topology
product topology
Let `X` and `Y` be topological spaces. The product topology on `X xx Y` is the topology having as basis the collection `B` of all sets of the form `U xx V`, where `U` is an open subset of `X` and `V` is an open subset of `Y`
if B is a basis for the topology of X and C is a basis for the topology
of Y , then the collection
D = {B × C  B ∈ B and C ∈ C}
is a basis for the topology of X × Y .
so D is a basis for X × Y .
projection function
Let π1 : X × Y → X be defined by the equation π1(x, y) = x; let π2 : X × Y → Y be defined by the equation π2(x, y) = y. The maps π1 and π2 are called the projections of X × Y onto its first and second factors, respectively.
The collection
S
= {π−1
1 (U)  U open in X}∪{π−1
2 (V)  V open in Y }
is a subbasis for the product topology on X × Y .
Let X be a topological space with topology T . If Y is a subset of X, the
collection TY = {Y ∩ U  U ∈ T }
is a topology on Y , called the subspace topology. With this topology, Y is called a
subspace of X; its open sets consist of all intersections of open sets of X with Y .
If B is a basis for the topology of X then the collection
BY = {B ∩ Y  B ∈ B}
is a basis for the subspace topology on Y
Let Y be a subspace of X. If U is open in Y and Y is open in X, then
U is open in X.
f A is a subspace of X and B is a subspace of Y , then the product
topology on A × B is the same as the topology A × B inherits as a subspace of X ×Y
Let X be an ordered set in the order topology; let Y be a subset
of X that is convex in X. Then the order topology on Y is the same as the topology Y
inherits as a subspace of X.
A subset A of a topological space X is said to be closed if the set X − A is open.
Let X be a topological space. Then the following conditions hold:
(1) ∅ and X are closed.
(2) Arbitrary intersections of closed sets are closed.
(3) Finite unions of closed sets are closed.
Let Y be a subspace of X. Then a set A is closed in Y if and only if
it equals the intersection of a closed set of X with Y .
Let Y be a subspace of X. If A is closed in Y and Y is closed in X,
then A is closed in X.
Given a subset A of a topological space X, the interior of A is defined as the union of
all open sets contained in A,
the closure of A is defined as the intersection of all
closed sets containing A
Let Y be a subspace of X; let A be a subset of Y ; let A¯ denote the
closure of A in X. Then the closure of A in Y equals A¯ ∩ Y .
Let A be a subset of the topological space X.
(a) Then x ∈ A¯ if and only if every open set U containing x intersects A.
(b) Supposing the topology of X is given by a basis, then x ∈ A¯ if and only if every
basis element B containing x intersects A.
Mathematicians often use some special terminology here. They shorten the state
ment “U is an open set containing x” to the phrase
“U is a neighborhood of x.”
If A is a subset of the topological space X and if x is a point of X, we say that x is a
limit point (or “cluster point,” or “point of accumulation”) of A if every neighborhood
of x intersects A in some point other than x itself. Said differently, x is a limit point
of A if it belongs to the closure of A − {x}. The point x may lie in A or not; for this
definition it does not matt
Let A be a subset of the topological space X; let A be the set of all
limit points of A. Then
A subset of a topological space is closed if and only if it contains all
its limit points.
In an
arbitrary topological space, one says that a sequence x1, x2, ... of points of the space
X converges to the point x of X provided that, corresponding to each neighborhood U
of x, there is a positive integer N such that xn ∈ U for all n ≥ N.
A topological space X is called a Hausdorff space if for each pair x1, x2
of distinct points of X, there exist neighborhoods U1, and U2 of x1 and x2, respectively,
that are disjoint.
Every finite point set in a Hausdorff space X is closed
Let X be a space satisfying the T1 axiom; let A be a subset of X.
Then the point x is a limit point of A if and only if every neighborhood of x contains
infinitely many points of A.
If X is a Hausdorff space, then a sequence of points of X converges
to at most one point of X.
Every simply ordered set is a Hausdorff space in the order topology.
The product of two Hausdorff spaces is a Hausdorff space. A subspace of a Hausdorff
space is a Hausdorff space.
continuous function
Let `X` and `Y` be topological spaces. A function `f : X > Y` is said to be continuous if
for each open subset `V` of `Y`, the set `f −1(V)` is an open subset of `X`.
prove that this definition of continuity is equivalent to the epsilon delta definition
continous function for a basis
given `f : X to Y` if the topology of the range space Y is given by a basis `cc B`, then to prove continuity of `f` it suffices to show that the inverse image of every basis element is open
Let X and Y be topological spaces; let f : X → Y . Then the
following are equivalent:
(1) f is continuous.
(2) For every subset A of X, one has f (A¯) ⊂ f (A).
(3) For every closed set B of Y , the set f −1(B) is closed in X.
(4) For each x ∈ X and each neighborhood V of f (x), there is a neighborhood U
of x such that f (U) ⊂ V.
If the condition in (4) holds for the point x of X, we say that f is continuous at
the point x.
homeomorphism
Let `X` and `Y` be topological spaces; let `f : X > Y` be a bijection. If both the function `f` and `f^{1} : Y > X` are continuous, then `f` is called a homeomorphism
imbedding
Now suppose that `f : X → Y` is an injective continuous map, where `X` and `Y` are topological spaces. Let `Z` be the image set `f(X)`, considered as a subspace of `Y` ; then the function `f_Z : X → Z` obtained by restricting the range of `f` is bijective. If `f_Z` happens to be a homeomorphism of `X` with `Z`, we say that the map `f : X → Y` is a topological imbedding, or simply an imbedding, of `X` in `Y` .
rules for constructing continuous functions
Let `X, Y` and `Z` be topological spaces, the the following statements all hold:

