Munkres Topology

Definition - Topology generated by a basis

A subset U is said to be open in X if for each element x in X, if x in U there is a B s.t. x in B contained in U

Definition of Haussdorf Space

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.

Topology

A set X has a topology T if T contains The empty set and X All unions of its sets All finite intersections of its sets

Definition of Convergence

A sequence converges to x if every neighborhood of x has an integer N such that U contains xn for all n>=N

Unit Circle

S1={x x y | x^2 + y^2 = 1 }

Discrete topology

Powerset on X

Limit Point

A point x is a limit point of a set A if every ε-neighborhood Vε(x) of x intersects the set A at some point other than x.

Lenma K-Topology and Lower Limit topology

Both are finer than the standard topology but they are not comparable.

U x V n U1 x V1

(U n U1) x (V n V1)

Open Rays and Why They Matter

(a, infinity), (-infinity,a) Open rays are open in the order topology because they are either infinite unions of basis elements of Order topology (a,x) or they take the form (a,xmax] and [xmin,a) if they exist. Basis elements of the order topology are just finite intersections of open rays thus open rays form a subbasis for the order topology

Intervals

(a,b) - open [a,b) -ho (a,b] -ho [a,b] - closed

Basis for a topology

A basis for a topology is a collection of subsets of X s.t. for every x in X, there is a basis element containing x. U(B) = X x in (B3 in B1 n B2) if x in B1 and x in B2

Boundary of A

A closure n Cl(X-A)

When is a subset closed limit point Theorem

A is a closed set if it contains all its limit points

Definition of a metric

A metric that takes X x X - > R. d(x,y) >=0, d(x,y)=0 iff x=y d(x,y)=d(y,x) Triangle inequality d(x,y)+d(y,z)>=d(x,z)

First countability axiom and countable basis at the point

A space X that has a countable basis at each of its points is said to satisfy the first countability axiom. A space X is said to have a countable basis at the point x if there is a countable collection {Un}n∈Z+ of neighborhoods of x such that any neighborhood U of x contains at least one of the sets Un.

Definition of Metrizable

A space is metrizable if there exists a metric d that induces a topology on X. X is now a metric space when it is metrizablel and combined with metric d.

Definition Subbasis

A subbasis only has the property that all its unions equals X Topology generated by subbasis is all unions of all finite intersections of elements of S -Finite Intersections of elements of S is a basis B1 = S1 n S2 n Sn B2 = S3 n S4 n Sm B1 n B2 = S n S n S = B3 in B1 n B2

Definition of Bounded

A subset A is bounded if there is a number M s.t. d(a1,a2)<=M for any a1, a2 in A

Topological Property of X

Any property of X that is entirely expressed in terms of the topology of X (that is, in terms of the open sets of X) yields, via the correspondence f , the corresponding property for the space Y . Such a property of X is called a topological property of X.

order topology

Assume X is simply ordered, and has more than one element. Order topology is that generated by a basis of [xmin, b) -DNE if xmin DNE (a,xmax] - DNE if xmax DNE (a,b) - can be disjointed if topo space is disjoint

Lenma Basis for Subspace Topology

B n Y for every y in U n Y we can find y in B if y in U s.t. B in U. Thus y in B n Y in U n Y. Thus B n Y is a basis

e-centered Ball definition

Basis element for a topology induced by metric d -> Bd(x,e)={y | d(x,y) < e)

Boundary and Interior of A

Bd n Int(A) = empty Closure = Bd u Int if x is in int(A). x is contained in an open set of A contained in A. Thus there is an open set of x that does not intersect X-A so x is not in the closure of X-A so Int(A) n closure(X-A) = empty. Cl(A) is every x that has a neighborhood that intersects A. if x is not in A then x lies in X-A so every neighborhood of x intersects X-A and A so its in the boundary. If x is in A and not in the boundary and every neighborhood of x intersects A then x belongs to atleast one open set that does not intersect X-A so x is in the interior. Thus if x is in the closure it is in either the boundary or interior. Let's say x is in boundary then x needs to be in closure of A by definition. Let's say x is in the interior then every neighborhood of x intersects A so x is in the closure. Equality.

