2nd April 2005

Munkres §24

Ex. 24.2 (Morten Poulsen). Let f : S^{1} → R be a continuous map. Define g : S^{1} → R by g(s) = f(s) − f(−s). Clearly g is continuous. Furthermore

g(s) = f(s) − f(−s) = −(f(−s) − f(s)) = −g(−s),

i.e. g is an odd map. By the Intermediate Value Theorem there exists s_{0} ∈ S^{1} such that g(s_{0}) = 0, i.e. f(s_{0}) = f(−s_{0}).

This result is also known as the Borsuk-Ulam theorem in dimension one. Thus there are no injective continuous maps S^{1} → R, hence S^{1} is not homeomorphic to a subspace of R, which is no surprise.

Ex. 24.4. [1, §17]. Suppose that X is a linearly ordered set that is not a linear continuum. Then there are nonempty, proper, clopen subsets of X:

• If (x, y) = ∅ for some points x < y then (−∞, x] = (∞, y) is clopen and ≠ ∅, X.

• If A ⊂ X is a nonempty subset bounded from above which has no least upper bound then

the set of upper bounds B = ⋂_{a∈A}[a, ∞) = ⋃_{b∈B}(b, ∞) is clopen and ≠ ∅, X. Therefore X is not connected [2, §23].

Ex. 24.8 (Morten Poulsen).

(a).

Theorem 1. The product of an arbitrary collection of path connected spaces is path connected.

Proof. Let {A_{j}}_{j∈J} be a collection of path connected spaces. Let x = (x_{j})_{j∈J} and y = (y_{j})_{j∈J} be two points in ∏_{j∈J} A_{j}

For each j ∈ J there exists a path γ_{j} : [0, 1] → A_{j} between x_{j} and y_{j}, since A_{j} is path connected for all j. Now the map γ : [0, 1] → ∏_{j∈J} A_{j} defined by γ(t) = (γ_{j}(t))_{j∈J} is a path between x and y, hence the product is path connected. □

(b). This is not true in general: The set S = {x × sin(x^{−1} | 0 < x < π^{−1} } is path connected, but \overline{S} = S ∪ (\{0\} × [−1, 1]) is not path connected, c.f. example 24.7.

(c).

Theorem 2. If f : X → Y is a continuous map and X path connected then f(X) is path connected.

Proof. Clearly it suffices to consider the case where f is surjective. Let y_{1} and y_{2} be two points in Y. Then there exists x_{1} and x_{2} in X such that f(x_{1}) = y_{1} and f(x_{2}) = y_{2}. Since X path connected there is a path γ : [0, 1] → X between x_{1} and x_{2}. Now δ = f ∘ g : [0, 1] → Y is a path between δ(0) = f(γ(0)) = y_{1} and δ(1) = f(γ(1)) = y_{2}, hence Y path connected. □

(d).

Theorem 3. Let {A_{j}}_{j∈J} be a collection of path connected spaces. If ⋂_{j∈J} A_{j} is nonempty then ⋃_{j∈J} A_{j} is path connected.

Proof. Let a and b be two points in ⋃_{j∈J} A_{j} and let c be an element of ⋂_{j∈J} A_{j}. Note that there exists s and t such that a ∈ A_{s} and b ∈ A_{t}. Clearly c ∈ A_{s} ∩ A_{t}. Since A_{s} and A_{t} are path connected there exists a path f : [0, 1] → A_{s} between a and c and a path g : [0, 1] → A_{t} between c and b.

Now h : [0, 1] → ⋃_{j∈J} A_{j} defined by

h(t) = \{ \begin{matrix}f(2t), & 0 ≤ t ≤ 1/2 \\ g(2t − 1), & 1/2 ≤ t ≤ 1 \end{matrix}

is a path, by the Pasting lemma, between a and b, hence ⋃_{j∈J} A_{j} is path connected.

□

Ex. 24.10 (Morten Poulsen). Let U be a nonempty, open and connected subspace of R^{2} and let x_{0} ∈ U. Furthermore let A be the set of points in U that can be joined to x_{0} by a path in U.

A open: Let a_{0} ∈ A ⊂ U. Since U open there is an open rectangle V (= (a, b) × (c, d)) such that a_{0} ∈ V ⊂ U. Since V clearly is path connected it follows that V ⊂ A, hence A open.

A closed: Let u_{0} ∈ U − A. If every open rectangle containing u_{0} intersects A then clearly u_{0} ∈ A, hence u_{0} is not a limit point of A. Thus no point of U − A is a limit point of A, hence A is closed.

Since U connected it follows that A = U, hence U path connected.

