# Math4225 Homework 8

7.2.1 The trivial topology is always connected. The discrete topology is connnected if and only if the set is either empty or contains one element.

7.2.2

Exercise 4.5.8

1. Connected. Any two non-empty open subsets intersect because any two topological bases intersect.

2. Connected. Any two non-empty open subsets intersect.

3. Connected. Any two non-empty open subsets intersect.

Exercise 4.5.9

1. Connected. Coarser than the usual topology, which is connected.

2. Not connected. R = (-∞, 0) ∪ [0, ∞) is a separation.

3. Not connected. R = (-∞, 0) ∪ [0, ∞) is a separation.

4. Connected. Any two non-empty open subsets intersect.

5. Connected. Any two non-empty open subsets intersect.

6. Connected. Any two non-empty open subsets intersect.

Exercise 4.6.3

1. Connected. Any two non-empty open subsets intersect because any two topological bases intersect.

2. Not connected. Z = (-∞, 3] ∪ [4, ∞) is a separation.

3. Connected. Any two non-empty open subsets intersect because any two topological bases intersect.

4. Connected. Any two non-empty open subsets intersect because any two topological bases intersect.

Since {1, 2, 3, 4} = {1, 2, 3} ∪ {4} is a separation, the topology is not connected. By intersecting with subsets, we get the separations for the subspaces {1, 2, 4}, {1, 3, 4}, and {2, 3, 4}. Therefore these subspaces are not connected. On the other hand, there is no nontrivial open and closed subsets in the subspace {1, 2, 3}. Therefore {1, 2, 3} is connected.

7.2.4 If a subset A contains two points. Choose c between the two points, then (A ∩ (-∞, c)) ∪ (A ∩ [c, ∞)) is a separation of A in the lower limit topology. By choosing c to be rational, this is also a separation of A in the Michael line. Therefore connected subsets are single points in both cases.

For any two points a, b in a subset A, assume a < b. Any open subset of A containing a must contain A ∩ (c, ∞) for some a > c. Therefore the open subset must also contain b. This shows that there are no disjoint open subsets in A, so that A is connected.

7.2.5 If X has a separation X = AB, then f(x) = 1 for xA and f(x) = 2 for xB is a continuous map onto two points with discrete topology. Conversely, if f: X → {1, 2} is a continuous map onto discrete two point space, then X = f-1(1) ∪ f-1(2) is a separation of X.

7.2.6 A separation in a coarser topology is still a separation in a finer topology. Therefore a coarser topology is not connected implies a finer topology is not connected. This is the same as: a finer topology is connected implies a coarser topology is also connected.

7.2.7 1. If AU and BV for some disjoint and open subsets U and V in X, then A = U ∩ (AB) and B = V ∩ (AB). Therefore both A and B are open subsets of AB.

2. Since A and B are disjoint open subsets of the subspace AB, by Exercise 5.3.17, there are disjoint open subsets U and V in X, such that A = U ∩ (AB) and B = V ∩ (AB). The two equalities are the same as AU and BV.

3. The counterexample in the second part of Exercise 5.3.17 also provides a counterexample for non-metric spaces. Let {1, 2, 3} have the topology { ∅, {1, 3}, {2, 3}, {3}, {1, 2, 3} }. Then the open subsets A = {1} and B = {2} form a separation of {1, 2}. However, any two open subsets of {1, 2, 3} respectively containing A and B must intersect.

7.2.8 Let U be a nonempty open and closed subset of ∪ Ai. Then for each i, UAi is open and closed subset (in the subspace topology of Ai). Since Ai is connected, we have either UAi = ∅ or AiU.

Let a be the commom point of Ai. Since U ≠ ∅, we must have some index i0, such that Ai0U. This implies aU, so that UAi ≠ ∅ for any index i. We thus conclude AiU for all i. This implies ∪ AiU, so that U = ∪ Ai.

We just proved that any nonempty open and closed subset of ∪ Ai must be the whole union. This implies ∪ Ai is connected.

7.2.10 Suppose A is not an interval, we claim that there is aA, such that (-∞, x) ∩ A ≠ ∅ and (x, ∞) ∩ A ≠ ∅. Once the claim is true, we have a separation A = [(-∞, x) ∩ A] ∪ [(x, ∞) ∩ A], so that A is not connected.

Assume our claim is not true. Then R - A = XY, where any xX satisfies (-∞, x) ∩ A = ∅ and any yY satisfies (y, ∞) ∩ A = ∅. X clearly have the property that x' < xX implies x'X. This implies that if a = supX (the supremum of all numbers in X), then X = (-∞, a}, where } could be either an open bracket ) or a closed bracket ]. By similar reason, we have Y = {b, ∞). Thus A = R - XY = {a, b} is an interval.

7.2.15 Consider the continuous function g(x) = f(x) - x defined on the interval [0, 1]. Since 0 ≤ f(x) ≤ 1 for any 0 ≤ x ≤ 1, we have g(0) = f(0) - 0 ≥ 0 and g(1) = f(1) - 1 ≤ 0. Since [0, 1] is connected, the continuous image g[0, 1] is also connected and must be an interval. In particular, any value between g(0) and g(1) can be reached. Since g(1) ≤ 0 ≤ g(0), we conclude that g(a) = 0 for some a. Then a is a fixed point of f.

