This post is the note from reading Bert Mendelson’s book
Introduction to Topology, 3rd Edition.
Most of the content of this note directly comes from the book.
The first chapter of the book about the set theory is omitted in
this note.
At the moment, the note is obviously unfinished and I am still reading the book.
Definition (Metric Space):⟨X,d⟩ is a metric space if X is the underlying set,
and d:X×X→R is a distance function, which satisfies
∀x,y∈X,d(x,y)≥0
∀x,y∈X,d(x,y)=0 only if x=y
∀x,y∈X,d(x,y)=d(y,x)
∀x,y,z∈X,d(x,z)≤d(x,y)+d(y,z)
The last one assert the transitivity of closeness: if x is close to y
and y is close to z, then x is close to z.
Example: ⟨R,d⟩ is a metric space on real numbers, where d(x,y)=∣x−y∣.
Theorem: Given a set of metric space ⟨X1,d1⟩, ⟨X2,d2⟩,
⋯, ⟨Xi,di⟩, we can obtain a metric space on
X=Πi=1nXi with d:X×X→R. Let
d(x,y)=i≤i≤nmaxdi(xi,yi)
Proof: By verifying the four requirements in the definition. □
Remark: By instantiating the above theorem,
⟨Rn,d:Rn×Rn→R⟩
is a metric space, where
d((x1,x2,⋯,xn),(y1,y2,⋯,yn))=i≤i≤nmax∣xi−yi∣
Remark: ⟨Rn,d’:Rn×Rn→R⟩
is a metric space, where
d’((x1,x2,⋯,xn),(y1,y2,⋯,yn))=Σi=1n(xi−yi)2
Note: d’ is the Euclidean distance function.
Definition (Continuity): A function f:R→R is continuous at point a∈R,
if given ϵ>0, there exists a δ>0, such that if
∣x−a∣<δ, then ∣f(x)−f(a)∣<ϵ for all x∈R.
The function is continuous if it is continuous at each point of R.
Definition (Continuity): Given two metric spaces ⟨X,d⟩ and ⟨Y,d’⟩,
a function f:X→Y is continuous at point a∈X,
if given ϵ>0, there exists a δ>0, such that if
d(x,a)<δ, then d’(f(x),f(a))<ϵ for all x∈X.
The function is continuous if it is continuous at each point of X.
Theorem: A constant function f:X→Y is continuous.
Proof: It is always the case that d’(f(x),f(a))=0. □
Theorem: The identity function f:X→X is continuous.
Proof: Choose δ=ϵ. □
Theorem (Composition of Continuous Functions): Let ⟨X,d1⟩, ⟨Y,d2⟩, and ⟨Z,d3⟩
be metric spaces. Let f:X→Y be continuous at a∈X, and g:Y→Z
be continuous at f(a)∈Y, then g∘f:X→Z is continuous at a∈X.
Corollary: Let ⟨X,d1⟩, ⟨Y,d2⟩, and ⟨Z,d3⟩
be metric spaces.
Let f:X→Y and g:Y→Z be continuous, then g∘f:X→Z is
continuous.
Definition (Neighborhoods): Let ⟨X,d⟩ be a metric space and a∈X. A set N⊂X
is called a neighborhood of a if there exists a δ>0 such that
B(a;δ)⊂N
The collection Na of all neighborhoods of a∈X is called
a complete system of neighborhoods of the point a.
Lemma (4.5): Let ⟨X,d⟩ be a metric space and a∈X. For every δ>0,
the open ball B(a;δ) is a neighborhood of point b∈B(a;δ).
Proof: We need to show that for all b∈B(a;δ), there exists
η>0 such that B(b;η)⊂B(a;δ).
Choose η<δ−d(a,b). Then ∀x∈B(b;η),
d(a,x)≤d(a,b)+d(b,x)<d(a,b)+η<d(a,b)+δ−d(a,b)=δ
Therefore, ∀x∈B(b;η), we have x∈B(a;δ). □
Lemma (4.6): Let f:⟨X,d⟩→⟨Y,d’⟩. f is continuous at point a∈X
iff for for all neighborhood M of f(a), there exists a corresponding
neighborhood N of a, such that
f(N)⊂M, or equivalently N⊂f−1(M)
Proof:
First assume that f is continuous at a.
