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2014 08 01 Products and Subspaces

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    Subspaces and products 1 / 14

    Subspaces and products

    2014-08-01 09:00

    http://goforward/http://find/http://goback/
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    Subspaces and products j Subspace topology 2 / 14

    1 Subspace topology

    2 Product topology

    3 Exercises

    4 Links

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    Subspaces and products j Subspace topology 2 / 14

    Denition

    Denition (Subspace topology)

    Let (X; fi) be a topological space and Y X. Then

    fiY = fY \ U j U 2 fig

    is a topology on Y, called subspace topology.

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    Subspaces and products j Subspace topology 2 / 14

    Denition

    Denition (Subspace topology)

    Let (X; fi) be a topological space and Y X. Then

    fiY = fY \ U j U 2 fig

    is a topology on Y, called subspace topology.

    Verify that fiY is indeed a topology on Y.

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    Subspaces and products j Subspace topology 2 / 14

    Denition

    Denition (Subspace topology)

    Let (X; fi) be a topological space and Y X. Then

    fiY = fY \ U j U 2 fig

    is a topology on Y

    , called subspace topology

    .

    Verify that fiY is indeed a topology on Y.

    Open sets relative to Y

    If Y is a subspace of X, and U Y, it could be that U isopen in Y but not open in X.

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    Subspaces and products j Subspace topology 3 / 14

    Base of the subspace topology

    Lemma

    Let B be a base for the topology of X. Then

    BY = fB\ Y j B 2 Bg

    is a base for the subspace topology on Y.

    S b d d t j S b t l 3 / 14

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    Subspaces and products j Subspace topology 3 / 14

    Base of the subspace topology

    Lemma

    Let B be a base for the topology of X. Then

    BY = fB\ Y j B 2 Bg

    is a base for the subspace topology on Y.

    Proof.

    S b d d t j S b t l 4 / 14

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    Subspaces and products j Subspace topology 4 / 14

    Example

    Subspace of the real line

    Subspaces and products j Subspace topology 4 / 14

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    Subspaces and products j Subspace topology 4 / 14

    Example

    Subspace of the real line

    Consider Y = [0; 1] X = R, where X has the standardtopology.

    Subspaces and products j Subspace topology 4 / 14

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    Subspaces and products j Subspace topology 4 / 14

    Example

    Subspace of the real line

    Consider Y = [0; 1] X = R, where X has the standardtopology.

    A base for the subspace topology on Y is the collectionof all sets of the form (a; b) \ Y.

    Subspaces and products j Subspace topology 4 / 14

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    Subspaces and products j Subspace topology 4 / 14

    Example

    Subspace of the real line

    Consider Y = [0; 1] X = R, where X has the standardtopology.

    A base for the subspace topology on Y is the collectionof all sets of the form (a; b) \ Y.

    They all are of one of the following forms: (a; b), [0; b),(a; 1], Y, ;.

    Subspaces and products j Subspace topology 4 / 14

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    Subspaces and products j Subspace topology 4 / 14

    Example

    Subspace of the real line

    Consider Y = [0; 1] X = R, where X has the standardtopology.

    A base for the subspace topology on Y is the collectionof all sets of the form (a; b) \ Y.

    They all are of one of the following forms: (a; b), [0; b),(a; 1], Y, ;.

    These are precisely the basic elements of the ordertopology on [0; 1].

    Subspaces and products j Subspace topology 5 / 14

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    p e p o j p e opo ogy 5 /

    Example

    Subspace and order topology not always the same

    Subspaces and products j Subspace topology 5 / 14

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    j gy /

    Example

    Subspace and order topology not always the same

    Consider now Y = [0; 1) [ f2g X = R.

    Subspaces and products j Subspace topology 5 / 14

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    Example

    Subspace and order topology not always the same

    Consider now Y = [0; 1) [ f2g X = R.

    In the subspace topology, the set f2g is open.

    Subspaces and products j Subspace topology 5 / 14

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    Example

    Subspace and order topology not always the same

    Consider now Y = [0; 1) [ f2g X = R.

    In the subspace topology, the set f2g is open.

    But in the order topology on Y, considering that 2 is

    max Y, any open set that contains 2 has to contain

    other points.

    Subspaces and products j Subspace topology 6 / 14

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    Order topology in intervals

    Lemma

    If X is an ordered set that has the order topology, and

    Y X is an interval, then the order topology and the

    subspace topology on Y are the same.

    Subspaces and products j Subspace topology 6 / 14

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    Order topology in intervals

    Lemma

    If X is an ordered set that has the order topology, and

    Y X is an interval, then the order topology and the

    subspace topology on Y are the same.

    Proof.

    Exercise

    Subspaces and products j Product topology 7 / 14

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    A base for the product topology

    Theorem (Base of the product topology)

    Let X and Y be topological spaces. Let B be thecollection of all subsets of X Y of the form U V ,where U is open in X and V is open in Y . Then B is abase for a topology on X Y .

    Subspaces and products j Product topology 7 / 14

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    A base for the product topology

    Theorem (Base of the product topology)

    Let X and Y be topological spaces. Let B be thecollection of all subsets of X Y of the form U V ,where U is open in X and V is open in Y . Then B is abase for a topology on X Y .

    Proof.

    Subspaces and products j Product topology 7 / 14

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    A base for the product topology

    Theorem (Base of the product topology)

    Let X and Y be topological spaces. Let B be thecollection of all subsets of X Y of the form U V ,where U is open in X and V is open in Y . Then B is abase for a topology on X Y .