7.2.16 If f: RusualRlower limit is continuous, by the connected property of Rusual, the image of f is a connected subset of Rlower limit. By Exercise 7.2.4, the image has to be a single point. In other words, f is constant.

7.2.17 The map g(x) = f(x) - f(-x): S1R satisfies g(-x) = - g(x). Therefore g(x) takes positive as well as negative values. Since S1 is connected, the image g(S1) is a connected subset of R that contains positive and negative numbers. Therefore the image must contain 0.

7.2.19 Fix aX. Then for any xX, the subset Ax = {a, x} is connected. By the second property in Theorem 7.2.3, X = ∪xX Ax is connected.

The same argument applied to four points, as long as X contains at least four points.

7.2.20 Suppose A1 ∪ … ∪ Ak is already known to be connected. Since (A1 ∪ … ∪ Ak) ∩ Ak+1 contains AkAk+1, which is nonempty, by the second property of Theorem 7.2.3, we see that Bk+1 = A1 ∪ … ∪ AkAk+1 is also connected.

Thus by induction, we have proved all Bk = A1 ∪ … ∪ Ak are connected. Since the intersection of all Bk is B1 = A1, which is nonempty, by the second property of Theorem 7.2.3 again, we see that the union ∪ Bk = ∪ Ak is connected.

7.2.21 For each fixed i, since AAi ≠ ∅, by the second property of Theorem 7.2.3, AAi is connected. Now since the intersection of AAi, for all i, is not empty (they all contain A), by the second property of Theorem 7.2.3 again, we conclude that ∪ i(AAi) = A ∪ ( ∪ Ai) is connected.

7.2.22 Let UYA be an open and closed subset in YA. Then UY is an open and closed subset of Y. Since Y is connected, we get either UY = ∅ or UY = Y.

Assume UY = ∅. Then UA. Since AB is a separation of X - Y, A is an open and closed subset of X - Y. By applying Exercise 5.3.12 to X, Y, A (in place of X, Y, Z), for both the closed and open cases, we conclude that the subset UA that is assumed to be open and closed in YA must be open and closed in X. Since X is connected and U is not the whole space, we conclude that U = ∅.

If UY = Y, then the argument can be repeated again with YA - U in place of U. The conclusion is that YA - U = ∅, which is the same as U = YA.

We conclude that the only open and closed subset of YA are ∅ and YA itself.

7.3.1 The trivial topology is path connected because any map γ: [0, 1] → (X, tirival topology) is continuous, and for any two points x and y, it is easy to construct a map satisfying γ(0) = x and γ(1) = y.

If the set contains at least two points, then the discrete toplogy is not path connected because it is not even connected: Any expression X = AB with A and B nonempty and disjoint is a separation int he discrete topology.

7.3.3 The first statement is no longer true. In Example 7.3.6, we have a path connected subset A and a subset B satisfying ABA, such that B is not path connected.

The second statement is still true. Let a be a common point of all Ai. Let x ∈ ∪ Ai. Then xAi0 for some index i0. Since a, xAi0, and Ai0 is path connected, we have a path in Ai0 connecting a to x. The path is also a path in ∪ Ai. Thus we proved that any point in ∪ Ai can be connected to a by a path in ∪ Ai. Then by Exercise 7.3.4, ∪ Ai is path connected.

The third and the fourth statements are still true. The reason is trivial.

7.3.4 For any two points y, zX, we have a path α joining x0 to y and another path β joining x0 to z. By connecting the pathes α and β together, we get a path connecting y to z.

7.3.5 1. Path connected. For v = (x, y) ∈ {(x, y): x4 + y4 = 1}, γ(t) = (±(1 - t4)1/4, t) [the sign is the same as the sign of x] for t between 1 and y is a path in the subset connecting v to (0, 1). By Exercise 7.3.4, the subset is path connected.

2. Not connected. Because x can be positive and negative but not 0, we have separation {(x, y): x4 - y4 = 1} = {(x, y): x4 - y4 = 1, x > 0} ∪ {(x, y): x4 - y4 = 1, x < 0} is a separation.

3. Path connected. For v = (x, y, z) ∈ {(x, y, z): x3 + y3 = z2}, γ(t) = (t2x, t2y, t3z) is a path in the subset connecting v to (0, 0, 0). By Exercise 7.3.4, the subset is path connected.

4. Path connected. Use the path constructed in 1..

5. Path connected. Let v1 = (x1, y1, z1), v2 = (x2, y2, z2) ∈ {(x, y, z): x3 + y3 = z2, x + y > 0}. Then γ1(t) = (x(t), y(t)) = (1 - t)(x1, y1) + t(x2, y2) = ((1 - t)x1 + tx2, (1 - t)y1 + ty2) is a path connecting (x1, y1) to (x2, y2) and still satisfies x(t) + y(t) = (1 - t)(x1 + x2) + t(y1 + y2) > 0. The path can be easily extended to a path in the subset connecting v1, v2 by introducing the third coordinate z(t) = (x3(t) + y3(t))1/2.

6. Not connected. Since xyz = 1, the coordinates cannot be zero. Thus {(x, y, z): x2 + y2 + z2 ≤ 4, xyz = 1} = {(x, y, z): x2 + y2 + z2 ≤ 4, xyz = 1, x > 0} ∪ {(x, y, z): x2 + y2 + z2 ≤ 4, xyz = 1, x < 0} is a separation.

