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Bobby Moretti Math 441 Homework # 11/19/
Munkres §23.1 Let T and T ′^ be two topologies on X. If T ′^ ⊇ T , what does connectedness of X in one topology imply about connectedness in the other?
Let U ∪V be any separation of T. Then U ∈ T ′^ and V ∈ T ′. Since these are disjoint nonempty sets whose union is all of X, they also form a separation of T ′. So if T ′^ is connected, so is T. The converse is not true. For example, let X be any set with at least two elements. Let T be the trivial topology on X, and let T ′^ be the discrete topology on X. There are only two open sets in T : the sets ∅ and X. These sets do not form a separation of X. So X under the topology T is connected. Now, certainly these two sets are contained in T ′, since every topology on X must contain them. But let A be any nonempty proper subset of X. We know that A exists, since X has at least two elements. Then every subset of X is a member of T ′, so it is certainly true that A ∈ T ′. But then X r A is certainly an open set, since X r A is a nonempty proper subset of X. But then
A ∪ (X r A)
is a separation of X. So T ′^ is not connected. Thus T being connected does not gaurantee that T ′^ is connected as well.
Munkres §23.2 Let { An } be a sequence of connected subspaces of X, such that An ∩ An+ 1 6 = ∅ for all n. Show that
⋃ An is connected.
Let Y =
⋃ An.
Suppose that there exists a separation of Y. Then there exist nonempty disjoint open sets U,V such that U ∪ V = Y. By lemma 23.2 we know that for every integer i, the set Ai either lies entirely within U or entirely within V. Consider the set A 1. Being an element of the sequence, it belongs entirely to either U or V. Suppose, without loss of generality, that it lies entirely within U (that is, A 1 ⊆ U, and A 1 ∩V = ∅). We claim that there exists an integer k such that Ak lies entirely within V. For if not, then Ak would lie entirely within U for all k. But then V would be empty, which is impossible. So there must be some k such that Ak− 1 lies entirely within U, and that Ak lies entirely within V. This contradicts the fact that Ak− 1 and Ak are disjoint. Thus no such separation can exist, and
⋃ An is connected.
Munkres, §23.4 Show that if X is an infinite set, it is connected in the finite com- plement topology.
Suppose that U is a nonempty set such that both U and its complement X rU are open in the finite complement topology. Then X rU is a finite set, since U is open. But the complement of X r U is the complement of a finite set. The complement of a finite subset of an infinite space must be infinite. It cannot be finite. Hence X r U is not open, which is a contradiction. Thus no such separation U,V can exist. Hence any infinite set X under the finite complement topology is connected.
Munkres, §23.11 Let p : X → Y be a quotient map. Show that if each set p−^1 ({ y }) is connected, and if Y is connected, then X is connected.
Suppose that X is not connected. Then we can separate X into disjoint nonempty open sets U and V. Consider the set of all y ∈ Y whose preimage in X is a subset of U:
U′^ =
y ∈ Y
∣ (^) p−^1 ({ y }) ⊆ U
We claim that U′^ is open in Y. For if x ∈ p−^1 (U′), then p−^1 ({ p(x) }) ⊆ U. But the point x gets mapped to the point p(x), so x is in the preimage of { p(x) }. Thus x ∈ U, so p−^1 (U′) ⊆ U. Conversely, if u ∈ U, then the subspace p−^1 ({ p(u) }) of X is connected. But U and V form a separation of X, so by lemma 23.2 in Munkres, we know that either p−^1 ({ p(u) }) lies entirely in U or entirely in V. But it cannot lie entirely in V , since u ∈ p−^1 ({ p(u) }). So we have
p−^1 ({ p(u) }) ⊆ U.
Hence U ⊆ p−^1 (U′). We have proved both inclusions, so U = p−^1 (U′). Using an identical argument, it is clear that
V = p−^1 (V ′),
where V ′^ =
y ∈ Y
∣ (^) p−^1 ({ y }) ⊆ V }^.
Now p is a quotient map. So, since p−^1 (U′) is open, U′^ is open in Y as well. Likewise, the set p−^1 (V ′) is open in X, so the set V ′^ is open in Y. We claim that U′^ and V ′^ form a separation of Y. For if z ∈ U′^ ∩V ′, then p−^1 ({ z }) ⊆ U ∩ V , and p−^1 ({ z }) is nonempty since p is surjective. But this is imposible, since U ∩V = ∅. And if y ∈ Y , then p−^1 ({ y }) ∈ U or p−^1 ({ y }) ∈ V , so y ∈ V ′^ or y ∈ U′. Thus U′^ ∪V ′^ = Y , and we have exhibited a separation of Y , which is impossible, since Y is connected. So there exists no separation of X. Thus X is connected.