Closure and Limit Point theorem

Closure = Itself U Limit Points Assume x is in itself U limit points. Then every U of x must intersect A at some point by definition. Thus x must also be in closure. Assume X is in closure. If x is in A then it's trivial that x is in itself. If x is not in a then every neighborhood of of x must intersect A at some point other than itself. Thus x is in limit. So x in closure must be in Itself U limit points

Theorem - Inverse projection subbasis

Collection of pi1^-1(U) union collection of pi2^-1(V) form a subbasis for the product topology. every elemt of S is open in T thus T' is in T every basis element of T'can be generated by the finite intersection of two S elements thus T in T'

Theorem of Constructing continous functions from X to Y

Constant function Inclusion function: Subspace of A to X Composites of Continous Restricted domain: f: X - Y is cont then f: A - Y is cont if A is subspace of X Restricting or Exapnding range: Local formulation of continuity: X->Y is continous if X can be written as a union of open sets Ua s.t. f(Ua) is continous for all a.

Theorem finite point set Haussdorf

Every finite point set in a Hausdorff space X is closed Suffices to show that every one-point set is closed. Take {x0} and {x}. There exists an open neighborhood of x that doesn't intersect {x0} so there are no limit points of {x0} other {x0} itself so {x0} is equal to its closure and thus it is closed

Haussdorf Product and Order and Subspace Theorem

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. U x V n U1 x V1 = (U n U1) x (V n V1) = empty - possible to create Y n (U) n Y n (V) = Y n U n V =Y n (empty) = empty

Subspace Topology

Given arbitrary subset A in X, open sets are of the form A intersect V where V is an open set in X. A is a subspace

Ordered Square Topology

I x I dictionary order

Thm Product Topo and Subspace

If A is a subspace of X and B is a subspace of Y. Product topology A x B is same as what A x B inherits as subspace of X x Y Their basis is the same

What is the diameter of set

If A is bounded and nonempty, A has a diameter of sup{d(a1,a2)}

Lenma Open Subspace

If U is open in Y and Y is open in X then U is open in X U = V n Y V open in X. U is open finite intersection of open X

Hausdorff Convergence Theorem

If X is a Hausdorff space, then a sequence of points of X converges to at most one point of X. Every neighborhood of x needs to intersect xn for all n>N. For convergence Assume x0 and x1 are both convergence points of xn. Then there must be an open neighborhood of x0 that contains an infinite amount of xn after n>N that some open neighborhood of x1 cannot. So x1 is not convergence since t.e a neighborhood that does not fit the definition.

Convex set theorem

If Y is a convex set in X that is simply ordered. Order topology on Y is same as topology Y inherits as subspace

Hausdorff in Product And Box Theorem

If each space Xα is Hausdorff space, then Xα is a Hausdorff space in both the box and product topologies.

Definition of homeomorphism

If f is bijective and f and f^-1 are both continous then f is homeomorphic.

Lenma - collection of open sets C is a basis if...

If for every open set of X, U, and for every element x in U we can find an open set in C s.t. x in C in U Show C is a basis. First is solved. It must be true that C contains every element of X C1 n C2 is an open set so by definition, there must be C3 s.t x in C3 in C1 n C2.

What is the limit of a sequence

In haussdorf space, the single convergence of a sequence xn

Lenma - Topology generated by a basis

Is a topology consistent of all unions of basis elements - All basis elements are open -> all unions of basis elements are open -> uB in C(U) -Every element of U has an element of B that contains it and is a subset of U. Then U is just a union of all Bx. So C(U) in uB C(U) = u(B)

Theorem of Subspaces and Products/Boxes

Let Aα be a subspace of Xα, for each α ∈ J . Then PI Aα is a subspace of PI Xα if both products are given the box topology, or if both products are given the product topology.

