comb space

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

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

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

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

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

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 $\backslash \{0\backslash \}\; \backslash times\; (0,1]$. E is also path connected and the closure of E is the comb space. As E $\backslash subset$ D $\backslash subset$ 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) × $\backslash R$ and B = (r, +∞) × $\backslash R$ 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 $\backslash \{0\backslash \}\; \backslash times\; (0,1]$

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

• {{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}}

Category:Topological spaces

Category:Trees (topology)