Mathematical Logic (Part II) December 22, 2014
1. Let N be the set of natural numbers. Define two distinct metrics on N (recall that distinct metrics can define the same topology).
Solution:
d1(x, x) = 0 and d1(x, y) = 1 for all x 6= y d2(x, y) = |x − y|.
2. Let R be the real line with the euclidean topology. Determine whether the following sets are open, closed, neither open nor closed:
(a) A = {5, 3, 7};
(b) B = {x ∈ Q : 0 ≤ x ≤ 1}, where Q is the set of rational numbers;
(c) C = {1/x : x ∈ R, x ≥ 1}.
Solution:
- A is closed, because A = {5} ∪ {3} ∪ {7} is union of singleton sets. Each singleton set {x} is closed because the complement R \ {x} is open. For every real y 6= x, the open interval (y − , y + ), where = |x − y|/2, is contained within R \ {x}.
- B is neither open neither closed. B is a countable set and it is not union of open intervals because each open interval is uncountable and contains irrational numbers. B is not closed because every irrational number in the interval [0, 1] belongs to the closure of B.
- C = (0, 1] is a semiopen interval; it is neither open (every interval containg 1 has points outside C) neither closed (the real 0 belongs to the closure of C).
3. Let R be the set of real numbers with the usual ordering. Let τ be the topology on R generated by the infinite intervals (x, +∞), where x ∈ R.
(a) Show that τ is different from the Alexandrov topology on R.
Solution:
The infinite interval [x, +∞) is Alexandrov open, but it is not τ open.
(b) Define two nontrivial (that is, 6= ∅, R) distinct closed subsets of the space (R, τ ).
Solution:
Recall that open sets generating the topology τ are upper sets. Then every closed set is a down-set. The following are two closed sets: (−∞, 5] and (−∞, 3].
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4. Let N be the set of natural numbers with the cofinite topology. Is the function f (x) = x2 continuous from N into N?
Solution:
Let A be a cofinite subset of N such that N \ A = {n1, . . . , nk}.
We have to show that f−1(A) is cofinite. If N \ A = {n1, . . . , nk} then f−1(A) = {r ∈ N : r2 ∈ {n/ 1, . . . , nk}} is cofinite!
5. Let N be the set of natural numbers with the usual ordering. Consider the Alexandrov topology on N and the product topology on N × N. Does the space N × N satisfy the separation axiom T0?
Solution:
By definition N × N = {(x, y) : x, y ∈ N}. A base of the product topology on N × N is given by the sets
U × V = {(x, y) : x ∈ U, y ∈ V }, where U and V are upper subsets of N.
Let (x, y) and (z, t) two distinct pairs of natural numbers. If x < z then the open set [z, +∞) × N = {(u, v) : u ≥ z, v ∈ N}
contains (z, t) but not (x, y). The same reasoning works if y < t or vice versa. Then the space is T0.
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