Theorem 2.47. Suppose B is a basis for a topology on X and Y C X. Then By := {BnY|| Be B} is a basis for the subspace topology on Y.
Theorem 2.47. Suppose B is a basis for a topology on X and Y C X. Then By := {BnY|| Be B} is a basis for the subspace topology on Y.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Question
Use the previous definition and theorem to solve Theorem 2.47
Transcribed Image Text:2.5. Basis for a Topology. Sometimes a topology on a set is too complicated to list all
the open sets. In that case, we use a smaller collection of open sets, called a basis to generate
the desired topology.
Exercise 2.40. You have already seen this idea in a few previous exercises, examples, and
problems. Which ones?
Definition 2.41. A set B is a basis for a topology on a set X provided B satisfies the
following axioms:
(1) UB = X.
(2) VB1, B2 E B, Vx € B1 N B2, 3B3 € B, x € B3 C B1 N B2.
The topology T on X generated by B is defined by
T:= {U C X | 3C C B,U = UC}.
In other words, U e T iff U is a union of elements of B.
Exercise 2.42. Show that: (1) The topology generated by a basis is a topology. (2) The
sets in the basis are open sets in that topology.
Proposition 2.43. Let X be a space with topology T and Ba subcollection of T such that
for all U E T, for all x E U, there exists B e B such that x E B CU. Then B is a basis
that generates the topology T on X.
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Definition 2.44. A set S is a subbasis for a topology on a set X provided US
The topology T on X generated by S is defined to be the collection of all unions of finite
intersections of elements of S.
Х.
Problem 2.45. Can you define the basis generated by a subbasis?
Exercise 2.46. Show that: (1) The topology generated by a subbasis is a topology. (2) The
sets in the subbasis are open sets in that topology.
{BnY |
Theorem 2.47. Suppose B is a basis for a topology on X and Y C X. Then By :=
BE B} is a basis for the subspace topology on Y.
Expert Solution
Step 1
Let be an open set in .
From the definition of subspace topology there exits an open set of such that .
Let . This implies since, (from the definition of intersection of sets).
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