In mathematics, particularly topology, a comb space is a particular subspace of R 2 {\displaystyle \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.

Topologist's comb
The intricated double comb for r=3/4.

Formal definition

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

( { 0 } × [ 0 , 1 ] ) ∪ ( K × [ 0 , 1 ] ) ∪ ( [ 0 , 1 ] × { 0 } ) {\displaystyle (\{0\}\times [0,1])\cup (K\times [0,1])\cup ([0,1]\times \{0\})}

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

( { 0 } × { 1 } ) ∪ ( K × [ 0 , 1 ] ) ∪ ( [ 0 , 1 ] × { 0 } ) {\displaystyle (\{0\}\times \{1\})\cup (K\times [0,1])\cup ([0,1]\times \{0\})}.

This is the comb space with the line segment { 0 } × ( 0 , 1 ) {\displaystyle \{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 } × ( 0 , 1 ] {\displaystyle \{0\}\times (0,1]}. E is also path connected and the closure of E is the comb space. As E ⊂ {\displaystyle \subset } D ⊂ {\displaystyle \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 {\displaystyle \mathbb {R} } and B = (r, +∞) × R {\displaystyle \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.
  • 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 } × ( 0 , 1 ] {\displaystyle \{0\}\times (0,1]}

See also

External links