7. Path connected. For v = (x, y, z, w) ∈ {(x, y, z, w): x2 + y2 + z2 - w2 = 1}, let x2 + y2 + z2 = r2, r > 0. In the 2-sphere of radious r, there is a path α connecting (x, y, z) to (0, 0, r). Then (α, w) is a path in the subset connecting v to (0, 0, r, w). Then in the hyperbola r2 - w2 = 1, there is a path β connecting (r, w) to (1, 0), so that (0, 0, β) is a path in the subset connecting (0, 0, r, w) to (0, 0, 1, 0). Combining (α, w) and (0, 0, β) together, we get a path connecting v to (0, 0, 1, 0).

8. Path connected. For v = (x, y, z, w) ∈ {(x, y, z, w): x2 + y2 - z2 - w2 = 1}, let x2 + y2 = r2, z2 + w2 = s2, r > 0, s ≥ 0. In the circles of radii r and s, there are paths α and β connecting (x, y) and (z, w) to (0, r) and (s, 0) respectively. Then (0, α, β, 0) is a path in the subset connecting v to (0, r, s, 0). Now in the hyperbola r2 - s2 = 1, there is a path γ connecting (r, s) to (1, 0), so that (0, γ, 0) is a path in the subset connecting (0, r, s, 0) to (0, 1, 0, 0). Combining (α, β) and (0, γ, 0) together, we get a path connecting v to (0, 1, 0, 0).

9. Not connected. Because x can be positive and negative, but cannot be 0. Thus {(x, y, z, w): x2 - y2 - z2 - w2 = 1} = {(x, y, z, w): x2 - y2 - z2 - w2 = 1, x > 0} ∪ {(x, y, z, w): x2 - y2 - z2 - w2 = 1, x < 0} is a separation.

10. Path connected. Since x02 ≤ 1 + x12 + … + xn2 implies (tx0)2t2 + (tx1)2 + … + (txn)2 ≤ 1 + (tx1)2 + … + (txn)2, for any v ∈ {(x0, x1, …, xn): x02 ≤ 1 + x12 + … + xn2}, γ(t) = tv is a path in the subset connecting v to (0, 0, …, 0).

11. Not connected. Since x02 ≥ 1 + x12 + … + xn2 implies x02 ≥ 1, we have separation {(x0, x1, …, xn): x02 ≥ 1 + x12 + … + xn2} = {(x0, x1, …, xn): x02 ≥ 1 + x12 + … + xn2, x0 > 0} ∪ {(x0, x1, …, xn): x02 ≥ 1 + x12 + … + xn2, x0 < 0}.

12. Not connected. Q × R = (Q ∩ (-∞, √2)) × R ∪ (Q ∩ (√2, ∞)) × R is a separation.

13. Connected. First, since R is connected and x × R, R × y share common point (x, y), we know x × RR × y is connected for any x and y. Then for each fixed x, x × RR × Q = ∪yQ (x × RR × y) is connected because the connected subsets share common point (x, 0). Finally, Q × RR × Q = ∪xQ (x × RR × Q) is connected because the connected subsets share common point (0, 0).

14. Connected. We have R2 - Q2 = (R - Q) × RR × (R - Q). By taking R - Q as Q in 13., the same argument shows that R2 - Q2 is connected. The general n dimensional case is similar.

15. Path connected.

16. Connected but not path connected. By 15. and the first property in Proposition 7.4. And there is no path between the point (0, 0) in the closure and points on the curve.

17. Path connected.

18. Connected but not path connected. By the same reason as 16..

19. Path connected. Same argument as in Example 7.18.

20. Not connected. O(n, R) = {orthogonal M: detM = 1} ∪ {orthogonal M: detM = -1} is a separation.

21. Path connected. Just take the part of the argument in Example 7.18 for the orthogonal matrix part.

7.3.6 Let x and y be two points in R2 - A. Consider all straight lines passing through x. There are uncountably many of them. Therefore there is at least one such line, say L1, such that L1 avoids all points in A, i.e., L1R2 - A.

Once L1 is found, consider all straight lines passing through y. By similar consideration, there is a straight line L2 among these, such that L1 avoids all points in A, and L2 is not parallel to L1. Then L1 and L2 are two straight lines in R2 - A that intersect somewhere. From this we find a path (consisting of two line segments) in R2 - A joining x to y. This implies that R2 - A is path connected, and is therefore connected.

7.3.7 The connectivity of X - {1 point} and X - {2 points} are topological properties.

For any n ≥ 2, Rn - {1 point} and Rn - {2 points} are always path connected, and are therefore connected.

R - {1 point} is not connected.

S1 has the property that S1 - {1 point} is connected and S1 - {2 points} is not connected.

7.3.10 For any two points a, b in the subset, let a < b. Then γ(t) = b for t ∈ [0, 1) and γ(1) = a gives a continuous path connecting the two points.

7.3.12 1. Path connected. The straight line construction works here. If f and g satisfy f(0) > f(1) and g(0) > g(1), then for any 0 ≤ s ≤ 1, (1 - s)f(0) + sg(0) > (1 - s)f(1) + sg(1).

2. Not connected. The subsets {f: f(0) > f(1)} and {f: f(0) < f(1)} are non-empty and open subsets that form a separation of {f: f(0) ≠ f(1)}.

