and 4. Let X be the space of reals with the cofinite topology (Example 2.1(d)), let A be the positive integers and B = {1,2}. Find the derived set, the closure, the interior, and the boundary of each of the sets A and B. 1-8
and 4. Let X be the space of reals with the cofinite topology (Example 2.1(d)), let A be the positive integers and B = {1,2}. Find the derived set, the closure, the interior, and the boundary of each of the sets A and B. 1-8
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Topology
Q4. Please find example as guide from 2nd photo
![(c) Let X be the space of real
let A = (0, 1]. Then every subset of X is both ope
A = cl(A) = int(A), and Fr(A) = 0.
(d) Let X be the complex plane with the usual topology, and let S =
{2:|2|<1}, and B = {z=(x,0): x ≥ 1}. Then
cl(S) = {2: |z| ≤ 1}UB, int (S) = A, and Fr(S) = {2:|2|=1}UB.
AUB, where A
(e) Let X be the space of real numbers with the usual topology, and
let A be the rational numbers. Then cl(A) = X, int(A) = 0), and
Fr(A) = X.
Definition
A subset A of a topological space X is said to be dense if cl(A) = X.
Thus the set of rational numbers is a dense subset of the space of real
mbers with the usual topology, and in a space with the trivial topology,
ery nonempty subset is dense.
ercises
1. Let X be the space of real numbers with the usual topology, and let N be
the integers. Find the derived set, the closure, the interior, and the boundary
of N.
2. Let X be the set of reals, and let T = {SCX:0€ X-S} U{X}. Show
that T is a topology for X and find the closure of the interval A = (1,2)
and of the interval B = (-1,1).
=
3. Let X be the set of positive integers. For each n E X, let Sn
{k € X:
k2n}. Show that T='{Sn: ne X} U {0} is a topology for X, and find
the closure of the set of even integers. Find the closure of the singleton set
A = {100}.
4. Let X be the space of reals with the cofinite topology (Example 2.1(d)), and
let A be the positive integers and B = {1,2}. Find the derived set, the
closure, the interior, and the boundary of each of the sets A and B.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2Fc8183e73-2717-4f33-81eb-d2fcf843da34%2Fw6wpm7n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(c) Let X be the space of real
let A = (0, 1]. Then every subset of X is both ope
A = cl(A) = int(A), and Fr(A) = 0.
(d) Let X be the complex plane with the usual topology, and let S =
{2:|2|<1}, and B = {z=(x,0): x ≥ 1}. Then
cl(S) = {2: |z| ≤ 1}UB, int (S) = A, and Fr(S) = {2:|2|=1}UB.
AUB, where A
(e) Let X be the space of real numbers with the usual topology, and
let A be the rational numbers. Then cl(A) = X, int(A) = 0), and
Fr(A) = X.
Definition
A subset A of a topological space X is said to be dense if cl(A) = X.
Thus the set of rational numbers is a dense subset of the space of real
mbers with the usual topology, and in a space with the trivial topology,
ery nonempty subset is dense.
ercises
1. Let X be the space of real numbers with the usual topology, and let N be
the integers. Find the derived set, the closure, the interior, and the boundary
of N.
2. Let X be the set of reals, and let T = {SCX:0€ X-S} U{X}. Show
that T is a topology for X and find the closure of the interval A = (1,2)
and of the interval B = (-1,1).
=
3. Let X be the set of positive integers. For each n E X, let Sn
{k € X:
k2n}. Show that T='{Sn: ne X} U {0} is a topology for X, and find
the closure of the set of even integers. Find the closure of the singleton set
A = {100}.
4. Let X be the space of reals with the cofinite topology (Example 2.1(d)), and
let A be the positive integers and B = {1,2}. Find the derived set, the
closure, the interior, and the boundary of each of the sets A and B.
![m/collab/ui/session/playback
BEBB 8
xX-R, T={USX: X-U is finite] {}
=co-finite top.
if SCR is finite, then S = p
of Casel Assume S=ØSR.
Bb
ex S: [1,2] 5' = 6
ex S= [0,00)
CIS IR
Suppose pes. Notice X=IRET
and X Contains p.
P.
So X meets S in a pt other than
So X meets & in a pt. other than Py
a contradiction. So there's no
PES!
Case 2 Assume Sp. So S= {x₁,..., XN). Suppose ots²
Le+ U=X-(S-{p3). Then X-U=S-{P3 which is finite.
Then UET. Also pell.
реи.
So U meets S in apt. other than p a contradiction,
U Contains no pt. of 5 (except possibly
since
at p).
O
kay
(ETS](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2Fc8183e73-2717-4f33-81eb-d2fcf843da34%2Fgxt1kyk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:m/collab/ui/session/playback
BEBB 8
xX-R, T={USX: X-U is finite] {}
=co-finite top.
if SCR is finite, then S = p
of Casel Assume S=ØSR.
Bb
ex S: [1,2] 5' = 6
ex S= [0,00)
CIS IR
Suppose pes. Notice X=IRET
and X Contains p.
P.
So X meets S in a pt other than
So X meets & in a pt. other than Py
a contradiction. So there's no
PES!
Case 2 Assume Sp. So S= {x₁,..., XN). Suppose ots²
Le+ U=X-(S-{p3). Then X-U=S-{P3 which is finite.
Then UET. Also pell.
реи.
So U meets S in apt. other than p a contradiction,
U Contains no pt. of 5 (except possibly
since
at p).
O
kay
(ETS
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