(Constant function) If `f : X → Y` maps all of `X` into the single point `y_0` of `Y` then `f` is continuous

(Inclusion) If `A` is a subspace of `X`, the inclusion function `j : A → X` is continuous

(Composites) If `f : X → Y` and `g : Y → Z` are continuous, then the map `g @ f : X → Z` is continuous.

(Restricting the domain) If `f : X → Y` is continuous, and if `A` is a subspace of `X`, then the restricted function `f_A : A → Y` is continuous.

(Restricting or expanding the range) Let `f : X → Y` be continuous. If `Z` is a subspace of `Y` containing the image set `f(X)`, then the function `g : X > Z` obtained by restricting the range of `f` is continuous. If `Z` is a space having `Y` as a subspace, then the function `h : X > Z` obtained by expanding the range of `f` is continuous

(Local formulation of continuity) The map `f : X > Y` is continuous if `X` can be written as the union of open sets `uuu_alpha` such that `f_{uuu_alpha}` is continuous for each `alpha`
Let X = A ∪ B, where A and B are closed
in X. Let f : A → Y and g : B → Y be continuous. If f (x) = g(x) for every
x ∈ A ∩ B, then f and g combine to give a continuous function h : X → Y , defined
by setting h(x) = f (x) if x ∈ A, and h(x) = g(x) if x ∈ B.
projection functions are continuous
The
projection functions `pi_1 : X xx Y to X` and `pi_2 : X xx Y to Y` are continuous
We'll start by showing that `pi_1` is continuous, so let `U` be open in `X` and now we can note that `pi_1^{1}(U) = U xx Y` since the second component never matters in the computation of `pi_1(x, y) = x`
By the very definition of the product toplogy, we can see it is generated by the basis consisting of `U xx V` open in `X` and `Y` respectively, we assumed `U` was open, `Y` is open in itself by the definition of toplogy, thus `U xx Y` is open in `X xx Y` since every basis element is also in the toplogy.
Thus we may conclude that `pi_1` is continuous, the proof for `pi_2` may be obtained symmetrically.
product functions are continuous
Let `f : A > X xx Y` be given by the equation `f(a) = (f_1(a) , f_2(a))`. Then `f` is continuous if and only if the functions `f_1: A > X` and `f_2: A > Y` are continuous. The maps `f_1` and `f_2` are called the coordinate functions of `f`
winking functions are continuous
Given `x_0 in X` and `y_0 in Y`, show that the maps `f: X > X xx Y` and `g: Y > X xx Y` defined by
`f(x) = x xx y_0` and `g(y) = x_0 xx Y`
are
imbeddings.
details
We'd like to show that `f` is an imbedding, that is that both `f^'` and `(f^{'})^{1}` are both continuous functions where `f'` is `f` with it's range restricted to `f(X)`.
Starting with `f^'` and acutally looking at the original `f` we can see that it can be defined as `f(x) = (id(x), c(x))` with the identity and constant function respectively. Looking at the rules for contructing continuous functions, we can see that `id: X > X` is simply an inclusion function from `X` into itself, and that `c: Y > y_0` is a constant funciton, and thus are both continuous, also since product functions are continuous we know that `f` is continuous, using rule 4, we can see that `f'` is also continuous.
Looking at `(f^')^{1}: X xx {y_0} > X`, we can see that this function simply drops the second component and thus `(f^')^{1} = pi_1` and since projection functions are continuous, we've shown that the inverse function is continuous.
With all that said, we now know that `f'` is a homeomorphism, which by definition means that `f` is an imbedding, which is what we needed to show.
Let J be an index set. Given a set X, we define a J tuple of elements
of X to be a function x : J → X. If α is an element of J , we often denote the value
of x at α by xα rather than x(α); we call it the αth coordinate of x. And we often
denote the function x itself by the symbol
(xα)α∈J ,
which is as close as we can come to a “tuple notation” for an arbitrary index set J . We
denote the set of all J tuples of elements of X by X J .
Let {Aα}α∈J be an indexed family of sets; let X =
α∈J Aα. The
cartesian product of this indexed family, denoted by
α∈J
Aα,
is defined to be the set of all J tuples (xα)α∈J of elements of X such that xα ∈ Aα for
each α ∈ J . That is, it is the set of all functions
x : J → 
α∈J
Aα
such that x(α) ∈ Aα for each α ∈ J .
Let {Xα}α∈J be an indexed family of topological spaces. Let us take as
a basis for a topology on the product space
α∈J
Xα
the collection of all sets of the form
α∈J
Uα,
where Uα is open in Xα, for each α ∈ J . The topology generated by this basis is called
the box topology.
Let
πβ :
α∈J
Xα → Xβ
be the function assigning to each element of the product space its βth coordinate,
πβ((xα)α∈J ) = xβ;
it is called the projection mapping associated with the index β.
Let 
β denote the collection