3. Not connected. The subsets {f: f(0) > |f(1)|} and {f: f(0) < -|f(1)|} are non-empty and open subsets that form a separation of {f: |f(0)| > |f(1)|}.

4. Path connected. The straight line construction connects any f satisfying |f(0)| ≥ |f(1)| to the constant 0 function.

7.3.14 1. Let X be an open subset of Rn. Then for any xX there is a small ball B(x, ε) ⊂ X. The ball B(x, ε) is path connected because it is convex.

2. For any yV = {yX: there is a path joining x to y}, there is a path γ joining x to y. Since X is locally path connected, there is a path connected open subset U containing y. Then for any zU, there is a path γ' in U joining y to z. By combining γ and γ' together, we find a path joining x to z. Thus UV. This proves that V is open.

Let zV. Since X is locally path connected, there is a path connected open subset U containing z. By Lemma 4.6.2, there is a point yVU. By definition of V, there is a path joining x to y. By U being path connected, there is a path joining z to y. Combining the two paths together, we have a path joining x to z. Thus zV. This proves that V is closed.

4. The subset V in 3. is open and closed. If X is connected, then we get V = X. This means that X is path connected.

7.3.15 1. For any yU, yx, define γ(t) = y for 0 ≤ t < 1, γ(1) = x. Then for any open subset V of X, we have

• xVUV ⇒ γ-1(V) = [0, 1], open in [0, 1]
• xV, yV ⇒ γ-1(V) = [0, 1), open in [0, 1]
• x, yV ⇒ γ-1(V) = ∅, open in [0, 1]

Therefore γ is a continuous path in U connecting x and y.

2. The number of open subsets in a finite topological space X is finite. For any xX, by taking the intersection of all open subsets containing x, we get the smallest open subset U containing x.

7.4.1 R is connected and path connected with respect to the usual topological basis B1. The whole space has one connected and path connected component. The claim holds for any topology coarser than the usual topology. Thus the whole space is one connected and path connected component with respect to B5, B6, B7, B9.

Suppose AR contains at least two points a and b. Assume a < b. Then A = (A ∩ (-∞, b)) ∪ (A ∩ [b, ∞)) is a separation of A with respect to B2 (note that aA ∩ (-∞, b) and bA ∩ [b, ∞))). Thus A is not connected. We conclude that the connected compoinents with respect to B2 are the single points. The claim also holds for any topology coarser or homeomorphic to B2. Then the connected (and path connected) components are also single points with respect to B3, B4, B10.

By the similar argument (separating A by using a rational number between a and b) the connected (and path connected) components are single points with respect to B8 and the Michael line.

Any continuous map from a connected space to R must have connected image. Therefore the image is contained in a connected components. Since the connnected compoments in the lower limit topology and the Michael line are single points, such continuous maps must be constants.

7.4.5 Let X1, …, Xk, be connected components of X. Then by Proposition 7.4.1, the components are closed and X = X1 ∪ … ∪ Xk, and the union is a disjoint one. Since th union is disjoint, we have X1 = X - X2 ∪ … ∪ Xk. Then since each Xi is closed, the finite union X2 ∪ … ∪ Xk is also closed. Thus we conclude that X1 is open.

Open and closed subsets in such space are unions of connected components.

7.4.7 Any path connected component is connected. Because connected components are maximal connected subsets, any connected subset is contained in a connected component.

7.4.8 1. Let A be a connected component. For any xA, by the locally connected assumption, there is a connected neighborhood N of x. Since A is the biggest connected subset containing x, we conclude that NA. This shows that A is open.

2. For any open subset U containing x, the assumption tells us that the connected component V of U that contains x is also open. Since V isalso an open subset of X, this shows the space is locally connnected.

3. Let V be a connected component of an open subset U of Y. By the second part, it suffices to prove that V is open. In other words, f-1(V) is open in X.

Let f(x) ∈ V. Then f(x) ∈ U. Since U is open in Y, f-1(U) is open in X. Since X is locally connected, f-1(U) contains a connected neighborhood N of x. The image f(N) is a connected subset of U that contains f(x). Since V is a connected component of U that contains f(x), we get f(N) ⊂ V. Therefore the neighborhood N of x is contained in f-1(V). This proves that f-1(V) is open in X.

7.4.9 By Exercise 7.4.8, it suffices to prove that any connected component is also path connected. By Exercise 7.3.14, it further suffices to prove that the connected component is locally path connected.

Let A be a connected component. For xA, an open subset of A containing x is of the form UA, where U is an open subset of X. By local path connected assumption, there is a path connected open subset V of X satisfying xVU. Since the subset V is also connected, and A is a connected component, we have VA. Thus V is also an open subset of A, and we just verified that A is locally path connected.

7.5.1 The condition for the Euclidean metric is ε > (√n)/2. The condition for the Lp-metric is ε > n1/p/2.

7.5.2 The first, the second, and the fifth collections are open covers. The third is not a cover because the integers are not covered. The fourth and the sixth contain subsets that are not open in the lower limit topology.

7.5.3 1. Open cover in L- and pointwise convergence topologies. Not open in L1-topology.

