M55: Exercise sheet 1

1. (a) Let (X, d) be a metric space. Show that if x ≠ y ∈ X then there are metric open

sets U, V ⊂ X with x ∈ U, y ∈ V and U ∩ V = ∅.

(b) Conclude that the following topologies are not metric: (i) the indiscrete topology on any set with more an one element; (ii) the Sierpinski 2-point topology; (iii) the Zariski topology on ℝ.

2. Let X = {a, b, c} be a set with three elements.

(a) List all the topologies 𝒯 on X for which {a} is the only open set with one element.

(b) List all topologies on X up to homeomorphism.

3. Suppose that 𝒯_{1} and 𝒯_{2} are topologies on the same underlying set X.

(a) Show that 𝒯_{1} ∩ 𝒯_{2} is a topology on X.

(b) Is 𝒯_{1} ∪ 𝒯_{2} necessarily a topology on X?

4. (a) Show that the identity map id : (X, 𝒯 ) → (X, 𝒯 ) is continuous.

(b) Show that any constant map f : (X, 𝒯 ) → (Y, 𝒮) is continuous.

5. Suppose X is a set with topologies 𝒯_{1} and 𝒯_{2}. If 𝒯_{1} ⊂ 𝒯_{2}, then we say that 𝒯_{2} is finer than 𝒯_{1} and that 𝒯_{1} is coarser than 𝒯_{2}. (Thus a finer topology has more open sets.)

(a) Show that id : (X, 𝒯_{2}) → (X, 𝒯_{1}) is continuous if and only if 𝒯_{2} is finer than 𝒯_{1}. Conclude that if 𝒯_{2} is strictly finer than 𝒯_{1}, then id is a continuous bijection that is not a homeomorphism.

(b) Show that the usual metric topology on ℝ is strictly finer than the cofinite topology.

6. Let 𝒯 be the Zariski (co-finite) topology on ℝ.

(a) Let p be a polynomial. Show that p : (ℝ, 𝒯 ) → (ℝ, 𝒯 ) is continuous. Hint: you may find it helpful to note that the composition of two polynomials is a polynomial.

(b) Show that sin : (ℝ, 𝒯 ) → (ℝ, 𝒯 ) is not continuous.

February 14, 2003

Home page: http://www.maths.bath.ac.uk/~feb/math0055.html

M55: Exercise sheet 1—Solutions

1. (a) Let ε = d(x, y) > 0 and set U = B_{ε/2}(x) and V = B_{ε/2}(y). Then U, V are certainly open and x ∈ U and y ∈ V. So the only issue is to see that U and V are disjoint. For this, suppose that z ∈ U ∩ V. Then the triangle inequality gives

d(x, y) ≤ d(x, z) + d(z, y) < ε/2 + ε/2 = d(x, y),

a palpable contradiction!

(b) None of the topologies listed have this property. Indeed, (i) the indiscrete topology has only one non-empty open set; (ii) any two non-empty open subsets in the Zariski topology have finite complement and then their intersection has finite complement (De Morgan!) and so cannot be empty; (iii) If X = {a, b} with {a} open and {b} not, then the only open set in the Sierpinski topology containing b is X which also conatins a.

2. (a) These are:

𝒯_{1} = {∅, {a}, X}
𝒯_{2} = {∅, {a}, {a, b}, X}
𝒯_{3} = {∅, {a}, {a, c}, X}
𝒯_{4} = {∅, {a}, {b, c}, X}
𝒯_{5} = {∅, {a}, {a, b}, {a, c}, X}.

Notice that 𝒯_{2} and 𝒯_{3} give homeomorphic spaces. Both have the same singleton set and whether we call the extra element in the set of size two, b or c makes no real difference.

(b) A systematic way to organise this is to find first all the topologies with no singleton set, then those with 1, 2 and 3 singleton sets. This gives

𝒯_{1} = {∅, X}
𝒯_{2} = {∅, {a, b}, X}
𝒯_{3} = {∅, {a}, X}
𝒯_{4} = {∅, {a}, {a, b}, X}
𝒯_{5} = {∅, {a}, {b, c}, X}
𝒯_{6} = {∅, {a}, {a, b}, {a, c}, X}
𝒯_{7} = {∅, {a}, {b}, {a, b}, X}
𝒯_{8} = {∅, {a}, {b}, {a, b}, {a, c}, X}
𝒯_{9} = {∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, X}.

