Middle East Technical University Spring 1994 Math 471 Elementary Topology and Geometry

POINT SET TOPOLOGY: GENERALITIES

Solutions

Topology in a set, its basis

Problem 1. (2 pts) Describe all topologies on the set {1, 2, 3}. Construct the least basises for them.

Solution: There exist 29 topologies splitting into 9 homeomorphism classes. The list below contains 1 representative from each class, the other ones one can get by permutations of 1, 2, 3 (the number of topologies in each class is given in square brackets). ∅ and X={1, 2, 3} are not included each time.

1) Antidiscreet (basis: X); 2) Discreet (basis: {1}, {2}, {3}); 3) {1}, {1, 2} [6]; 4) {1}, {2}, {1, 2} [3]; 5) {1}, {2, 3} [3]; 6) {1}, {1, 2}, {1, 3} [3]; 7) {1} [3]; 8) {1, 2} [3]; 9) {1}, {2}, {1, 2}, {1, 3} [6] (basis: all sets exept of the third one). Minimal basises in the cases 3–8 include all sets specified (exeptional case: 4) and X (exeptional case: 6).

Problem 2.

(1) (2 pts) Prove that the set of all arithmetic sequences is a basis for some topology in ℕ.

(2) (6 pts) How to use this topology to prove the infinity of the number of

Primes ?

Hint: prove that if the number of Primes is finite, then the set {1} is open (?!).

Solution:

(1): For A_{i} = {a_{i} + kd_{i} | k ∈ ℕ}, i = 1, 2 the intersection A_{1} ∩ A_{2} either empty or

equal to {a + kd | k ∈ ℕ}, where a = min(A_{1} ∩ A_{2}), d = LCM(a_{1}, a_{2}).

(2): Note that the sets F_{p} = {n ∈ ℕ | p divides n} are closed, becouse

ℕ − F_{p} = ⋃_{1⩽k⩽p−1} U_{k}, where U_{k} = {n ∈ ℕ | n = k mod p}.

So, if the number of Primes was finite, then ℕ − {1} = ⋃_{all primes p |} F_{p} would be closed ?!

Metric spaces, metric topology

Problem 3. (2 pts) Construct a metric space and two balls in it such that the ball with the smaller radius contains the ball with the bigger one and not coinsides with it.

What is the least number of points in the space with the above property ?

Solution: Example is \{1, 2, 3\} with the standard metric: B_{\frac{3}{2}} (1) = \{1, 2\} ⫋ B_{\frac{4}{3}} (2) = {1, 2, 3}. Three points in an example is, obviously, minimum.

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Problem 4. (3 pts) Prove that the following maps ℝ^{n} × ℝ^{n} → ℝ_{+} are metrics:

(1) (x, y) → max_{i=1...n} |x_{i} − y_{i}|,

(2) (x, y) → ∑_{i=1...n} |x_{i} − y_{i}|.

(3) Describe balls and spheres in the above spaces in the case n = 2.

(4) Are these metrics equivalent ?

Solution: For (1)–(2) the nontrivial part is the triangle inequality:

max_{i=1...n} |x_{i} − y_{i}| + max_{i=1...n} |y_{i} − z_{i}| ⩾ max_{i=1...n}(|x_{i} − y_{i}| + |y_{i} − z_{i}|) ⩾ max_{i=1...n} |x_{i} − z_{i}|,

∑_{i=1...n} |x_{i} − y_{i}| + ∑_{i=1...n} |y_{i} − z_{i}| ⩾ ∑_{i=1...n}(|x_{i} − y_{i}| + |y_{i} − z_{i}|) ⩾ ∑_{i=1...n} |x_{i} − z_{i}|.

(3): Balls are squares with sides parallel to the coordinate axices in case (1), and to the bisectrices of axices in case (2). (4): Yes, since ρ_{1}(x, y) ⩽ ρ_{2}(x, y) ⩽ nρ_{1}(x, y), where ρ_{i} is the metric of item (i).

Problem 5. (3 pts) Let ρ: X × X → ℝ be a metric.

(1) Prove that (x, y) ↦ \frac{ρ(x,y)}{1+ρ(x,y)} is also a metric.

(2) Prove that the new metric is equivalent to ρ.

Solution: (1): The nontrivial part is the triangle inequality. Let a = ρ(x, y), b = ρ(y, z), c = ρ(x, z). Since a + b > c and \frac{x}{1+x} = 1 − \frac{1}{1+x} is increasing function, we have: \frac{c}{1+c} < \frac{a+b}{1+a+b} = \frac{a}{1+a+b} + \frac{b}{1+a+b} < \frac{a}{1+a} + \frac{b}{1+b}.

(2): Denote the new metric ρ′. Then min(\frac{1}{2}, \frac{1}{2}ρ(x, y)) ⩽ ρ′(x, y) ⩽ ρ(x, y) Hence, B′_{min( \frac{1}{2} , \frac{r}{2} )} ⊂ B_{r}(x) ⊂ B′_{r}(x), so, any ball in one of these metrics contains some, small enouph, ball in the other one.

Induced Topology

Problem 6. (2 pts) Describe the topology induced on ℕ from (1) ℝ with the usual topology,

(2) ℝ with Zarissky topology,

(3) ℝ with the ”arrow” topology.

Solution: (1): Discreet topology; (2): closed sets are finite ones and ℕ itself; (3): open sets

are U_{n} = {k ∈ ℕ | k ⩾ n}, n ∈ ℕ, and ∅.

Position of points in respect to a set

Problem 7. (2 pts) Prove that Fr(A) = Fr(X − A). Find a formular for Fr(A), which contains A and X − A symmetrically.

Solution: Fr(A) = Cl(A) − Int(A) = Cl(A) ∩ (Int(A))^{C} = Cl(A) ∩ Cl(X − A).

Problem 8. (3 pts) Is it true, that Int(A ∩ B) = Int A ∩ Int B; Int(A ∪ B) = Int A ∪ Int B ? Give an example when one of the above equalities fails. Find some inclusion, which is valid for all A and B.

Solution: Int(A ∩ B) ⊂ Int A ∩ Int B, since the left side is open and is contained both in A and B. The opposite inclusion holds, since Int A ∩ Int B is an open set, contained both in A and B, hence, in their intersection too.

The second equality is false, consider, e.g., A = ℚ, B = ℝ − ℚ. But, the inclusion Int A ∪ Int B ⊂ Int(A ∪ B) is true, since Int A, Int B are, obviously, contained in A ∪ B, and, hence, in its interior.

Dense sets

Problem 9. (1 pts) Prove that the intersection A ∩ B of two open dense sets A, B ⊂ X is also dense.

Solution: If U ⊂ X, is open, then V = B ∩ U is open too and nonempty, since B is dence.

Then A ∩ V is nonempty, and (A ∩ B) ∩ U = A ∩ (B ∩ U) = A ∩ V ≠ ∅, q.e.d.

Problem 10. (2 pts) Describe the sets which are dense in the spaces with the discrete and with the antidiscrete topologies. The same question for ℝ with the ”arrow” topology and with Zarissky topology.

Solution: For discreete X: only X. For antidiscreete X: any A ≠ ∅. For ”arrow”: A is dence iff sup A = ∞. For Zarissky topology: iff A is an infinite set.

Problem 11. (2 pts) Let f: X → Y be a continious surjective map and A ⊂ X is dense in X. Prove that f(A) is dense in Y.

Solution: Let U ⊂ Y be open, then f(A) ∩ U = f(f^{−1}(U) ∩ A) is nonempty, since F^{−1}(U) is open.