2. Open cover in L- and pointwise convergence topologies. Not open in L1-topology.

3. Not cover.

4. Open cover in L- and pointwise convergence topologies. Not open in L1-topology.

5. Open cover in L- and pointwise convergence topologies. Not open in L1-topology.

6. Open cover in L-topology. Not open in L1- and pointwise convergence topologies.

7. Open cover in L1- and L-topologies. Not open in pointwise convergence topology.

7.5.4 Since any open cover in a coarser topology is also open cover in the finer topology, we see that compactness of the finer topology implies compactness of the coarser topology. The converse is not true, because the discrete topology (the finest topology) on an infinite set is not compact.

7.5.6 The subset {1/n: nN} is not compact with respect to the basis {(a, +∞)} because the open cover {(1/n, +∞)} has no finite subcover. Moreover, since the usual topology, the lower limit topology, and the upper limit toplogy are finer, the subset is also not compact with respect to these topologies. Finally, the subset is compact with respect to the basis {(-∞, a)} because for any open cover U, the open subset in U that contains 1 already covers the subset.

The subset {1/n: nN} ∪ {0} is compact with respect to the lower limit topology. The reason is that any open cover U must contain some open subset U (in the lower limit topology) such that [0,1/N) ⊂ U for some N. We further find open subsets U1, U2, …, UNU, such that 1/nUn for any 1 ≤ nN. Then {U, U1, U2, …, UN} is a finite subcover.

Since the usual topology, {(a, +∞)} and {(-∞, a)} are coarser than the lower limit topology, the subset {1/n: nN} ∪ {0} is also compact with respect to these topologies. Finally, the subset is not compact in the upper limit topology because the open cover {(1/(n+1), 1/n]} ∪ {(-1, 0]} has no finite subcover.

The subset Z is not compact with respect to {(a, +∞)} because the open cover {(a, +∞)} has no finite subcover. Then the subset is not compact with respect to the finer topologies, including the usual topology, the lower limit topology, and the upper limit toplogy. The subset is not compact with respect to {(-∞, a)} by the similar reason.

All subsets are compact with respect to the finite complement topology.

7.5.7 1. Suppose the refinement V of U has a finite subcover {V1, V2, …, Vn}. By the definition of refinement, for any Vi, there is UiU containing Vi. Then {U1, U2, …, Un} is a finite collection in U that still covers X.

2. Let U be a topological basis of X and let U be an open cover of X. Then for any xX, we have xUx for some UxU. Since Ux is open, we have xBxVx for some BxB. Then the collection V = {Bx: xX} is an open cover of X, consisting of some subsets in a topological basis B. Moreover, by the construction, V is a refinement of U.

3. The claim is a combination of the first two parts.

7.5.8

Exercise 4.5.8

1, 2, 3. Compact. Since the only open subset containing 1 is N, any open cover must contain the whole space as one open subset.

Exercise 4.5.9

1. Not Compact. The open cover {(-n, n)} has no finite subcover.

2, 3. Not Compact. The open cover {[-n, n)} has no finite subcover.

4. Compact. The only open subset containing 0 is R.

5. Not compact. The open cover {(-n, 2n)} has no finite subcover.

6. Compact. If x < an, then the only open subset containing x is R.

Exercise 4.6.3

1. Not Compact. The open cover {[-n, n]} has no finite subcover.

2, 4. Compact. The only open subset containing 0 is Z.

3. Not Compact. The open cover {[-6n, 6n]} has no finite subcover.

Exercise 7.3.5

1. Compact. Closed because the subset if f-1(1) for the continuous map f(x, y) = x4 + y4: R2R. Bounded because |x| < 1, |y| < 1.

4. Compact. Closed because the subset if f-1(0 × [0,1]) for the continuous map f(x, y, z) = (x3 + y3 - z2, x4 + y4): R2R. Bounded because |x| < 1, |y| < 1, which implies |z| < √2.

6, 10. Compact. Closed and bounded.

2, 3, 5, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18. Not Compact. Not bounded.

19. Not Compact. Not bounded if n > 1. Diagonal matrices with r, r-1, 1, …, 1 along the diagonal are in SL(n, R).

20, 21. Compact. Closed because MMTM and det are continuous maps. Bounded because the colume vectors have length 1.

7.5.9 By considering whether the open cover {(-n, ∞)} has finite subcover, we find that a compact subset must have bounded infimum. If infAA, then the open cover {(infA + n-1, ∞)} has no finite subcover. Therefore we further find that it is necessary to have infAA for A to be compact.

Conversely, if infA is bounded and lies in A, then for any open cover of A, the open subset that contains infA must be of the form (x, ∞) for some x < infA. Therefore the one open subset (x, ∞) from the cover already covers A.

7.5.10 Since U covers [0, 1], we have 0 ∈ U for some UU. Since U is open, we have [0, ε] ⊂ U for some ε > 0. This shows that the subset A is not empty.

Denote a = supA. Then aU for some UU. Since U is open, we have (a - ε, a + ε) ⊂ U for some ε > 0. On the other hand, we have xA satisfying x > a - ε. Then [0, x] is covered by finitely many from U. Combined with UU that contains (a - ε, a + ε), we see that [0, a] ⊂ [0, x] ∪ (a - ε, a + ε) is covered by finitely many from U. This proves that aA.

In the argument above, if a < 1, we may choose ε > 0 so that a + ε < 1. Then we also see that [0, a + ε/2] ⊂ [0, x] ∪ (a - ε, a + ε) is covered by finitely many from U. This proves that a + ε/2 ∈ A, constradicting to a = supA. Therefore we must have a = 1.