Since we are only interested in topologies up to homeomorphism, it does not matter which point comprises the single set when there is only one of these, or which point fails to comprise a singleton set when there are two of these. Notice that 𝒯_{4} differs from 𝒯_{5} by having the singleton set contained in the set with two elements. Therefore these topologies are genuinely different.

3. (a) We check the axioms:

(i) Certainly, ∅, X ∈ 𝒯_{i}, i = 1, 2, so that ∅, X ∈ 𝒯_{1} ∩ 𝒯_{2}. (ii) If A, B ∈ 𝒯_{1} ∩ 𝒯_{2} then A, B ∈ 𝒯_{i}, i = 1, 2 whence A ∩ B ∈ 𝒯_{i}, i = 1, 2 giving A ∩ B ∈ 𝒯_{1} ∩ 𝒯_{2}. (iii) If A_{α} ∈ 𝒯_{1} ∩ 𝒯_{2}, α ∈ I, then all A_{α} are in each T_{i}, i = 1, 2, so that ⋃_{α∈I} A_{α} ∈ 𝒯_{i}, i = 1, 2. That is, ⋃_{α∈I} A_{α} ∈ 𝒯_{1} ∩ 𝒯_{2}.

(b) The problem is that the union or intersection of a set in 𝒯_{1} and a set in 𝒯_{2} might lie in neither 𝒯_{i}. With that in mind we cast around for a counter-example. Question 2 gives us a good supply of simple topologies to think about and so I take X = {a, b, c} and 𝒯_{1} = {∅, {a}, X} and 𝒯_{2} = {∅, {b}, X}. Now 𝒯_{1} ∪ 𝒯_{2} = {∅, {a}, {b}, X} which is not a topology since it does not contain {a, b} = {a} ∪ {b}.

4. (a) Almost nothing to prove here: for U ∈ 𝒯 , id^{−1}(U) = U ∈ 𝒯 as required.

(b) Let f be constant with value y ∈ Y. Let V ∈ 𝒮. There are two cases: if y ∈ V, then f^{−1}(V) = X ∈ 𝒯 ; if y ∉ V then f^{−1}(V) = ∅ ∈ 𝒯 . Either way, f^{−1}(V) is open so that f is continuous.

5. (a) For any subset U ⊂ X, id^{−1}(U) = U. Thus id : (X, 𝒯_{2}) → (X, 𝒯_{1}) is continuous if and only if whenever U ∈ 𝒯_{1}, id^{−1}(U) = U ∈ 𝒯_{2}. That is, if and only if 𝒯_{1} ⊂ T_{2}. Thus, if 𝒯_{2} is strictly finer than 𝒯_{1}, id : (X, 𝒯_{2}) → (X, 𝒯_{1}) is a continuous bijection whose inverse (also id) is not continuous (X, 𝒯_{1}) → (X, 𝒯_{2}).

(b) First note that any Zariski open set is metric open: if A is Zariski open with complement

x_{1} < ⋯ < x_{n} then

A = (−∞, x_{1}) ∪ (x_{1}, x_{2}) ∪ ⋯ ∪ (x_{n}, ∞)

which is certainly metric open. Thus the metric topology is finer than the Zariski topology. It is strictly finer because the interval (0, 1), for example, is metric open but not Zariski open: its complement is infinite!

6. (a) Let A ⊂ ℝ be a non-empty Zariski open set with complement x_{1}, …, x_{n}. Let q(x) = (x − x_{1}) … (x − x_{n}) so that q is a polynomial with A = q^{−1}(ℝ \ {0}). Then p^{−1}(A) = p^{−1}(q^{−1}(ℝ \ {0})) = (q ∘ p)^{−1}(ℝ \ {0}) with complement (q ∘ p)^{−1}({0}). Since q ∘ p is polynomial, this last is either finite or all of ℝ so that p^{−1}(A) is Zariski open.

(b) Note that ℝ \ {0} is Zariski open while sin^{−1}(ℝ \ {0}) = ℝ \ πℤ which is not Zariski open (it has infinite complement).