# Comb space

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

## Formal definition

Consider with its standard topology and let *K* be the set . The set *C* defined by:

considered as a subspace of equipped with the subspace topology is known as the comb space. The deleted comb space, D, is defined by:

- .

This is the comb space with the line segment 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 . E is also path connected and the closure of E is the comb space. As E D 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**R**^{2}) 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/*n*,*z*) 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*) × and*B*= (*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.

- Suppose there is a path from

## See also

## References

- James Munkres (1999).
*Topology*(2nd ed.). Prentice Hall. ISBN 0-13-181629-2.

- Kiyosi Itô (ed.). "Connectedness". Encyclopedic Dictionary of Mathematics. Mathematical Society of Japan. Cite journal requires
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