Comb space

{{Short description|Pathological topological space}} In mathematics, particularly topology, a '''comb space''' is a particular subspace of <math>\R^2</math> that resembles a comb. The comb space has properties that serve as a number of counterexamples. The topologist's sine curve has similar properties to the comb space. The '''deleted comb space''' is a variation on the comb space. right|thumb|Topologist's comb right|thumb|The intricated double comb for r=3/4.

==Formal definition==

Consider <math>\R^2</math> with its standard topology and let ''K'' be the set <math>\{1/n ~|~ n \in \mathbb N\}</math>. The set ''C'' defined by:

:<math>(\{0\} \times [0,1] ) \cup (K \times [0,1]) \cup ([0,1] \times \{0\})</math>

considered as a subspace of <math>\R^2</math> equipped with the subspace topology is known as the comb space. The deleted comb space, D, is defined by:

:<math>(\{0\} \times \{1 \}) \cup (K \times [0,1]) \cup ([0,1] \times \{0\}) </math>.

This is the comb space with the line segment <math>\{0\} \times (0,1)</math> deleted.

==Topological properties==

The comb space and the deleted comb space have some interesting topological properties mostly related to the notion of connectedness.

* The comb space, C, is path connected and contractible, but not locally contractible, locally path connected, or locally connected. * The deleted comb space, D, is connected: *:Let E be the comb space without <math>\{0\} \times (0,1] </math>. E is also path connected and the closure of E is the comb space. As E <math>\subset</math> D <math>\subset</math> the closure of E, where E is connected, the deleted comb space is also connected. *The deleted comb space is not path connected since there is no path from (0,1) to (0,0): *:Suppose there is a path from ''p'' = (0, 1) to the point (0, 0) in ''D''. Let ''f'' : [0, 1] → ''D'' be this path. We shall prove that ''f'' −1{''p''} is both open and closed in [0, 1] contradicting the connectedness of this set. Clearly we have ''f'' −1{''p''} is closed in [0, 1] by the continuity of ''f''. To prove that ''f'' −1{''p''} is open, we proceed as follows: Choose a neighbourhood ''V'' (open in '''R'''2) about ''p'' that doesn't intersect the ''x''–axis. Suppose ''x'' is an arbitrary point in ''f'' −1{''p''}. Clearly, ''f''(''x'') = ''p''. Then since ''f'' −1(''V'') is open, there is a basis element ''U'' containing ''x'' such that ''f''(''U'') is a subset of ''V''. We assert that ''f''(''U'') = {''p''} which will mean that ''U'' is an open subset of ''f'' −1{''p''} containing ''x''. Since ''x'' was arbitrary, ''f'' −1{''p''} will then be open. We know that ''U'' is connected since it is a basis element for the order topology on [0, 1]. Therefore, ''f''(''U'') is connected. Suppose ''f''(''U'') contains a point ''s'' other than ''p''. Then ''s'' = (1/''n'', ''z'') must belong to ''D''. Choose ''r'' such that 1/(''n'' + 1) < ''r'' < 1/''n''. Since ''f''(''U'') does not intersect the ''x''-axis, the sets ''A'' = (−∞, ''r'') × <math>\R</math> and ''B'' = (''r'', +∞) × <math>\R</math> will form a separation on ''f''(''U''); contradicting the connectedness of ''f''(''U''). Therefore, ''f'' −1{''p''} is both open and closed in [0, 1]. This is a contradiction. * The comb space is homotopic to a point but does not admit a strong deformation retract onto a point for every choice of basepoint that lies in the segment <math>\{0\} \times (0,1] </math>

==References== * {{cite book | author = James Munkres | author-link = James Munkres | year = 1999 | title = Topology | edition = 2nd | publisher = Prentice Hall | isbn = 0-13-181629-2 }} *{{Cite encyclopedia|title=Connectedness|encyclopedia=Encyclopedic Dictionary of Mathematics|publisher=Mathematical Society of Japan|editor=Kiyosi Itô}}

==External links== * {{mathworld |urlname=CombSpace |title=Comb space }}

Category:Topological spaces Category:Trees (topology)