Munkres, §24. (a) Show that no two of the spaces ( 0 , 1 ), ( 0 , 1 ], [ 0 , 1 ] are homeomorphic. (b) Suppose that there exist imbeddings f : X → Y and g : Y → X. Show by means of an example that X and Y need not be homeomorphic. (c) Show that Rn^ and R are not homeomorphic if n > 1.
Munkres, §24. (a) Let X and Y be ordered sets in the order topology. Show that if f : X → Y is order preserving and surjective, then f is a homeomorphism. (b) Let X = Y = R+. Given a positive integer n, show that the function f (x) = xn^ is order preserving and surjective. Conclude that its inverse, the nth root function, is continuous. (c) Let X be the subspace (−∞, − 1 )∪[ 0 , ∞) of R. Show that the function f : X → R defined by setting f (x) = x + 1 if x < −1, and f (x) = x if x ≥ 0, is order preserving and surjective. Is f a homeomorphism? Compare with (a).
(a) Let’s first show that f is injective. Since < is an order relation, by property (1), if a 6 = b, then either a < b or b < a. Hence either f (a) < f (b) or f (b) < f (a), since f preserves order. Thus by property (2), we have f (a) 6 = f (b) in either case. So f is injective. We are given that it is surjective. Hence f is bijective, and f −^1 preserves order (if f −^1 does not preserver order, it is trivial to see that f does not). To show that f and f −^1 are continuous, it will suffice to examine basis open sets. So let U be a basis open set in X. Then U = (a, b) for some points a 6 = b ∈ X. Hence f (U) =
f (a), f (b)
These are clearly all open sets in Y. Likewise, if V is a basis open set in Y , then V = (c, d) for some points c 6 = d ∈ Y. And since f −^1 is order preserving, we know that f −^1 (V ) =
f −^1 (c), f −^1 (d)
which is open in X. Thus f is a homeomorphism. (b) We prove that the nth power function is order preserving by using induction on n. For n = 1, we have that x < y implies x < y, which is obviously true. For n = k, suppose that xk^ < yk. Then xkx < ykx by a basic property of the real numbers. But by the same property, ykx < yky. Putting this together, we have xk+^1 < ykx < yk+^1 , as desired. So the nth power function is order preserving. It is surjective since every point x ≥ 0 has a unique nth root, that is, a unique y ∈ R such that yn^ = x, for every positive n (this is another basic property of R). So by part (a), the nth power function is a homeomoprhism of R, and thus its invers, the nth root function, is continuous. (c) The function f : X → R cannot be a homeomorphism, for R is connected, while X is clearly not. The problem is that the topology that X inherits from R in the subspace topology is not the same as the restriction of the ordering of R to X. For example, the set [ 0 , ∞) is open in the subspace topology, but it is not open in the order topology on X inherited from R.
Munkres, §24.10 Show that if U is an open connected subspace of R^2 , then U is path connected. [Hint: Show that given x 0 ∈ U, the set of points that can be joined to x 0 by a path in U is both open and closed in U.]
Fix any x 0 ∈ U. Let C = { u ∈ U | there exists a path between x 0 and u }. If u ∈ C, then u ∈ U, and since U is open, we can find a ball around u of radius δ such that B(u, δ ) ⊆ U. Now if x ∈ B(u, δ ), then there exists a path from x to u (a straight line segment will do). Join this path to the path connecting x 0 and u, which exists since u ∈ C (we can always use the pasting lemma to construct such a path). Thus x can be joined to x 0 by a continuous path. So x ∈ C. Thus B(u, δ ) ⊆ C, hence C is open in U. Now we have that
U rC = { u ∈ U | there is no path connecting x 0 and u }.
Pick any v ∈ U rC. Then v ∈ U, so there exists a real number ε such that B(v, ε) ⊆ U. Suppose x is in B(v, ε). Now we claim that the points x and x 0 cannot be joined by a path. For if they could be joined by a path f , joining the straight line segment from x to v would result in a path from v to x 0 , which is impossible, for v ∈ U r C, which means that no continuous path can connect v and x 0. Thus x ∈ U r C, and hence B(v, δ ) ⊆ U rC. So U rC is open; consequently, C is closed. Now U is a connected space. So the only set in U that is simultaneously closed and open is U itself. So C = U. We have shown that for any x 0 ∈ U, the set of points that can be connected to it by a path in U is equal to U itself. So U is path connected.