7.5.11 Since the lower limit topology is finer than the usual topology, a subset compact in the lower limit topology is also compact in the usual topology. Moreover, if a = lim anA for some strictly increasing anA, then { [an, an+1): nN } ∪ { (-∞, a1), [a, ∞) } is an open cover of A in the lower limit topology with no finite subcover. Therefore the second condition is also necessary for A to be compact in the lower limit topology.

Conversely, suppose A satisfies the two conditions. Consider a cover { [ai, bi) } of A. For each [ai, bi), we choose a new open interval (in the usual topology) according to two cases.

• If aiA, then we take (ai, bi).
• If aiA, then by the second condition, there is εi > 0, such that (ai - εi, ai) ∩ A = ∅. In this case, we take (ai - εi, bi).

By the construction, the new collection { (ai, bi): aiA } ∪ { (aj - εj, bj): ajA } still covers A. Since A is compact in the usual topology, we have

A ⊂ (ai1, bi1) ∪ … ∪ (aim, bim) ∪ (aj1 - εj1, bj1) ∪ … ∪ (ajn - εjn,, bin)

for finitely many ai*A and aj*A. Then by (aj* - εj*, aj*) ∩ A = ∅, we may delete (aj* - εj*, aj*) from the right side and still get

A ⊂ (ai1, bi1) ∪ … ∪ (aim, bim) ∪ [aj1, bj1) ∪ … ∪ [ajn, bjn)
⊂ [ai1, bi1) ∪ … ∪ [aim, bim) ∪ [aj1, bj1) ∪ … ∪ [ajn, bjn).

This proves that A is covered by finitely many from the original cover.

7.5.12 If f is a continuous function on a compact space X, then f(X) is a compact subset of R. In other words, f(X) is bounded and closed. The boundedness implies that a = inff(X) and b = sup f(X) exist as finite numbers. The closedness implies that a, bf(X). Therefore inff(X) = a = f(x0) and supf(X) = b = f(x1) for some x0, x1X.

If f is lower-semicontinuous, that means f is a continuous as a map from X to R with topology {(a, ∞)}. Therefore if X is compact, then f(X) is compact with respect to the special topology on R. By Exercise 7.5.9, the subset f(X) has bounded munimum, and the minimum lies in f(X). In other words, f reaches its minimum.

7.5.13 By Exercise 2.4.10, for fixed x, d(x, a) is a continuous function of aA. Then by Exercise 7.5.12, the function reaches its minimum on the compact subset A. In other words, d(x, A) = infaAd(x, a) = d(x, a0) for some a0A.

If A and B are compact subsets, then by Exercise 2.4.15 and similar argument we also have d(A, B) = d(a, b) for some aA and bB.

7.5.14 This is obtained by taking A = B = X in Exercise 7.5.13.

7.5.15 For each r > 0, the collection of balls of radius r is an open cover. Then the space is covered by finitely many balls Br,1, Br,2, …, Br, nr of radius r. By taking r = 1/k for natural numbers k, we get a countable collection B = {B1/k, i: k is a natural number, 1 ≤ in1/k} of balls with the following property. For any x and r > 0, there is a ball BB (taking k > 2/r), such that xBB(x, r). The property shows that B is a topological basis in the metric topology.

7.5.16 Since the only open subsets of Y that contains b are Y and Y - a, any open cover of Y - a = X ∪ {b} must contain either Y or Y - a. Therefore either {Y} or {Y - a} is a finite subcover of the open cover. This shows that Y - a is compact.

The intersection (Y - a) ∩ (Y - b) = X is homeomorphic to X, which may not be compact.

7.5.17 Note that adding (X × {1}) ∪ {a} and Y to the product topology still gives a topology. In fact, if U is a closed subset of Z, then {open subsets in Z} ∪ {U ∪ {a}, Z ∪ {a}} is a topology on Z ∪ {a}.

1. The only open subsets containing a are the whole space and (X × {1}) ∪ {a}. This implies that (X × {1}) ∪ {a} is compact.

2. For any xX, an open subset containing (x, 1) must be either U × {1}, or U × {1, 2}, or (X × {1}) ∪ {a}, or Y, where U is open in X. In all the cases, the intersection of the open subset with (X × {1}) ∪ {a} is not empty. Therefore (x, 1) is in the closure of (X × {1}) ∪ {a}.

An open subset containing (x, 2) must be either (X × {1}) ∪ {a} or Y, where U is open in X. In all the cases, the intersection of the open subset with (X × {1}) ∪ {a} is not empty. Therefore (x, 2) is in the closure of (X × {1}) ∪ {a}.

An open subset containing a must be either U × {1, 2} or Y. In all the cases, the intersection of the open subset with (X × {1}) ∪ {a} is not empty. Therefore a is in the closure of (X × {1}) ∪ {a}.

3. Let U be an open cover of X with no finite subcover. Then {U × {1, 2}: UU} ∪ {(X × {1}) ∪ {a}} is an open cover of Y with no finite subcover.

7.5.18 Let X be a Hausdorff space and let A, B be compact subsets. Then A, B are closed subsets by the second property of Theorem 7.5.2. Therefore AB is a closed subset of a compact space A. By the first property of Theorem 7.5.2, AB is also closed.