Let M be a neighborhood of f(a). We must to show that
there exists a neighborhood N of a and f(N)⊂M.
1) By the definition of neighborhoods, there exists ϵ such that
B(f(a),ϵ)⊂M.
2) Since f is continuous, there exists a δ>0 such that
f(B(a;δ))⊂B(f(a);ϵ) (theorem 4.2).
3) Now, let N be B(a;δ). We can conclude that
f(N)=f(B(a;δ)⊂B(f(a);ϵ)⊂M
To prove the other direction, assume that ∀ neighborhood M of f(a),
there exists a neighborhood N of a, such that f(N)⊂M.
Let ϵ>0 be the threshold of such neighborhood M=B(f(a);ϵ).
To prove f is continuous, we must show that there exists δ>0
such that f(B(a;δ))⊂B(f(a);ϵ) (theorem 4.2).
By the hypothesis, there is neighborhood N of a such that f(N)⊂M.
By the definition of neighborhood, there exists δ such that
B(a;δ)⊂N. Therefore, we have
f(B(a;δ))⊂f(N)⊂M=B(f(a);ϵ)
□
Lemma (4.7): Let f:⟨X,d⟩→⟨Y,d’⟩. f is continuous at point a∈X
iff for each neighborhood M of f(a), f−1(M) is a neighborhood of a.
Theorem (4.8): Let ⟨X,d⟩ be a metric space.
∀a∈X, there exists at least one neighborhood of a.
∀a∈X and its neighborhood N, a∈N.
∀a∈X, if N is its neighborhood and N’⊃N, then N’ is also a neighborhood of a.
∀a∈X and its neighborhoods N and M,
N∩M is also a neighborhood of a.
∀a∈X and each neighborhood N of a,
there exists a neighborhood O of a, such that O⊂N and
O is a neighborhood of points in O.
Definition (Basis of Neighborhoods): Let a be a point in metric space X. A collection Ba
of neighborhoods of a is called a basic for the neighborhood
system at a, if every neighborhood N of a contains some element
B of Ba.
First, recall the definition of limit on real line.
Definition (Limit):a∈R is the limit of the sequence a1,a2,⋯
if given ϵ>0, there exists a positive integer N such that,
for all n>N, we have ∣a−an∣<ϵ.
We can also say the sequence converges to a, and write limnan=a.
Definition (Limit, generalized): Let ⟨X,d⟩ be a metric space. Let a1,a2,⋯ be a sequence
of points in X. A point a∈X is the limit of the sequence if
nlimd(a,an)=0
Corollary (5.3): Let ⟨X,d⟩ be a metric space and a1,a2,⋯ be a sequence of points in X.
∃a∈X,limnan=a⇔ for each neighborhood V of a,
∃N∈N,∀n>N,an∈V.
Proof:
Let V be a neighborhood of a. By definition of neighborhoods, there is
ϵ such that a∈B(a;ϵ)⊂V. Then if limnan=a,
there exists an ineger N, such that d(a,an)<ϵ whenever n>N.
Therefore an∈V.
Given ϵ>0 and B(a;ϵ) is a neighborhood of a.
By premises, we have N such that n>N,an∈B(a;ϵ).
Then d(a,an)<ϵ, and therefore limnan=a. □
If S is a set of infinite points, and given a proposition P, there is
a finite subset X⊂S, such that ∀x∈X,P(x), then
we can say the P is true for almost all the elements in S.
Therefore, limnan=a if for each neighborhood V of a,
almost all points an are in V. In other wors, for each V,
there are at most finite elements in sequence an are not in V.
Theorem (5.4): Let ⟨X,d⟩ and ⟨Y,d’⟩ be two metric space. A function
f:X→Y is continuous at a point a∈X if and only if,
if limnan=a for a sequence of points in X, limnf(an)=f(a).
Proof: TODO.
If limnan=a, then by a substitution, limnf(an)=f(a)
is equivalent to limnf(an)=f(limnan). In other words,
the continuous function commutes with the limit operation.
Definition (bounds): Let X be a set of real numbers.
A number b is an upper bound of A if ∀x∈A,x≤b.
