Flashcards
§3 Creating New Topological Spaces
Cards in this deck (22)
- 1Definition 3.1: Subspace topologyLet X be any topological space, with topology T. Let Y ⊆ X.
Define \(T_Y\)={U∩Y: U open in X}
= {U∩Y: U ∈ T}
Then \(T_Y\) is a topology on Y, called the \(\underline{subspace-topology}\) on Y (inherited from X), and makes Y a \(\underline{subspace}\) of X.
(more precisely, Y,\(T_Y\) is a subspace of (X,T)) - 2Follow form Defn 3.1Thus: a set V⊆ Y is open in Y iff V=U ∩ Y for some U∈ X.
- 3Definition 3.2: Standard topologyLet Y be and subset of \(R^n_{std}\). The subspace topology on Y inherited from \(R^n_{std}\) is called the \(\underline{standard-topology-on-Y}\).
- 4Definition 3.3 Closed set in subspace top.If Y is a subspace of X, then a set C⊆Y is called \(\underline{closed-in-Y}\) if it is closed in the subspace topology on Y. This means that (Y-C) is open in Y.
i.e. (Y-C) = U∩Y for some U that is open in X. - 5Theorem 3.4C⊆Y is closed in Y iff C=D∩Y for some D that is closed in X.
- 6Corollary from Thm 3.4If C is closed in Y & Y is closed in X, then C is closed in X.
- 7Theorem: \(Cl_X\)(B) & \(Cl_Y\)(B)If (Y,\(T_Y\)) is a subspace of (X,T), then any B⊆Y, \(Cl_Y\)(B)= \(Cl_X\)(B)∩Y,
where \(Cl_Y\)(B) = closure of B in space Y
& \(Cl_X\)(B) = closure of B in space X. - 8Theorem 3.5If \(\underline{B}\)is a basis for space (X,T) & Y⊆X, then \(B_Y\) = {B∩Y : B ∈\(\underline{B}\) is a basis for the subspace topology \(T_Y\) on Y.
- 9Subspace relation is transitiveIf Y is a subspace of X & W is a subspace of Y, then W us a subspace of X.
i.e. the topology on W inherited from X = the topology on W inherited from Y. - 10Theorem 3.7: The product toplogy\(B_{XY}\) is a basis on X×Y for a topology on X×Y called the \(\underline{product-topology}\), making X×Y a \(\underline{product-space}\).
- 11Theorem 3.8If \(\underline{C}\) is a basis for X & \(\underline{D}\) is a basis for Y, then
\(\underline{E}\) = {C×D : C∈\(\underline{C}\) & D ∈ \(\underline{D}\)} is a basis for the product space X×Y. - 12Theorem 3.9Given A⊆X & B⊆Y, so A×B ⊆ X×Y, then the subspace top. on A×B, as a subspace of the product space X×Y is equal to the product of the subspace topology (A,\(T_A\)) on A inherited from X, with the subspace topology (B,\(T_B\)) on B inherited from Y.
- 13"Slicing up" X×YGiven a point x∈X, let {x}×Y = {(x,y) : y∈Y}
{x}×Y is a "copy of Y".
Have a bijection {x}×Y ⟷ Y
Then X×Y = \(\underset{x∈X}{∪}\){x}×Y
Similarly, X×Y = \(\underset{y∈Y}{∪}\)X×{y} - 14Theorem 3.10If A⊆X & B⊆Y, then \(Int_{X×Y}\)(A×B) = \(Int_X\)(A)×\(Int_Y\)(B).
Also have \(Cl_{X×Y}\)(A×B) = \(Cl_X\)(A)×\(Cl_Y\)(B). - 15Quotient TopologiesGiven a space X, a set Y & a function f:X⟶Y onto Y, can we use f to "transfer" the topology from X to Y
Motivation: f may identify/glue/collapse points of X. - 16Review: Equivalence relations + partitionsLet R be an \(\underline{equivalence-relation}\) on set X. Then R ⊆ X×X = {x,y) : x,y∈X}
may write xRy when (x,y)∈R. - 17Review: Equivalence classThe \(\underline{equivalence-class}\) of an x∈X is [x\(]_R\)= {y∈X : xRy}
- 18Review: Equivalence Facts(1) [x\(]_R\) = [y\(]_R\) iff xRy
(11) x ∈ [x\(]_R\) (as xRx)
(111) If [x\(]_R\) ≠ [y\(]_R\), then [x\(]_R\) ∩ [y\(]_R\) = ∅ - 19Review: PartitionA partition of X is a collection P of non-empty mutually disjoint (a.k.a pairwise disjoint) subsets of X where union is X.
[more in proofs page] - 20Definition: Quotient MapThe \(\underline{quotient-map}\) \(f_R\):X⟶X\R is defined by \(f_R\)(x) = [x\(]_R\) ∈ X\R.
\(f_R\) is surjective, i.e. onto X\R
i.e ∀ C ∈ X\R ∃ x ∈ X s.t. C = \(f_R\)(x) - 21Surjective Images as QuotientsLet f:X⟶Y be surjectice i.e. ∀ y∈Y ∃ x∈X s.t. y=f(x)
Recall: the f-image of X is f(X)={f(x):x∈X}
Define \(R_f\) ⊆ X×X by xRz iff f(x)=f(z) ∀ x,z∈X
The \(R_f\) is an equivalence relation on X & [x\(]_{R_f}\) = {z∈X : f(x) = f(z)}
As f is surjective then [x\(]_{R_f}\)= {z∈X : f(z) = y}
=f-1(y) where f-1(y) is called the point inverse. - 22Pre-images of SubsetsGiven f:X⟶Y and subset U of Y, we have f-1(U) = {x∈X : f(x)∈U} = {x : ∃ y∈U (y=f(x))}
which is the pre-image of U under f.
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- Diploma/Certificado de qualificação profissional de nivel 3 Mathematics (Topology (MATH313)) Flashcards on §3 Creating New Topological Spaces, created by Amy Wogan on 27/04/2016.
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