Let X be a Hausdorff space and let A be a compact subset. Then A is a closed subset by the second property of Theorem 7.5.2. Therefore the closure of A is A itself, and is compact.

If X in Exercise 7.5.16 is Fréchet, then Y is also Fréchet. Therefore the intersection of compact subsets in a Fréchet space is not necessarily compact.

7.5.21 By the proof of the second property in Theorem 7.5.2, for any bB, there are open subsets Ub and Vb, such that AUb and bVb. The collection {Vb: bB} is an open cover of B. Since B is compact, we have BVb1Vb2 ∪ … ∪ Vbn = V. On the other hand, U = Ub1Ub2 ∩ … ∩ Ubn is a open subset containing A and is disjoint from V.

7.5.22 Let AX × Y be a closed subset. To show the projection of A to X is closed, we consider xX that is not in the projection. Then x × Y is disjoint from A. Since A is closed, for any yY, there are open subsets Uy and Vy, such that (x, y) ∈ Uy × Vy and Uy × Vy is disjoint from A. The collection {Vy: yY} is an open cover of Y. Since Y is compact, we have YVy1Vy2 ∪ … ∪ Vyn = U. Moreover, V = Uy1Uy2 ∩ … ∩ Uyn is a open subset containing x. Finally, U × VUy1 × Vy1Uy2 × Vy2 ∪ … ∪ Uyn × Vyn is disjoint from A. Therefore U is an open subset containing and x is disjoint from the projection of A in X. This proves the projection is closed.

7.5.23 By Exercise 7.1.14, the continuity implies the graph is closed. Conversely, suppose the graph is closed. Let AY be a closed subset. Then X × A is a closed subset of X × Y. The intersection (X × A) ∩ Γf is also a closed subset. By Exercise 7.5.22, the projection of (X × A) ∩ Γf to X is also closed. Since the projection is simply the preimage f-1(A), we see that A closed implies f-1(A) closed. In other words, f is continuous.

7.5.24 Let U be an open cover of X. For any yY, the preimage f-1(y) is also covered by U. Since f-1(y) is compact, we have f-1(y) ⊂ Uy1Uy2 ∪ … ∪ Uyn = Uy for finitely many Uy1, Uy2, …, Uyn in U. Since f maps closed subsets to closed subsets, Vy = Y - f(X - Uy) is an open subset of Y containing y. Moreover, we have

Y - Vy = f(X - Uy) ⇒ X - f-1(Vy) = f-1(Y - Vy) ⊃ X - Uyf-1(Vy) ⊂ Uy.

Now {Vy: yY} is an open cover of Y. Since Y is compact, we get YVy1Vy2 ∪ … ∪ Vym for finitely many open subsets in the open cover. This implies

X = f-1(Y) ⊂ f(Vy1) ∪ f(Vy2) ∪ … ∪ f(Vym).

Since each f(Vyi) ⊂ Uyi and Uyi is covered by finitely many from U, we conclude that X is covered by finitely many from U.

7.5.25 The reason for T to be a topology is the same as the finite complement topology (applied to each "slice" X × y.

1. For UY, we have πY-1(U) = X × U = X × Y - F for F = X × (Y - U). Since X is infinite, the subset F has the property that FX × y is finite for any y if and only if Y = U. This proves that the quotient topology is trivial.

2. The open subsets in the subspace πY-1(y) = X × y are (X × Y - F) ∩ X × y = X × y - FX × y. Therefore the subspace topoloogy is the finite complement topology, which we know is compact.

7.5.26 1. A collection C of a closed subsets corresponds to a collection U = {X - C: CC} of open subsets. The compactness would means the following: if U covers X, then there is a finite subcollection of U that also covers X. Note that the statement is the same as the following: if any finite subcollection of U does not cover X, then U does not cover X.

A finite subcollection {X - C1, X - C2, …, X - Cn} covers X means exactly C1C2 ∩ … ∩ Cn = ∅. The whole collection U = {X - C: CC} covers X means exactly ∩CCC = ∅. Therefore we conclude that the compactness is equivalent to the following: If C1C2 ∩ … ∩ Cn ≠ ∅ for any finitely many C1, C2, …, CnC, then ∩CCC ≠ ∅.

2. Fix one K0 in the collection K. Then by considering K' = {KK0: KK}, everything happens inside K, a compact Hausdorff space. Therefore K' is a collection of closed subsets in K, and the conclusion follows from the first part.

3. The collection K' = {K - U: KK} is a colletiton of compact subsets. Moreover, ∩KKKU implies that ∩K'K'K' = ∅. Therefore by the second part, we have K'1K'2 ∩ … ∩ K'n = ∅ for finitely many subset in the collection K' . If K'i = Ki - U, then this means K1K2 ∩ … ∩ KnU.

7.5.27 1. We have ∅ ∈ TT+. By taking K = ∅, which is compact, we get X+ = (X - ∅)+T+.

To see T+ is closed under union, we need to consider U ∪ (∪i (X - Ki)+) = U ∪ (X - ∩i Ki)+ = (X - (X - U) ∩ (∩i Ki))+. Th eproblem boils down to the following: If U is open and Ki are compact, then (X - U) ∩ (∩i Ki) is compact. Since X is Hausdorff, the compact subsets Ki are also closed. Moreover, X - U is closed. Therefore (X - U) ∩ (∩i Ki) is a closed subset of the compact subset Ki, and must be compact.

