Topology/Comb Space

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.

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.

Topological properties
The comb space and the deleted comb space satisfy some interesting topological properties mostly related to the notion of local connectedness (see next chapter).

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 space (next chapter).

4. Let us prove our claim in 2. Suppose there is a path from p = (0, 1) to a point q in D, q ≠ p. Let ƒ:[0, 1] → D be this path. We shall prove that ƒ&minus;1{p} is both open and closed in [0, 1] contradicting the connectedness of this set. Clearly we have ƒ&minus;1{p} is closed in [0, 1] by the continuity of ƒ. To prove that ƒ&minus;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. Then there is a basis element U containing ƒ&minus;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&minus;1{p} is open. Suppose ƒ(U) contains a point (1/n, z) other than p. Then (1/n, z) 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 = (&minus;&infin;, r) &times; $$\R$$


 * B = (r, +&infin;) &times; $$\R$$

will form a separation on f(U); contradicting the connectedness of f(U). Therefore, f&minus;1{p} is both open and closed in [0, 1]. This is a contradiction.

NOTE TO READER
ATTEMPT QUESTIONS 2.c), 2.d) AND 3 IMMEDIATELY AFTER STUDYING THE NEXT SECTION.

Exercises
1. a)* Prove that the comb space is compact without using the Heine Borel theorem. b) HENCE show that the set K = {1/n | n is a natural number} U {0}    is compact (Hint: Prove that if X  X  Y is a product space, and Y is compact, then the projection onto the first co-ordinate is a closed map (i.e, maps closed sets in X  X  Y onto closed sets in X). Then if C is the comb space, C is a closed subset of I X I (I = [0,1]) given the product topology. Assume that I = [0,1] is compact and use a theorem from the section on compactness) c) Show that the deleted comb space is not compact

2*. a) Let A be a connected subset of R. Show that if x is in A, y is in A with x < y, then the whole interval [x,y] is a subset of A. b) Show that a compact subset of R necessarily contains both its supremum and infimum (Hint: If A is a compact subset of R, A is closed. Prove that both the supremum of A and infimum of A belong to the closure of A and hence to A.) c) Show that every closed interval in R is locally connected. d) Show that the comb space cannot be imbedded in R (Hint: Suppose it could be imbedded in R and let A be the subset of R that the comb space, C, is homeomorphic to. By noting that the comb space is path connected and hence connected, and that A must be compact (since C is homeomorphic to A and C is compact by exercise 1.a)), show that A has to be a closed interval. Therefore, A is locally connected by exercise 2.c) e) Can the deleted comb space be imbedded in R? Justify your answer.

3. a) Prove that an open subspace of a locally connected space is locally connected. b) Let X be locally homeomorphic to Y; that is there is a map f from X to Y that satisfies the following property:

For each point x of X, there is a neighbourhood V of x that is homeomorphic to an open subset of Y under the map f (i.e, the map f restricted to V is the homeomorphism)

Prove that if Y is locally connected, so is X (Hint: Use part a))

c) Let C be the comb space. Prove that C is not a manifold (a manifold is a Hasudorff topological space X that has a countable base for its topology and is locally homeomorphic to R^n for some integer n). (Hint: Use part b) and note that a subspace of a Haudorff space is Haudorff, and that a subspace of a space having a countable basis for its topology also has a countable basis for its topology).