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Solution to Midterm 1: Math 124 Summer 2023

[1] (2 × 6 = 12 points) Give precise definitions of the following conceptsand Theorems. Statements equivalent to definitions are acceptable.

(1) A continuous map from a topological space X to a topological space Y .

(2) A compact topological space.

(3) A Hausdorff topological space.

(4) A limit point of a subset A of a topological space X.

(5) Bolzano-Weierstrass Theorem.

(6) Heine-Borel Theorem.

Solution: See textbook for definitions.

[2] (3 × 15 = 45 points) Answer the following questions by yes or no. Answers are : (1) to (5) and (14), No. The rest, Yes.

(1) A torus with a finitely many distinct points removed is compact.

No. It is not compact. It is bounded, but not a closed subset of R 3 .

(2) Closed subsets in a Hausdorff space are compact.

No. R n is Hausdorff, and a closed subset may not be bounded.

(3) Compact subsets of a topological space are closed.

No. In Hausdorff spaces, compact subsets are closed. But not in general cases. For example, in finite complement topology on R, every subset is compact, but not necessarily closed. See the last problem [6] in this quiz.

(4) Let Y ⊂ R 2 be the graph of y = x 2 for x ≥ 0. That is Y = {(x, x2 ) | x ≥ 0}. Then Y is a compact subset of R 2 .

No. It is not bounded.

(5) The set of all rational numbers contained in the closed interval [0, 1] is compact.

No. The set is a bounded set, but not closed in R.

(6) Topologist’s closed sine curve is a subset of R 2 given by {(x,sin 1x) | 0 < x ≤ 1} ∪ {(0, y) | −1 ≤ y ≤ 1}. This subset is a compact subset of R 2 .

Yes. It is a bounded and closed subset of R 2 . It is also connected, by the way.

(7) Let f : X → Y be a continuous onto map from a compact space X to a space Y .

Then Y is necessarily compact.

Yes. This is one of the basic theorems discussed in class.

(8) Hawaiian earring is a subset of R 2 given by S n≥1 {(x, y) | (x − 1n) 2 + y 2 = ( 1n ) 2}. This is a union of circle converging to the origin. The Hawaiian earring is a compact topological space.

Yes. It is a bounded closed subset of R 2 .

(9) A compact subset of R n is bounded.

Yes. This is a basic theorem for compact subsets in R n .

(10) Let h be a homeomorphism from the boundary of a closed 2-dimensional disc D2 to the boundary of a M¨obius strip M. Identify the boundary points of D2 and M by h. The resulting space is a projective plane P 2 .

Yes. This is one way to think of the projective plane. To see this, remove a small disc from the center of the disc model of the projective plane. We get an annulus with identification along the outer boundary. Apply cut and paste using a horizontal cut. You get a M¨obius band.

(11) Let Y be a subspace of a topological space X. Let A be a subset of Y . If A is open in Y and Y is open in X, the A is necessarily an open subset of X.

Yes. Since A is open in Y , it is of the form A = Y ∩ U, where U is open in X. Since Y is also open in X, the intersection T ∩ U is also open in X. Hence A is open in X.

(12) The Klein bottle and the projective plane are compact topological spaces.

Yes. They are bounded closed subsets of R 4 .

(13) Let X be a Hausdorff space and let A, B be disjoint compact subsets. Then there exist disjoint open subsets U and V of X such that A ⊆ U and B ⊆ V .

Yes. For each point x ∈ B, apply problem [5] below to find open subsets Ux and Vx such that A ⊆ Ux and x ∈ Vx. Now cover B by open sub-sets {Vx}x∈B. Since B is compact, we can choose a finitely many of them Vx1 , . . . , Vxn . Let U = Ux1 ∩ · · · ∩ Uxn and V = Vx1 ∪ · · · ∪ Vxn . Then U and V are disjoint open neighborhoods of A and B.

(14) The subset A = {(x, y) ∈ R 2 | y = 0} of R 2 is an open subset of R 2 .

No. A does not contain any open ball centered at any point of A.

