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Milds # Comb space

In mathematics, particularly topology, a comb space is a particular subspace of $\mathbb {R} ^{2}$ 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.

## Formal definition

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

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

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.

1. The comb space is an example of a path connected space which is not locally path connected.

2. 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.

3. 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 ƒ : [0, 1] → D be this path. We shall prove that ƒ −1{p} is both open and closed in [0, 1] contradicting the connectedness of this set. Clearly we have ƒ −1{p} is closed in [0, 1] by the continuity of ƒ. To prove that ƒ −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 ƒ −1{p}. Clearly, f(x) = p. Then since f −1(V) is open, there is a basis element U containing x such that ƒ(U) is a subset of V. We assert that ƒ(U) = {p} which will mean that U is an open subset of ƒ −1{p} containing x. Since x was arbitrary, ƒ −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, ƒ(U) is connected. Suppose ƒ(U) contains a point s other than p. Then s = (1/nz) must belong to D. Choose r such that 1/(n + 1) < r < 1/n. Since ƒ(U) does not intersect the x-axis, the sets A = (−∞, r) × $\mathbb {R}$ and B = (r, +∞) × $\mathbb {R}$ 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.

4. The comb space is homotopic to a point but does not admit a deformation retract onto a point for every choice of basepoint.