorphic to the unit disk. I'm using Willard's General Topology and Hocking and Young's Topology. Any suggestion would be appreciated.
Asked
Active
Viewed 612 times
5
-
1You should be more specific about the kind of characterization you are looking for. The disk can be characterized among surfaces pretty easily. Also, the segment is far from being the only metric continuum with 2 non cut points, so I am not sure about the meaning of "characterization" in your question. – Benoît Kloeckner Dec 18 '13 at 11:43
-
2The OP may be referring to characterizations such as in http://mathoverflow.net/questions/123760 or http://mathoverflow.net/questions/76134 . – Emil Jeřábek Dec 18 '13 at 12:03
-
1The reading of the above questions made me realize that my last sentence is based on an error of mine, sorry. I stick to the first part of my comment, though. – Benoît Kloeckner Dec 18 '13 at 14:50
-
After reading about the topological characterization of the spaces homeomorphic to the unit interval and spaces hoeomorphic to the unit cirlce, it was only natural to ask about characterization of the unit disk along the same lines. I thought about the following: Let K be a metric continuum. Denote by B the boundary of K and by P(a,b) a path from a to b in K. Suppose that for any a, b on B, K-P(a,b) is not connected, then K is homeomorphc to the unit disk. I searched for literature on this subject, and I couldn't find any. I was hoping I would find sources @BenoîtKloeckner – user116340 Dec 18 '13 at 22:34
-
The specifics you give in comments should have been in the question; but I don't know what the "boundary" of a topological space is in general. Moreover, under any reasonable interpretation of this word, your attempted characterization fails to distinguish between the disc and any not too trivial tree. – Benoît Kloeckner Dec 19 '13 at 09:27
3 Answers
2
A topological characterisation of the disk is given by Zippin (see R L Wilder, "Topology of Manifolds", 1949). Here is the statement.
Theorem [Zippin]: Suppose that $D$ is a Peano continuum, containing a subspace $C$ homeomorphic to $S^1$. Suppose that:
- $D$ contains at least one arc spanning $C$.
- Every arc spanning $C$ separates $D$.
- No proper subset of an arc spanning $C$ separates $D$.
Then $D$ is homeomorphic to the unit disk, and $C$ is its boundary.
Here an arc is an embedding of the unit interval. Also, an arc spans $C$ if it meets $C$ exactly in its endpoints.
Sam Nead
- 26,191
1
It seems clear from the formulation that the question is about a topological characterisation of the unit disc, presumably the closed one. The correct characterisation of the one-dimensional case can be found in the classic "Dynamic topology" by Whyborn and Duda: a compact, connected, second countable space for which every point with the exception of two specified ones (the endpoints) is a cut point. A perhaps not very satisfactory characterisation of the closed unit disc is then that the space be homeomorphic to the product of two such spaces.
user6891
- 106
1
Here is an excerpt from a mathscinet review of the paper Johannes de Groot - Topological characterization of metrizable cubes (1972):
A connected $T_1$ space of dimension $n=1,2,⋯,\infty$ is homeomorphic to $I_n$ ($I_{\infty}$ being the Hilbert cube) if and only if it has a countable subbase that is both comparable and binary. A subbase $\sigma$ for the closed sets of a space is comparable if any two members of $\sigma$ disjoint from a given member of $\sigma$ are comparable (one contains the other). If every linked subcollection of $\sigma$ (i.e., each pair of the subcollection has a nonempty intersection) has a nonempty intersection, then $\sigma$ is said to be binary.
If course, $I_2$ is homeomorphic to the closed unit disk. The papers itself was fun to read. Highly recommend it!
erz
- 5,385