↑ arvau

Wikipedia / Comb space

In mathematics, particularly topology, a comb space is a particular subspace 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.

Formal definition

with its standard topology and let K be the set. The set C defined by:

equipped with the subspace topology is known as the comb space. The deleted comb space, D, is defined by:

  1. .

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:

    1. . E is also path connected and the closure 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):

    1. 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 f1{p} is both open and closed in [0,1] contradicting the connectedness of this set. Clearly we have f1{p} is closed in [0,1] by the continuity of f. To prove that f1{p} is open, we proceed as follows: Choose a neighbourhood V (open in R2) about p that doesn't intersect the x–axis. Suppose x is an arbitrary point in f1{p}. Clearly, f(x) = p. Then since f1(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 f1{p} containing x. Since x was arbitrary, f1{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) and B = (r, +) will form a separation on f(U); contradicting the connectedness of f(U). Therefore, f1{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

See also

References