M55: Exercise sheet 5

1. Let X be a set. What is the coarsest topology on X for which all singleton sets are closed?

2. Show that a metric space is normal.

3. Let A, B be compact subsets of a topological space X. Show that A ∪ B is compact.

4. (a) Let (F_{n})_{n∈ℕ} be a family of non-empty closed subsets of a compact topological space X such that, F_{n+1} ⊂ F_{n}, for all n ∈ ℕ. Prove that

⋂_{n∈ℕ}F_{n} ≠ ∅.

(b) Let f : X → X be a continuous map of a compact Hausdorff topological space into itself. Show there is a non-empty closed subset A of X such that f(A) = A.

5. Let X be a set equipped with the co-finite topology. Show that any subset of X is compact. Deduce that ℝ with the Zariski topology has compact subsets which are not closed.

6. Show that a compact Hausdorff space is normal (and so T_{4}).

7. Recall the Moore plane Γ.

(a) Show that any subset of the x-axis is closed in Γ.

(b) Show that Γ is regular.

(c) Show that Γ is not normal^{1}.

8. Show that there is no Hausdorff topology on [0, 1] which is strictly coarser than the metric topology. Hint: Contemplate the identity map…

March 21, 2003

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

1This is sufficiently difficult that I offer a small prize for the first correct solution!

M55: Exercise sheet 5—Solutions

1. The co-finite topology is coarsest topology for which all singleton sets are closed: indeed, any topology for which singleton sets are closed has all finite sets closed (they are finite unions of singleton sets!) and so contains the co-finite topology.

2. Let C, D ⊂ X be disjoint closed subsets of a metric space. For x ∈ C, x ∈ X \ D which is open so there is ε_{x} > 0 with B_{ε_{x}}(x) ⊂ X \ C. Similarly, for y ∈ D, there is ε_{y} > 0 such that B_{ε_{y}}(y) ⊂ X \ C. Now define open sets by

U = ⋃_{x∈C} B_{ε_{x}/2}(x) V = ⋃_{y∈D} B_{ε_{y}/2}(y).

Then C ⊂ U ⊂ X \ D and D ⊂ V ⊂ X \ C and all we have to do is see that U and V are disjoint. However, if z ∈ U ∩ V, there is x ∈ C and y ∈ D such that d(x, z) < ε_{x}/2 and d(z, y) < ε_{y}/2. The triangle inequality now gives that d(x, y) < max{ε_{x}, ε_{y}}. Without loss of generality, suppose ε_{x} ≥ ε_{y}. Then we have y ∈ B_{ε_{x}}(x) ∩ D: a palpable contradiction.

3. Let {U_{α}}_{α∈I} be an open cover for A ∪ B. Thus each U_{α} is open in X and ⋃_{α∈I} U_{α} ⊃ A ∪ B. Thus {U_{α}}_{α∈I} is also an open cover of A and so there is a finite set J_{A} ⊂ I with ⋃_{α∈J_{A}} U_{α} ⊃ A. Similarly, there is another finite set J_{B} ⊂ I with ⋃_{α∈J_{B}} U_{α} ⊃ B. Now set J = J_{A} ∪ J_{B}. Then ⋃_{α∈J} U_{α} ⊃ A ∪ B whence {U_{α}}_{α∈J} is a finite subcover for A ∪ B and so A ∪ B is compact.

4. (a) Suppose, for a contradiction, that ⋂_{n∈ℕ} F_{n} = ∅. Taking complements and using De Morgan’s Laws gives us X = ⋃_{n∈ℕ}(X \ F_{n}) so that {X \ F_{n}}_{n∈ℕ} is an open cover of X. Thus there is a finite subcover X \ F_{n_{1}}, …, X \ F_{n_{k}} with n_{1} < ⋯ < n_{k}. Thus ⋃_{i=1}^{k}(X \ F_{n_{i}} = X or taking complements ⋂_{i=1}^{k} F_{n_{i}} = ∅. But, since each F_{n+1} ⊂ F_{n}, ⋂_{i=1}^{k} F_{n_{i}} = F_{n_{k}} ≠ ∅. This is a contradiction!

(b) Let f^{n} denote the n-times composition of f with itself:

f^{n} = \underbrace{f ∘ ⋯ ∘ f}_{n times}

and set F_{n} = f^{n}(X). Now X is compact and f^{n} continuous so that F_{n} is compact and therefore closed since X is Hausdorff. Further, F_{n+1} = f^{n}(f(X)) ⊂ f^{n}(X) = F_{n} so we are in the situation of part (a). So set A = ⋂_{n∈ℕ} F_{n} which is non-empty by part (a). I claim that f(A) = A. For this, first note that if f(F_{n}) = F_{n+1} so that

f(A) ⊂ ⋂_{n∈ℕ} f(F_{n}) = ⋂_{n∈ℕ} F_{n+1} = A.

For equality, let a ∈ A and contemplate f^{−1}({a}) ∩ F_{n}: these are closed since {a} is closed and f is continuous (X is Hausdorff), nested since the F_{n} are and non-empty: for each n, a ∈ F_{n+1} and so there is an x_{n} ∈ F_{n} such that f(x_{n}) = a. Thus we are once again in the situation of part (a) and so can conclude that

∅ ≠ ⋂_{n∈ℕ}(f^{−1}({a}) ∩ F_{n}) = f^{−1}({a}) ∩ A.