Theorem Let f : A → α∈J PI Xα be given by the equation f (a) = ( fα(a))α∈J ,

Let PI Xα have the product topology. Then the function f is continuous if and only if each function fα is continuous. Suppose f is continious then fb = pi dot f which is composite of two continous functions Suppose f(b) is continous then. To show f is continous, we need to show that the inverse of f under every basis element is open. This is true if the inverse of f under ever subbasis element pi_B^(-1)(U)) is open. But f^-1(pi_B^-1) = f_b(-1). Which is continious, so this is open. So f is continous as intersection of open sets are open and unions of open sets are open

Theorem of a subset of a hausdorff space

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. Assume x is a limit point. Every neighborhood of x intersects A at some point other than itself. But then let's say that there is an open neighborhood of x that intersects A at finitely many points. Then A-{x} is also intersected at finitely many points. U n (A-{x})={x1,x2,...xn}. X-(U n (A-{x})) would then be an open set of X containing x. But then (U n X-(U n (A-{x}))) is an open neighborhood that doesn't intersect A - {x}. But then x is not a limit point of A so it's a contradiction

Closed Set Conditions Thm

Let X be a topological space o and X are closed arb int are closed finite un are closed

The sequence lenma with metrizability

Let X be a topological space; let A ⊂ X. If there is a sequence of points of A converging to x, then x ∈ A¯; the converse holds if X is metrizable if a sequence of point of A converge to x then every neighborhood of x contains x_n for n>=N and every neighborhood intersects A. So x is a limit point. Converse. X is metrizable Assume x in A- then a neighborhood of x of form Bd(x,1/n) intersects A at any point call this x_n. x_n converges to x. because for any neighborhood Bd(x,e), choose N s.t. 1/N < e then x_n all lie in Bd(x,e) for n>=N.

Expansion of closure in subspace thm

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.

Closure in a subspace Thm

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 .

Closure of a set within a subspace theorem

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 .

Closure of a set

S bar, contains all of the points of S and it's limit points. intersection of all closed sets containing s

Finer topologies from balls lenma

Let d and d' be two metrics on the set X; let T and T' be the topologies they induce, respectively. Then T' is finer than T if and only if for each x in X and each e > 0, there exists a δ > 0 such that Bd'(x,e) in Bd(x,δ)

Theorem for metric of product topology on R^w

Let d¯(a, b) = min{|a − b|, 1} be the standard bounded metric on R. If x and y are two points of Rω, define D(x,y)=sup{d^-(x_i,y_i)/i}}. Then D is a metric that induces the product topology on Rω. Properties of a metric are satisfied trivially except for triangle inequality. notice that d-(x,z)/i <= d-(x,y)/i + d-(y,z)/i <= D(x,y) + D(y,z). If all d-(x,z)

Theorem of converging functions

Let f : X → Y . If the function f is continuous, then for every convergent sequence xn → x in X, the sequence f (xn) converges to f (x). The converse holds if X is metrizable.. If f is continuous then let V be a neighborhood of f(x). U= f^(-1)(V) is an open set of X containing x and since we know xn converges to x, it follows that there exists N s.t. x_n for n>=N is contained in U. and f(x_n) in V for n>=N thus converges. Take the previous lenma as help for proving converse. Prove f(cl(A)) in cl(f(A)). Lets say x is in the closure of A. Then x has a converging sequence xn associated with it and f(x) has a converging sequence f(x_n) associated in it. thus if x is in closure of A then f(x) is in the closure f(A). so f(cl(A)) in cl(f(A)) as needed for continousnous .

Theorem 21.1 continuity under the view of metrics

Let f : X → Y ; let X and Y be metrizable with metrics dX and dY , respectively. Then continuity of f is equivalent to the requirement that given x ∈ X and given e> 0, there exists δ > 0 such that dx(x,y) -> dy(f(x),f(y))<e

Theorem of closure of products

Let {Xα} be an indexed family of spaces; let Aα ⊂ Xα for each α. If Pi Xα is given either the product or the box topology, then Pi cl(A) = cl(Pi(A)) Let x be a part of of Pi cl(A). then every x_a is in cl(A_a). Thus one can choose a basis element U = Pi(U_a) that contains x and where each U_a intersects must A_a at some point y_a. Then ther exists y=(y_a) is in the intersection of U of A. This is true for any U of x. So x is in cl(Pi A_a) Let x be a part of cl(PI A_a). We want to prove x_b is in cl(A_b) for all b and all x_b. Suppose V_b contains x_b. The inverse map of V_b is an open set. This open set is a cartesian product of X_A for all index except for b which is just V_b. This cartesian product naturally intersects A since it contains x and x is in the closure of A and thus V_b intersects A_b non emptily at some point . V_b is arbitrary so it's true for all V_b of b and all b so.

