Comb space

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In mathematics, particularly topology, a comb space is a subspace of $\R^2$ that looks rather like a comb. The comb space satisfies some rather interesting properties and provides interesting counterexamples. The topologist's sine curve satisfies similar properties to the comb space. The deleted comb space is an important variation on the comb space.

1 Formal definition
2 Topological properties
3 See also
4 References

[edit] 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:

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

is just the comb space with the line segment $\{0\} \times (0,1)$ deleted.

[edit] Topological properties

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

1. We shall note that the comb space is clearly path connected and hence connected. Also, if we deleted the set (0 X [0,1]) out of the comb space, we obtain a new set whose closure is the comb space. Since this ‘new set’ is connected, and the deleted comb space, D, is a superset of this ‘new set’ and a subset of the closure of this new set, the deleted comb space is also connected.

2. However, the deleted comb space is not path connected since there is no path from (0,1) to (0,0).

3. The comb space is an example of a path connected space which is not locally path connected; see the page on locally connected spaces.

4. Let us prove our claim in 2. Suppose there is a path from p = (0, 1) to a point q 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. Then there is a basis element U containing ƒ⁻¹{p} such that ƒ(U) is a subset of V. We know that U is connected since it is a basis element for the order topology on [a, b]. Therefore, ƒ(U) is connected. We assert that ƒ(U) = {p} so that ƒ⁻¹{p} is open. Suppose ƒ(U) contains a point (z, 1/n) other than p. Then (z, 1/n) must belong to D. Choose r such that 1/(n − 1) < r < 1/n. Since ƒ(U) doesn’t intersect the x-axis, the sets:

A = (−∞, r) × R

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.

[edit] See also

[edit] References

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