To see T+ is closed under finite intersection, we need to consider U ∩ (X - K)+ = U ∩ (X - K) = U - K and (X - K1)+ ∩ (X - K2)+ = (X - K1K2)+. In the first case, since X is Hausdorff, the compact subset K is closed, so that U - K is an open subset of X. In the second case, the union K1K2 of two compact subsets is still compact.

2. Any open cover U = {Ui} ∪ {(X - Kj)+} of X+ must include an open subset in T+ that contains +. Such open subset is of the form (X - K)+ for some compact subset K of X. Then U' = {Ui} ∪ {X - Kj} covers K. Since X is Hausdorff, the compact subsets Kj are closed, so that U' is a cover of the K by open subsets of X. Since K is a compact subset of X, we have

KUi1 ∪ … ∪ Uim ∪ (X - Kj1) ∪ … ∪ (X - Kjm).

This implies

X+Ui1 ∪ … ∪ Uim ∪ (X - Kj1)+ ∪ … ∪ (X - Kjm)+ ∪ (X - K)+.

7.5.29 The key for X+ to be Hausdorff is the separation of xX and the point +. This means that for any xX, we have disjoint U and (X - K)+ containing x and + respectively. The condition is the same as there are open U and compact K, such that xUK. This also means exactly that the compact subset K is a neighborhood of x. We get (1) ⇔ (4).

Both (2) and (3) are stronger than (4). So we need to show (4) ⇒ (3) and (1) ⇒ (2).

Assuming (4). Since X is Hausdorff, K is also closed. Therefore we actually have xUUK. The closure U is a closed subset of the compact subset K, so that the closure is also compact. This proves (2).

Assuming (1). Let xV for open V. Then x is not in the closed subset X+ - V = (X - V)+. Since X+ is compact Hausdorff, the two are separated. Therefore we have xU and (X - V)+ ⊂ (X - K)+, where U is open, K is compact, and U and (X - K)+ are disjoint. Thus xUKV. This shows that V contains a compact neighborhood K of x.

7.5.30 Since Y is Hausdorff, the single point + is closed, so that X is an open subset of Y, and the open subsets of Y not containing + are exactly the open subsets of X. It remains to show that the open subsets V of Y containing + must be of the form (X - K)+ for some compact subset K of X.

The condition Y = X is the same as + is a limit point of X. In other words, any open subset V of Y containing + must contain some xX. This means V = Y - K = (X - K)+ for a closed subset K of Y satisfying KX, KX. Since Y is compact, K is also compact.

Conversely, for any compact subset K of X, since Y is Hausdorff, K is closed. Then (X - K)+ = Y - K is an open subset of Y.

Note: The assumption Y = X actually implies that X cannot be compact.

7.6.1 A limit point is still a limit point in weaker topology. Therefore if a topology is limit point compact, then any weaker topology is still limit point compact.

7.6.2 This is because any infinite subset contains a countable subset, and any limit point of a smaller set is a limit point of a bigger set.

7.6.3 By Exercise 4.5.3, the limit points of an infinite subset in the finite complement topology is the the whole space.

7.6.4 By Exercise 4.5.5, any nonempty subset has limit point.

7.6.5 We already know compact implies limit point compact. We need toprove the converse. Assume A is limit point compact in the lower limit topology. By Exercise 7.5.11, we need to show that A is compact in the usual topology and lim anA for any strictly increasing anA.

First, A is also limit point compact in the weaker usual topology. Since usual topology is a metric topology, this implies that A is compact in the usual topology.

Second, if lim anA for some strictly increasing anA, then the sequence form an infinite subset of A with no limit points. Therefore A cannot be limit point compact. This verifies the second condition.

7.6.6 Suppose AR is limit point compact in the lower limit and upper limit topologies. Then by Exercise 7.6.5, it is compact in both topologies. By the second part of Exercise 7.5.11, for any a, we have A ∩ (a - ε, a) = ∅ and A ∩ (a, a + ε) = ∅ for some ε > 0. In other words, we have A ∩ (a - ε, a + ε) = ∅ or {a} for some ε > 0. Combining this with the compactness of A (taking {(a - εa, a + εa)} as an open cover), we conclude that A is finite.

7.6.9 Take U = {(x, y): (1 + x2)y > -1} and V = {(x, y): (1 + x2)y < 1}.

7.6.10 Suppose εi > 0 is an increasing sequence of Lebesgue numbers and ε = lim εi. For any fixed x and any i, we have B(x, εi) ⊂ U for some U in the cover. If the cover is finite, then we can find a fixed U from the cover and a subsequence εik, such that B(x, εik) ⊂ U. Since U is fixed, we get B(x, ε) = ∪k B(x, εik) ⊂ U. This proves that B(x, ε) is always contained in some U in the cover. In other words, ε is a Lebesgue number.

The argument does not work if the cover is infinite. A counterexample is given by the cover { (-1 + n-1, ∞) } ∪ { (-∞, 1 - n-1) } of R in the usual topology. For any ε = 1 - n-1, we have B(x, ε) ⊂ (-1 + n-1, ∞) for x > 0 and B(x, ε) ⊂ (-∞, 1 - n-1) for x ≤ 0. However, for ε = 1, B(0, 1) is not in any from the cover.