Use the previous definition and theorem to solve Theorem 2.47 2.5. Basis for a Topology. Sometimes a topology on a set is too complicated to list allthe open sets. In that case, we use a smaller collection of open sets, called a basis to generatethe desired topology.Exercise 2.40. You have already seen this idea in a few previous exercises, examples, andproblems. Which ones?Definition 2.41. A set B is a basis for a topology on a set X provided B satisfies thefollowing 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 byT:= {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) Thesets 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 thatfor all U E T, for all x E U, there exists B e B such that x E B CU. Then B is a basisthat generates the topology T on X.We can do it even cheaper!Definition 2.44. A set S is a subbasis for a topology on a set X provided USThe topology T on X generated by S is defined to be the collection of all unions of finiteintersections 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) Thesets 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.

Answered: Theorem 2.47. Suppose B is a basis for…

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.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.7: Distinguishable Permutations And Combinations

Problem 30E

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Question

Use the previous definition and theorem to solve Theorem 2.47

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.
We can do it even cheaper!
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 $U$ be an open set in $Y$ .

$\Rightarrow$ From the definition of subspace topology there exits an open set $V$ of $X$ such that $U = V \cap Y$ .

Let $x \in U$ . This implies $x \in Y$ since, $U = V \cap Y$ (from the definition of intersection of sets).

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