Coordinate Functions

Mapping of f1: A -> x or f2: A-> Y if f=f1,f2 and f: A -> X x Y

Metric Topology

Metric Topology generated by basis of all Bd(x,e) for all e>0 given a metric. An Open set is in the metric topology iff for every x in U, there is a Bd(x,e) in U. Clearly the union of B is X. Let's assume y is in ball 1, and in ball 2, then it is possible to draw smaller balls 1' and ball 2' in y st ball 1' in 1 and 2' in 2. The smaller of these two ' balls must then be in both ball 1 and ball 2. Thus there exists B3 in the intersection of B1 and B2.

Theorem of Basis on Product and Box

Product Basis is PI B_alpha where B_alpha is a basis element on X_alpha for finite amount of alpha and then is just X_alpha for all others Box Basis is PI B_alpha where B_alpha is a basis element on X_alpha.

projection mapping

Projection pi_b where pi_b((x_alpha) alpha in J)=X_b

Topological Space

Space X connected to Topology T

Pasting Lemma

Suppose f: X->Y is a map. Let X be the finite union of closed sets Fi. If the restriction of f to Fi is continuous for all i, then f is continuous. Also holds if Fi are open where f=f_i if x is in f_i domain let h: X -> Y. h^-1(V)=nf_i^(-1)(U) if each U is closed then each f_^i(-1)(U) is closed and thus h^-1(V) is closed.

Definition - Projection

Surjective Mapping of X x Y to X or Y pi^-1(U of X) = U x Y

Finer, Coarser

T is in T' - T' is finer/larger. Every open set of T is an open set of T'.

Lenma - When bases are comparable

T' is finer than T is equivalent to saying that for every x in X, and every basis element containing x in T, one can always find a basis element containing x in T' s.t. it's contained in the basis element of T. 1.(2-1) Every open set of T is a union of basis elements of T. But it is also an open set of basis elements of T'. Thus every open set in T is an open set in T'. 2. 1 -2. Every open set of T is an open set of T'. All of B are open sets of T'. Thus All of B have B' elements s.t. B' in B by definition.

What is the T1 Axiom

T1 axiom is the idea that a finite point is closed

Theoreom of Comparison of Product and Box Topology

The basis of the Box Topology is the cartesian product of finite or infinite open sets on each X_a. The basis for the product topology is the cartesian product of a finite number of open sets on each X_a and an infinite amount of X_a after some finite number of indices. Proof. 1 comes from definition of Box topology. 2. The basis from the subbasis for the product topology is formed of finite intersections of the sub basis. finite intersections of pi_b1^(-1)(U_1) will result in Cartesian products of open sets in X_a but since these intersections are finite, any indice not cited in this intersection is then just X_A in the preimage. There has to be an infinite amount of X_A for infinite products since subbasis only forms basis under finite intersections of its elements. Both are the same if the cartesian product is finite .

Bounded standard metric Theorem

The bounded standard metric is d=min(d(x,y),1) and induces the same topology as d. Triangle Inequality holds, D(x,y)=d(y,x), d(y,y) = 0. Notice All balls of form e<1 contains a ball of the standard metric also with e<1 and vice versa. Same topology induced

Definition of Cartesian Product

The cartesian product of a family of sets {A_alpha}_{alpha in J } is the set of of all Jtuples (x_alpha) elements of X such that x_alpha is in A_alpha for each alpha in J. The set of all functions x: J -> UA_alpha

Theorem on topologies induced by Rn due to square and euclidean metric

The topologies on Rn induced by the euclidean metric d and the square metric ρ are the same as the product topology on Rn.

Standard Topology

The topology on the real line generated by basis elements of open intervals on R of form (a,b) where a<b

K-Topology

The topology on the real line generated by basis elements of open intervals on R of form (a,b)-K where a<b and K ={1/n}

lower limit topology

The topology on the real line generated by basis elements of open intervals on R of form [a,b) where a<b

Uniform Metric and Uniform Metric topology definition

The uniform metric is the metric p-=max{d-(xa,ya)} for a indices in J. This induces the uniform metric topology R^j bounded by a metric of 1.