Thus there is some x ∈ a such that f(x) = a and so f(A) = A as required.

5. Let A ⊂ X and {U_{α}}_{α∈I} be an open cover for A. Let U_{α_{0}} be a non-empty element of the cover. Then X \ U_{α_{0}} and so A \ U_{α_{0}} is finite. Write A \ U_{α_{0}} = {a_{1}, …, a_{n}} and for each i between 1 and n, choose U_{α_{i}} containing a_{i}. Then U_{α_{0}}, …, U_{α_{n}} covers A and so is a finite subcover. Thus A is compact. When ℝ is equipped with this topology, there are many non-closed sets (for example, ℤ) but all of these are compact by what we have just proved.

6. Let C, D be disjoint closed (and therefore compact) subsets of X and fix x ∈ C. Then, for each y ∈ D, there are disjoint open sets U_{y}, V_{y} with x ∈ U_{y} and y ∈ V_{y}. Then {V_{y}}_{y∈D} is an open cover of D and so there is a finite subcover V_{y_{1}}, …, V_{y_{n}}. Now set U_{x} = ⋂_{1≤i≤n} U_{y_{n}} and W_{x} = ⋃_{1≤i≤n} V_{y_{n}}. Then x ∈ U_{x}, D ⊂ W_{x} and U_{x}, W_{x} are disjoint open sets. In particular, {U_{x}}_{x∈C} is an open cover of C so that there is a finite subcover U_{x_{1}}, …, U_{x_{m}}. Set U = ⋃_{1≤i≤m} U_{x_{i}} and V = ⋂_{1≤i≤m} W_{x_{i}} to get disjoint open sets with U ⊃ C and V ⊃ D whence X is normal and so (being Hausdorff) T_{4}.

7. (a) First notice that the x-axis is closed in Γ: its complement is the union of all the ε-balls in {(x, y) : y > 0}. Second, recall that the induced topology on the x-axis is the discrete topology so that all subsets of the x-axis are closed with respect to the subspace topology on the x-axis. Thus any subset of the x-axis is the intersection of the x-axis with a closed subset of Γ and so is itself closed in Γ.

(b) Let C ⊂ Γ be closed and p ∈ Γ \ C. Now Γ \ C is open so there is a basic open set B with p ∈ B ⊂ Γ \ C. If p is in the strict upper half plane then we may take B to be an ε-ball about p and then set U = B_{ε/2}(p), V = {q ∈ Γ : ∥q − p∥ > ε/2} to get disjoint open sets with p ∈ U and C ⊂ V. If, on the other hand, p lies on the x-axis, then B is of the form A ∪ {p} with A an ε-ball tangent to the x-axis at p. Let A′ be the ball of radius ε/2 tangent to the x-axis at p and let D be its (metric) closure. Then U = A′ ∪ {p} and V = Γ \ D are disjoint open sets with p ∈ U and C ⊂ V.

(c) This is quite tough: one must find disjoint closed sets C, D in Γ such that whenever U, V are open sets with C ⊂ U and D ⊂ V then U ∩ V ≠ ∅. We take C = ℚ and D = ℝ \ ℚ: these are disjoint subsets of the x-axis and so closed by part (a). Suppose that U, V are disjoint open sets with U ⊃ C and V ⊃ D. Then, for p ∈ ℝ, there is a ball tangent to the x-axis at p of radius r(p) > 0 which lies entirely in U or V according to whether p is rational or not. Now for the clever bit: define subsets A_{n} ⊂ ℝ, n ∈ ℕ, by

A_{2n} = {q ∈ ℚ : r(q) > 1/n}
A_{2n−1} = {p ∉ ℚ : r(p) > 1/n}.

The complete metric space ℝ is the union of the A_{n} and so the Baire Category Theorem tells us that at least one A_{n} is not nowhere dense, that is, there is some n for which the closure of A_{n} contains an interval (a, b). Without loss of generality, assume that n is even (and so consists of rationals) and choose some irrational p ∈ (a, b). Then there is a sequence (q_{k}) in A_{n} converging to p. I claim that the r(p)-ball tangent to p must intersect some r(q_{k})-ball tangent to q_{k} which contradicts the disjointness of U, V. For this, choose 0 < y < min{1/n, r(p)} so that (p, y) is in the r(p)-ball tangent to p. Then, as k → ∞,

∥(p, y) − (q_{k}, 1/n)∥ → |y − 1/n| < 1/n.

Thus, for k large enough, (p, y) lies in the 1/n-ball tangent to the x-axis at q_{k} and so in the r(q_{k})-ball tangent to the x-axis at q_{k}. This completes the proof.

8. Denote the metric topology on [0, 1] by 𝒯_{d} and suppose that 𝒯 is a coarser Hausdorff topology on [0, 1]. Then id : (X, 𝒯_{d}) → (X, 𝒯 ) is a continuous bijection from a compact space to a Hausdorff one and so is a homeomorphism. Thus 𝒯_{d} = 𝒯 .