Comb space

Template:Short description In mathematics, particularly topology, a comb space is a particular subspace of ${ℝ}^2$ 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.

Topologist's comb

The intricated double comb for r=3/4.

Formal definition

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

$(\{0\}\times [0,1])\cup (K\times [0,1])\cup ([0,1]\times \{0\})$

considered as a subspace of ${ℝ}^2$ equipped with the subspace topology is known as the comb space. The deleted comb space, D, is defined by:

$(\{0\}\times \{1\})\cup (K\times [0,1])\cup ([0,1]\times \{0\})$.

This is the comb space with the line segment $\{0\}\times (0,1)$ 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:

  Let E be the comb space without $\{0\}\times (0,1]$. E is also path connected and the closure of E is the comb space. As E $\subset$ D $\subset$ 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 R²) 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) × $ℝ$ and B = (r, +∞) × $ℝ$ 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 $\{0\}\times (0,1]$

See also

• Connected space
• Hedgehog space
• Infinite broom
• List of topologies
• Locally connected space
• Order topology
• Topologist's sine curve

References

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• Template:Cite encyclopedia