β = {π−1
β (Uβ)  Uβ open in Xβ},
and let 
denote the union of these collections,

= 
β∈J

β.
The topology generated by the subbasis 
is called the product topology. In this topol
ogy
α∈J Xα is called a product space.
The box topol
ogy on
Xα has as basis all sets of the form
Uα, where Uα is open in Xα for
each α. The product topology on
Xα has as basis all sets of the form
Uα, where
Uα is open in Xα for each α and Uα equals Xα except for finitely many values of α
Suppose the topology on each space Xα is given by a basis Bα. The
collection of all sets of the form
α∈J
Bα,
where Bα ∈ Bα for each α, will serve as a basis for the box topology on
α∈J Xα.
The collection of all sets of the same form, where Bα ∈ Bα for finitely many
indices α and Bα = Xα for all the remaining indices, will serve as a basis for the
product topology
α∈J Xα.
Let Aα be a subspace of Xα, for each α ∈ J . Then
Aα is a
subspace of
Xα if both products are given the box topology, or if both products are
given the product topology.
If each space Xα is Hausdorff space, then
Xα is a Hausdorff space
in both the box and product topologies.
Let {Xα} be an indexed family of spaces; let Aα ⊂ Xα for each α. If
Xα is given either the product or the box topology, then
A¯α = Aα.
Let f : A →
α∈J Xα be given by the equation
f (a) = ( fα(a))α∈J ,
where fα : A → Xα for each α. Let
Xα have the product topology. Then the
function f is continuous if and only if each function fα is continuous
metric on a set
A metric on a set `X` is a function `d : X xx X to RR` having the following properties:
 `d(x, y) ge 0` for all `x, y in X`; equality holds if and only if `x = y`
 `d(x, y) = d(y, x)` for all `x, y in X`
 `d(x, y) + d(y,z) ge d(x,z)` for all `x, y, z in X`
Given a metric `d` on `X`, the number `d(x, y)` is often called the distance between `x` and `y`
metric ball
Given `epsilon in RR^{gt 0}`, consider the set `B_d(x, epsilon) = {y : d(x, y) < epsilon}` of all points `y` whose distance from `x` is less than `epsilon`. It is called the `epsilon`ball centered at `x`. Sometimes we omit the metric `d` from the notation and write this ball simply as `B(x, epsilon)`, when no confusion will arise.
topology induced by a metric
If `d` is a metric on the set `X`, then the collection of all `epsilon`balls `B_d (x, epsilon)`, for `x in X` and `epsilon in RR^{gt 0}`, is a basis for a topology on `X`, called the metric topology induced by `d`
If `X` is a topological space, `X` is said to be metrizable if there exists a metric `d` on the set `X` that induces the topology of `X`. A metric space is a metrizable space `X` together with a specific metric d that gives the topology of `X`
In `RR^n` define the metric `d^'(x, y) = abs(x_1  y_1) + ... + abs(x_n  y_n)` show that `d^'` is a metric that induces the standard topology of `RR^n`
(TODO: show that it's a metric first of all)
Since we know that `d^'` is a metric, then now we can check to see if it induces the usual standard topology of `RR^n`. To show this we'll show that the toplogy induced is equal to the standard toplogy of `RR^n`
One method to show that two topologies are equal is to satisfy the requirements of topology equality for topologies generated by a basis
Let `p in RR^n` and let `p in B_(d^')(x, epsilon)` for some `x in RR^n` and `epsilon in RR^(gt 0)` be a basis element for the topology induced by `d^'`
Recall that `B_(d^') (x, epsilon)` is a metric ball and thus is equal to `{y in RR^n : d(x, y) = abs(x_1  y_1) + ... + abs(x_n  y_n) < epsilon}`
Let X be a metric space with metric d. A subset A of X is said to be
bounded if there is some number M such that
d(a1, a2) ≤ M
for every pair a1, a2 of points of A. If A is bounded and nonempty, the diameter of A
is defined to be the number
diam A = sup{d(a1, a2)  a1, a2 ∈ A}.
Let X be a metric space with metric d. Define d¯ : X × X → R by
the equation
d¯(x, y) = min{d(x, y), 1}.
Then d¯ is a metric that induces the same topology as d.
The metric d¯ is called the standard bounded metric corresponding to d.