(15) Suppose a continuous map f : X → Y is 1:1 and onto. If f is a closed map, then f is a homeomorphism.

Yes. When f is closed, it maps closed sets to closed sets. Since f is bijective, it maps complements of sets to complements. Hence f maps open sets to open sets. Hence f −1 is continuous.

[3] (8+5+5=13 points) Let X be a topological space, and let A and B be subsets of X. The closure of A is defined as the union of A with all of its limit points.

A = A ∪ {limit points}.

(1) The closure of A is the smallest closed subset containing A.

We show X − A is open. Let x ∈ X − A. Since x is a not a limit point of A, there exists an open neighborhood Ux of x containing no points of A. Hence x is not a limit point of A. Also note that every point of Ux is not a limit point of A, either. This Ux ⊆ (X − A). Hence the complement X − A is open. Thus A is a closed subset. Let A ⊆ V be a closed subset containing A. Then the complement of V is open and has no points of A. So no points of V can be limit points. Hence all limit points of A must be already contained in V . Thus, A ⊆ V . This means that A is the smallest closed subset containing A.

(2) Show that if A ⊆ B, then A ⊆ B.

Proof: We use a characterization of the closure as the smallest closed subset containing the given set. If A ⊆ B then A ⊆ B. Since B is a closed subset containing A, it also contains the closure of A, A ⊆ B.

Alternate proof. Since A ⊆ B ⊆ B, we have A ⊆ B. Now let x ∈ X be a limit point of A. Then every neighborhood of x intersects with A non-trivially at a point different from x. Since A ⊆ B, this same neighborhood intersects with B at a point different from x. Hence x is also a limit point of B. Thus, A ⊆ B.

(3) Show that A ∪ B = A ∪ B in X.

We prove both inclusion relations. Since A, B ⊆ A ∪ B ⊆ A ∪ B, we have A, B ⊆ A ∪ B. Hence A ∪ B ⊆ A ∪ B. For the other inclusion relation, note that A∪B ⊆ A∪B and the right hand side is a closed set. Hence it contains the closure of the left hand, A ∪ B ⊆ A ∪ B. This proves both inclusion relations.

An alternate proof of the second inclusion relation A ∪ B ⊆ A ∪ B. We have A∪B ⊆ A∪B by (1). We show that a limit point of A∪B is contained in A ∪ B. Suppose not. Then x is in the complement of A ∪ B, and since the complement is open, there exists an open neighborhood U of x not intersection with A nor B. Thus U does not contain any points from A or B. Thus x cannot be a limit point of A∪B. This is a contradiction. Hence x must be contained in A ∪ B.

[4] (4+6+6+6=22 points) Let X be the set of real numbers R with finite complement topology.

(1) Describe open sets of finite complement topology on the set R.

They are ∅, R or those subsets U of R such that R − U is a finite set.

(2) Is X a Hausdorff topological space? Explain.

No. Any two non empty open sets have nontrivial intersection.

(3) Show that X is a compact topological space.

Let {Uλ} be an open cover of X. and let Uλ0 be a nonempty member of the family. Let X − Uλ0 consist of points {x1, . . . , xr}. Since we have an open cover, for each i, there exists an open set Uλi in the family containing xi. Then {Uλ0 , Uλ1 , . . . , Uλr } is a finite open cover of X.

This proof shows that any topological space with finite complement topology is always compact.

(4) Show that every subset of X is a compact subset. Do the same argument as above. Or induce a topology on this subset from X. Then it is a topological space with finite complement topology. Hence by (3), it is compact.

[5] (10 points) Let X be a Hausdorff space and let A be a compact subset and let x / ∈ A.

Then prove that there exist disjoint open neighborhoods of A and x.

For each point y ∈ A, choose disjoint open neighborhoods Uy and Vy of y and x using Hausdorff property. Cover A by open sets {Uy}y∈A. Since A is compact, there exists a finite subcover consisting of Uy1 , . . . , Uyn . Then let U = Uy1 ∪ · · · ∪Uyn and V = Vy1 ∩ · · · ∩ Vyn . Then U and V are disjoint open subsets containing A and x, respectively.