Ex. 24.11 (Morten Poulsen). Let A be a subspace of X.

A connected ⇏ Int A and Bd A connected: If A = [0, 1] then Bd A = {0, 1} is not connected. If A = \overline{B(−1 × 0, 1)} ∪ \overline{B(1 × 0, 1)} ⊂ R^{2}, then Int A = B(−1 × 0, 1) ∪ B(1 × 0, 1) is not connected.

Int A connected ⇏ A connected: If A = (0, 1) ∪ {2} then Int A = (0, 1) is connected, but A is not connected.

Bd A connected ⇏ A connected: A = Q is not connected but Bd A = R is connected.

Int A and Bd A connected ⇏ A connected: One example is A = Q. An example with nonempty interior is

A = ([0, 1] × [0, 1]) ∪ ({0, 1} × [1, 2]) ∪ (([0, 1] ∩ Q) × {2}) ⊂ R^{2}

where

Bd A = ({0, 1} × [0, 2]) ∪ ({0, 1, 2} × [0, 1])

and

Int A = (0, 1) × (0, 1).

both are connected but A is not connected.

Ex. 24.12 (The long line). The idea is that the two linear continua S_{Ω} ×[0, 1) and Z_{+} ×[0, 1) = [1, ∞), or rather the long line L = (S_{Ω} × [0, 1)) − {a_{0} × 0} and the real line R = (Z_{+} × [0, 1)) − {1×0}, should have a great deal in common. L satisfies the conditions of a 1-dimensional manifold but 2nd countability. The long line is normal [Ex 32.8] but not metrizable [Ex 50.5].

(a) and (b). Easy.

(c). I do the hint first. Let a > a_{0} be an element of S_{Ω} which has no immediate predecessor (there are uncountably many such elements [Ex 10.6]). The set of predecessors S_{a} = {b ∈ S_{Ω} | b < a} = {b_{1}, b_{2}, …} is countable and the sets (b_{n}, a] = (b_{n}, a + 1) is a neighborhood basis at a. Since b_{1} is not an immediate predecessor, there is an element a_{1} ∈ (b_{1}, a]. Since sup{a_{1}, b_{2}} is not an immediate predecessor, there is an element a_{2} ∈ (sup{a_{1}, b_{2}}, a]. Proceeding inductively, we find a sequence of elements a_{n} < a such that a_{n} > a_{n−1} and a_{n} > b_{n} for all n. Then a_{n} is an increasing sequence and since a_{n} > b_{n}, b_{n−1}, …, b_{1} for all n, the sequence a_{n} converges to a.

Let now J be the set of points a such that [a_{0} × 0, a × 1) has the order type of [0, 1). I claim that J is inductive. Suppose that S_{a} ⊂ J. If a has an immediate predecessor a_{1} then [a_{0} × 0, a × 1) = [a_{0} × 0, (a − 1) × 1) ∪ [a × 0, a × 1) has the order type of [0, 1) by (a). Otherwise, [a_{0} × 0, a × 1) has the order type of [0, 1) by the hint and (b) (let x_{0}, x_{1}, x_{2}, … be the sequence a_{0} × 0, a_{1} × 1, a_{2} × 1 …).

(d,e). For every point a × t of S_{Ω} × [0, 1), the intervals [a_{0} × 0, a × 1) and [a_{0} × 0, a × t) have the order type of [0, 1) by (c) and (a). Then (a_{0} × 0, a × t) has the order type of (0, 1). Since intervals are convex, the subspace topology on (a_{0} × 0, a × t) is the order topology [Thm 16.4] so (a_{0} × 0, a × t) is homeomorphic to (0, 1). From this we see that any two points in L are contained in an interval homeomorphic to (0, 1) and therefore there is continuous path between them.

(f). Suppose that L is 2nd countable. Then also S_{Ω} − {a_{0}} is 2nd countable since this property is preserved under open continuous maps [Ex 16.4, Ex 30.12]. But S_{Ω} − {a_{0}} is not 2nd countable for it does not contain countable dense subsets. (Every countable subset of S_{Ω} − {a_{0}} ⊂ S_{Ω} is bounded [Thm 10.3] so that the complement contains an interval of the form (α, Ω) which is non-empty, in fact, uncountable [Ex 10.6], cf. [Ex 30.7].)

References

[1] Jesper M. Møller, General topology, http://www.math.ku.dk/~moller/e03/3gt/notes/gtnotes.pdf.

[2] James R. Munkres, Topology. Second edition, Prentice-Hall Inc., Englewood Cliffs, N.J., 2000. MR 57 #4063