    Proof.

    Denition (Product topology)

    The topology just mentioned is called the product

    topology on the Cartesian product X Y.

    Subspaces and products j Product topology 8 / 14

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    Product by a base

    Theorem (Product topology given by a base)

    Let X be a topological space with base B, and Y be aspace with basis C. Then the set

    D = fU V j U 2 B; V 2 Cg

    is a base for the product topology on X Y .

    Subspaces and products j Product topology 8 / 14

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    Product by a base

    Theorem (Product topology given by a base)

    Let X be a topological space with base B, and Y be aspace with basis C. Then the set

    D = fU V j U 2 B; V 2 Cg

    is a base for the product topology on X Y .

    Proof.

    Subspaces and products j Product topology 9 / 14

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    Examples

    Example (Product topology on R2)

    A base for the product of two standard topologies on R is

    the collection of all rectangles of the form (a; b) (c; d).It follows that the product topology is the same as the

    metric topology on R.

    Subspaces and products j Product topology 9 / 14

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    Examples

    Example (Product topology on R2)

    A base for the product of two standard topologies on R is

    the collection of all rectangles of the form (a; b) (c; d).It follows that the product topology is the same as the

    metric topology on R.

    Example (Product of Sierpinski spaces)

    Describe the points and the open sets on the productX X, when X is a Sierpinski space.

    Subspaces and products j Product topology 10 / 14

    P j ti

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    Projections

    Denition (Projections)The maps 1: X Y ! X and 2: X Y ! Y are calledprojections of X Y.

    Subspaces and products j Product topology 10 / 14

    P j ti

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    Projections

    Denition (Projections)The maps 1: X Y ! X and 2: X Y ! Y are calledprojections of X Y.

    Inverse images of projections

    If U is open in X, then `11 (U) =

    Subspaces and products j Product topology 10 / 14

    P j ti

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    Projections

    Denition (Projections)The maps 1: X Y ! X and 2: X Y ! Y are calledprojections of X Y.

    Inverse images of projections

    If U is open in X, then `11 (U) =

    Subspaces and products j Product topology 10 / 14

    Projections

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    Projections

    Denition (Projections)The maps 1: X Y ! X and 2: X Y ! Y are calledprojections of X Y.

    Inverse images of projections

    If U is open in X, then `11 (U) = U Y, which is open in

    X Y. Similarly, if V is open in Y, the set

    `1

    2 (V

    ) = X V, is open in X Y.

    Subspaces and products j Product topology 10 / 14

    Projections

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    Projections

    Denition (Projections)The maps 1: X Y ! X and 2: X Y ! Y are calledprojections of X Y.

    Inverse images of projections

    If U is open in X, then `11 (U) = U Y, which is open in

    X Y. Similarly, if V is open in Y, the set

    `1

    2 (V

    ) = X V, is open in X Y.

    We have

    `11 (U) \

    `12 (V) = U V

    Subspaces and products j Product topology 11 / 14

    Subbase of the product

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    Subbase of the product

    Theorem (Subbase)

    The collection S = f 11 (U)gU2fiX [ f 12 (V)gV2fiY is asubbase for the product topology on X Y .

    Subspaces and products j Product topology 12 / 14

    Subspace and products

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    Subspace and products

    Theorem (Subspace and product topology)

    Let A; B be subspaces of X; Y respectively. Then the

    product topology on A B is the same as the topology on

    A B as subspace of the product space X Y .

    Subspaces and products j Product topology 12 / 14

    Subspace and products

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    Subspace and products

    Theorem (Subspace and product topology)

    Let A; B be subspaces of X; Y respectively. Then the

    product topology on A B is the same as the topology on

    A B as subspace of the product space X Y .

    Proof.

    Exercise

    Subspaces and products j Exercises 13 / 14

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    Show that if X is a topological space, and A Y X,then the topology of A as a subspace of Y is the same as

    a subspace of X (Y is a subspace of X).

    Subspaces and products j Exercises 13 / 14

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    Show that if X is a topological space, and A Y X,then the topology of A as a subspace of Y is the same as

    a subspace of X (Y is a subspace of X).

    A map f : X ! Y between topological space is said to beopen if f(U) is open for every U X which is an open set.Prove that the projection X: X Y ! X is an open set.

    Subspaces and products j Exercises 13 / 14

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    Show that if X is a topological space, and A Y X,then the topology of A as a subspace of Y is the same as

    a subspace of X (Y is a subspace of X).

    A map f : X ! Y between topological space is said to beopen if f(U) is open for every U X which is an open set.Prove that the projection X: X Y ! X is an open set.

    Let I denote the closed interval [0; 1] R as a subspaceof the usual topology on R. Compare the product

    topology on I I, the dictionary order topology on I I,

    and the product Id I, where Id denotes discrete topology.

    Subspaces and products j Links 14 / 14

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    Subspace topology - Wikipedia, the free encyclopedia

    Subspaces and products j Links 14 / 14

    http://en.wikipedia.org/wiki/Subspace_topologyhttp://en.wikipedia.org/wiki/Product_topologyhttp://en.wikipedia.org/wiki/Subspace_topology
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    Subspace topology - Wikipedia, the free encyclopedia

    Product topology - Wikipedia, the free encyclopedia

    http://en.wikipedia.org/wiki/Subspace_topologyhttp://en.wikipedia.org/wiki/Product_topologyhttp://en.wikipedia.org/wiki/Product_topologyhttp://en.wikipedia.org/wiki/Subspace_topology

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