Continue in this fashion, removing the middle third from all of the segments of set Ak, thus obtaining Ak+1. 00 (a) Let C ດີ . Ak. Show that C0. (Hint: Use the Finite Intersection Property, Lay, Theorem 3.5.7) k=1 =
Continue in this fashion, removing the middle third from all of the segments of set Ak, thus obtaining Ak+1. 00 (a) Let C ດີ . Ak. Show that C0. (Hint: Use the Finite Intersection Property, Lay, Theorem 3.5.7) k=1 =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Please see attached images...
![Let I =
[0, 1] be the unit interval. Construct a sequence of closed sets A₁ A₂ A3... in the following
way:
From I remove the middle third interval, leaving you with A₁
=
[0][¹]
U
From each of the segments of A₁, remove the middle third, leaving you with
27
A₂
[0]
0,
U
[¹].
2
3'9
Continue in this fashion, removing the middle third from all of the segments of set Ak, thus obtaining
Ak+1.
∞0
(a) Let C = Ak. Show that C 0. (Hint: Use the Finite Intersection Property, Lay, Theorem 3.5.7)
k=1
=
U](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e7c908-505f-4c35-8ac4-e541f2ab69b2%2F69b42d8a-0c31-40b2-96eb-6daf2aa106fc%2F6lfa8xt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Let I =
[0, 1] be the unit interval. Construct a sequence of closed sets A₁ A₂ A3... in the following
way:
From I remove the middle third interval, leaving you with A₁
=
[0][¹]
U
From each of the segments of A₁, remove the middle third, leaving you with
27
A₂
[0]
0,
U
[¹].
2
3'9
Continue in this fashion, removing the middle third from all of the segments of set Ak, thus obtaining
Ak+1.
∞0
(a) Let C = Ak. Show that C 0. (Hint: Use the Finite Intersection Property, Lay, Theorem 3.5.7)
k=1
=
U
Transcribed Image Text:3.5.7 THEOREM
Let = {Ka: ae} be a family of compact subsets of R. Suppose
that the intersection of any finite subfamily of Fis nonempty. Then
n{Ka: α € } # Ø.
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