Theorem of Box, Product, and R^j topology

The uniform topology on RJ is finer than the product topology and coarser than the box topology; these three topologies are all different if J is infinite. Basis for product topology is Ua for finite a and Xa for infinite a. Xa cases are trivial as all balls would be within Xa. For every x in U. Each xa has a y_a it is farthest from and set epsilon_a such that each ball is still contained in each e_a. Then set our e as the min of all these e.s Then the ball in the p- metric lies within U. Thus we have found a ball for each arbitrary x and arbitrary U. The ball makes sure that all p(x,z) distances are less than some universal e. so that the max distance allowed is e. But take a series of basis elements of form (x-1/2 e, x + 1/2e). Every y in this basis element lies within the ball. So we've found a basis element within the ball for any ball and any x.

Comparable Topologies

Topologies where one is finer than the other

Definition of Box topology

Topology generated by a basis of P(U_alpha) where U_alpha is open in X_alpha for each alpha in J

Definition Product Topology

Topology generated by a basis of form U x V where bla bla

Definition of Product Topology

Topology generated by a subbasis of the union of collections of {pi_b^-1(Ub)} where Ub is open in X_b. Here, pi_\alpha X_a is the product space.

Finite Complement Topology

Topology on which ever complement of an open set is finite(Every closed set is finite). Proof: (X-nA)=U(X-A), X-uA=n(X-A) both are finite by definition. Union of finite sets are finite . Intersections of finite sets are finite. The intersection could generate the empty set. Both X and Empty are in the finite complement topology(MUST CHECK)

indiscrete/trivial topology

Topology only of X and Empty

Dictionary Order Topology on R x R

Topology with basis of (axb, cxd) or ax(b,d). Intervals of the second type also form a basis(checkable)

What must YOU check for something to be a topology

X and Empty are in it(DONT FORGET)

closed set

X-A is open

Closure in a subspace

Y is sub X A closed in Y if A in Y and A is a closed set in Y

A U B x C U D

contains (A x C) U ( B x D)

Theorem Let X and Y be topological spaces; let f : X → Y . Then the following are equivalent

f is continous f(cl(A)) subset cl(f(A)) For every closed B in Y f^-1(B) is closed in X For every neighborhood V of f(x) there is a neighborhood U of x s.t. f(U) subset V

Image rules with unions and intersections

f(A u B) = f(A) u f(B) f(A n B) subset f(A) n f(B). - only equal if f is injective

Condition for homeomorphism

f(U) is continous iff U is continous

Theorem of continious of Maps of Products

f: A -> X x Y is continous iff f1: A -> X, f2:A -> Y are both continous and f=f1,f2

Definition of continuous function

f: X->Y is continous if f^-1(V) = U where V is open in Y and U is open in X suffices to show inverse of every basis element is open suffices even more to show inverse of every subabsis elements are open

Inverse image rules with unions and intersections

f^-1 u = u f^-1 f^-1 n = n f^-1

Definition of a Jtuple of elements

function x: J->X. If alpha is an element of J, then x(alpha) = x_alpha or the alphath coordinate of x

Theorem Basis D for Product Topology

if B is a basis for X and C is a basis for Y B x C is a basis for X x Y

Topological Imbedding

if f : X-Y is injective and continous. Let Z be the image space of f(X) on Y. h: X-Z is bijective and continous. If h is homeomorphic then f is an imbedding of X in Y

Definition of norm, euclidean metric, square metric

norm -> |x| = sqrt(\sum_x_i^2 ) euclidean metric =sqrt(\sum_(x_i-y_i)^2 ) square metric = max[xi-y_i]

Interior set

the union of all open sets contained in A

Theorem of x belonging to a closure

x is in the closure of A iff every neighborhood of x intersects A x is in the closure of A iff every basis element of x intersects A x is not in the closure of A iff there is a neighborhood of that doesnt intersect A Assume x not in A-. Then X-A- is an open set containing x that does not intersect A as A is contained in A-. Assume there is a neighborhood of x U that doesn't intersect A. Then X-U is a closed set that contains A. thus X-U contains A-. Thus x cannot be in the closure of A.