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22M: Fall 07 J. Simon
Exam 1 Solutions
Problem 1.
Here are four topologies on the set R. For each pair of topologies, determine whether one is a refinement of (i.e. contains) the other. [Justify your claims.]
[note: So you have
2
= 6 comparisons to make.] A The usual (i.e. standard) topology. B The discrete topology. C The lower-limit topology (recall R with this the topology is denoted Rℓ). D The counter-finite (i.e. finite-complements) topology.
Answers:
a Counter-finite is strictly coarser than Standard. Proof: Each finite set in R is closed in the standard topology, so each set whose complement is finite is open in the standard topology. However, an open interval (1, 2) is open in the standard topology; but its complement is infinite, so the interval (1, 2) is not open in the finite-complements topology. b Standard is strictly coarser than lower-limit. Proof: First show that each basis element for the standard topology is open in the lower-limit topology: (a, b) = ∪{[x, b) | b > x > a}. On the other hand, a basis set [a, b) for the lower limit cannot be a union of basis sets for the Standard topology since any open interval in R containing point a must contain numbers less than a. c Lower-limit is strictly coarser than Discrete. Proof: In the Discrete topology, every set is open; so the Lower-limit topology is coarser-than-or-equal-to the Discrete topology. On the other hand, the singleton set { 0 } is open in the discrete topology but is not a union of half-open intervals.
The remaining comparisons follow (by transitivity) from the three above; the four topologies are linearly ordered by proper inclusion.
Problem 2.
Let (X, T ) be a topological space, and let S be a sub-basis for T.
Prove: If S is countable, then there exists a countable basis for T.
Proof. The set of all finite intersections of elements of S is a basis B for T. We shall define a countable set and a function from that countable set onto B; we have a theorem that the image of a countable set is countable, so we will be able to conclude that B is countable. For each n ∈ Z+, let Sn = (S × S ×... × S) n times. A finite cartesian product of countable sets is countable, so each Sn is countable. Let S = ∪∞ n=1Sn. A countable union of countable sets is countable, so S is countable. Now define a surjective function φ : S → B by
φ(S 1 , S 2 ,... Sk) = S 1 ∩ S 2 ∩... Sk.
Problem 3.
Let (X, T ) be a topological space, with A, B ⊆ X. Prove:
A ∪ B = A ∪ B.
Proof. There are various ways to prove this; here is one.
We have the following theorems:
- The closure of a set is closed.
- The union of two closed sets is closed.
- The closure of a set is defined to be the intersection of all closed sets containing the given set.
- and from the last statement above, if P ⊆ Q then P ⊆ Q.
So we can argue as follows:
A ⊆ A , and B ⊆ B =⇒ A ∪ B ⊆ A ∪ B.
Since A ∪ B ⊆ the closed set A ∪ B, the closure of A ∪ B must be contained in that closed set; i.e. A ∪ B ⊆ A ∪ B.
Conversely, A ⊆ A ∪ B =⇒ A ⊆ A ∪ B
and likewise for B.
Proof. Let B = {x ∈ X | f (x) > g(x) }. We will show B is open. All open intervals (a, b) and open rays (a, ∞), (−∞, b) are open sets in the order topology on Y , and these sets form a basis; the rays form a sub-basis. Let x be any point in B. We want to find a neighborhood of x contained in B. So we need to find an open neighborhood U(x) such that for each w ∈ U, f (w) > g(w). As in the proof that an order topology has the Hausdorff property, we claim there exist disjoint sub-basic neighborhoods [i.e. open rays] U of f (x) and V of g(y). [If there exists w ∈ (g(y), f (x)) then use (−∞, w) and (w, ∞). If the open interval (g(y), f (x) is empty, then use the rays (−∞, f (x)) and (g(y), ∞).] Then f −^1 (U) ∩ g−^1 (V ) is a neighborhood of x on which f is strictly larger than g.
For part (b), note that the function h is just f on the closed set A and g on the dual closed set C where g(x) ≤ f (x). On the intersection A ∩ C, we have f (x) = g(x). So by the “pasting lemma”, this function is well-defined and continuous.
Problem 6.
Suppose X, Y are topological spaces, and f : X → Y is a continuous function. In the space X × Y (with the product topology) we define a subspace G called the “graph of f ” as follows:
G = {(x, y) ∈ X × Y | y = f (x)}.
Prove: G is homeomorphic to X.
(Be sure to state clearly whatever theorem(s) you use in doing this proof.)
Proof. Define φ : X → G by φ(x) = (x, f (x)). As the cartesian product of two continuous functions, φ is continuous. Check that φ iss a bijection [1-1 is because f is a function; surjective is by definition of the “graph of f ”.] The function φ−^1 is just the restriction to G of the projection map πX : X × Y → X. We have a previous theorem that πX is continuous and a theorem that the restriction of a continuous function to a subspace is continuous. So φ−^1 : G → X is continuous.
