AMATH/PMATH 331 Assignment 3
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
AMATH/PMATH 331
Assignment 3
Due Friday February 24 at 11:59 pm (end of day).
Set-Up: As you may now realize, all the theorems we developed so far depended on one intricate concept, namely that of a distance. Adopting a distance is a way to measure how elements are distributed within a space, so that we are able to say a point is closer to another, or that a sequence is “approaching” some limit, and so on. The most generic and minimalistic way for providing such a frame-of-work is through defining a “topology”. Then, a space equipped with an actual distance function will be a special case, in which the topology is derived from the balls defined via that distance. Let’s see:
Definition 0.1. Let X be a set and τ be set of subsets of X . The pair (X,τ) is called a topological space if
(1) ∅,X e τ .
(2) τ is closed under arbitrary unions, and closed under finite intersections. That is, a (possibly infinite) union of elements of τ is again in τ; and an intersection of finitely many elements from τ is again in τ .
In that case we say that τ is a topology defined on X .
For a very small example, given X = {a,b}, the Sierpinski Topology on X is defined by τ = {∅,X,{a}}.
Definition 0.2. Let X be a set and d ∶ X × X → R be a scalar-valued function. The pair (X,d) is called a metric space if the following are true for all x,y,z e X
(1) d(x,y) ≥ 0, and d(x,y) = 0 if and only if x = y .
(2) d(x,y) = d(y,x).
(3) d(x,z) ≤ d(x,y) + d(y,z) (the Triangle Inequality).
The function d is usually called a distance or a metric. The topology induced by a metric is simply defined through open balls, and a set is (then) open if it is a union of open balls, or, rephrased differently, a set U is open if every element of it can be contained in an open ball falling entirely inside U .
Questions:
(1) In a metric space M, a subset A ⊂ M is said to be dense if 
= M . Prove that each of the following statements is a characterization of A being dense:
(a) Every point of M is a limit point of some sequence in A.
(b) For all x e M and ϵ > 0, Be (x) ∩ A ≠ ∅.
(c) For every nonempty open set U in M , U ∩ A ≠ ∅.
(d) The complement Ac = M − A has an empty interior.
(2) A subset A of M is said to be nowhere dense if int(
) = ∅. Prove that {x} is nowhere dense if and only if x is not an isolated point of M! Recall that a point is isolated from a set if there exists an open ball containing it that is not intersecting with that particular set.
(3) Negating the definition of continuity of a function f ∶ R → R, we can formally define the set of discontinuities D(f) as follows:
D(f) = {a ∶ 3ϵ > 0 Aδ > 0 3x s.t ∣x − a∣ < δ and ∣f(x) − f(a)∣ ≥ ϵ}.
Now, define the oscillation of f on a bounded interval I by ω(f;I) = sup{∣f(x) − f(y)∣ ∶ x,y e I}, and accordingly define oscillation at a point to be
ω(a) = inf{ω(f;I) ∶ I is open and a e I} = inf sup {∣f(x) − f(y)∣}. I open x,y ∈I
a∈I
Prove that D(f) has to be a countable union of closed sets in R.
(4) Sets of the above type, namely countable unions of closed sets in R, are called Fσ sets. The F stands for ferm´e (FR: closed) and the σ stands for somme (FR: sum). Taking the complement of such an Fσ set we get a G6 set, which is a countable intersection of open sets. In the latter the G is for Gebiet (DE: region) and the δ is for Durchschnitt (DE: intersection).
Use the Baire Category Theorem to show that a dense G6 has to be uncountable.
(5) Show that Q cannot be written as a countable intersection of open sets (Hint: use the Baire Category Theorem).
(6) Show that R − Q cannot be realized as D(f) for any function f ∶ R → R.
2023-02-25