A number c is a lower bound of A if ∀x∈A,c≤x.
If A has both an upper bound and a lower bound, then A is said to be
bounded.
An upper bound b∗ is a least upper bound of A, if for all
upper bound b of A, b∗≤b. A lower bound c∗ is a
greatest lower bound of A, if for all lower bound c of A,
c≤c∗.
Definition (Completeness postulate): Completeness postulateis a properties of the real number systems, stating that
if a non-empty set A of real numbers has an upper bound, then it also has
a lub; and if a non-empty set B of real numbers has a lower bound, then it
also has a glb .
Lemma (5.6): Let b be a glb of a non-empty set A. For each ϵ>0,
exists an element x∈A such that x−b<ϵ.
Proof: By contradiction. □
Corollary (5.7): Let b be a glb of a non-empty set A of real numbers. There
is a sequence an of real numbers such that an∈A and
limnan=b.
Definition (Distance between an element and a set): Let ⟨X,d⟩ be a metric space. Let a∈X and A
be a non-empty subset of X.
The glb of {d(a,x)∣x∈A} is called the distance
between a and A and is denoted by d(a,A).
Definition (Open Set): A subset O of a metric space is open if O is a neighborhood of each of its points.
We shall also use "open balls" to defined "open sets".
Theorem (6.2): A subset O of a metric space ⟨X,d⟩ is an open set
iff it is a union of open balls.
Proof:
For the first direction, assume that O is open.
By the definition of openness, for each point a∈O,
there exists a ball B(a;δa)⊂O.
Therefore O=∪a∈OB(a;δa).
For the other direction, assume that O is a union of open balls,
whose centers form an indexed set I.
Then O=∪a∈IB(a;δa).
If x∈O, then there exists a∈I s.t. x∈B(a;δa).
B(a;δa) is a neighborhood of x, and sicne B(a;δa)⊂O,
O is also a neighborhood of every x.
□
Theorem (6.3): Let f:⟨X,d⟩→⟨Y,d’⟩. f is continuous
iff for each open set O of Y, the subset f−1(O) is an open subset
of X.
Proof:
For the first direction, we need to show that f−1(O)
is a neighborhood of every point in it.
To show this, let a∈f−1(O), then f(a)∈O and O
is a neighborhood of f(a).
Since f is continous, by Theorem 4.7, f−1(O) is a neighborhood of a.
For the other direction, assume that ∀O⊂Y, f−1(O) is open.
We need to show f is continous.
Let a∈X and M be a neighborhood of f(a).
There exists B(f(a),ϵ)⊂M, since B(f(a),ϵ) is open,
f−1(B(f(a),ϵ)) is open and is a neighborhood of a.
Therefore, f is continous.
□
Theorem (6.4): Let ⟨X,d⟩ be a metric space.
The empty set is open
X is open
If O1,O2,⋯,On are open,
O1∩O2⋯∩On is also open.
If O1,O2,⋯,On are open,
O1∪O2⋯∪On is also open.
Definition (Closedness): A subset F of a metric space is close if its complement C(F) is open.
A set can be both open and closed. For example, given a metric
space ⟨X,d⟩, ∅ and X are open, therefore
their complement X and ∅ are closed. So, ∅
and X are both open and closed.
Definition (Limit Point): Let A be a subset of metric space X. A point b∈X
is a limit point of A if every neighborhood of b contains
a point of A different from b.
Theorem (6.7): Given a metric space ⟨X,d⟩, a set F⊂X is closed
iff F contains all its limit points.
Proof:
For the first direction, assume that F is cosed and therefore C(F)
is open. Let F’ denote the set of limit points of F. We need to proof
that F’⊂F by showing ∀x∈/F,x∈/F’.
Choose a point b∈F, since C(F) is open, there exists a
δ s.t. B(b;δ)⊂C(F). Hence, b∈/F’ and
F’⊂F.
For the other direction, assume that F’⊂F and therefore C(F)⊂C(F’).
We need to proof F is closed by showing C(F) is open.
Since F’⊂F, if b∈C(F), then b∈/F’.
C(F) is a neighborhood of each points, since B(b;δ)⊂C(F) for some δ. □