6. Let X be the space of real numbers with the usual topology, and let Q be the subspace of rational numbers. Let A =.{re Q: 0 < r < 1}, B = {re Q: 0 < r ≤ √2}, C = {re Q:-√2 ≤r ≤ √2). For each of the sets A, B, and C, tell whether in the subspace topology it is open, closed, or neither.
6. Let X be the space of real numbers with the usual topology, and let Q be the subspace of rational numbers. Let A =.{re Q: 0 < r < 1}, B = {re Q: 0 < r ≤ √2}, C = {re Q:-√2 ≤r ≤ √2). For each of the sets A, B, and C, tell whether in the subspace topology it is open, closed, or neither.
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
Topology: Please follow structured example in photo to solve Q26
Transcribed Image Text:26. Let X be the space of real numbers with the usual topology, and let Q
be the subspace of rational numbers. Let A = {r EQ: 0 < r < 1},
B = {re Q:0 < r ≤ √2}, C = {re Q:-√2 ≤ r ≤ √2}. For each
of the sets A, B, and C, tell whether in the subspace topology it is open,
closed, or neither.
![s.bbcollab.com/collab/ui/session/playback
ex Let (X,T) = (IR, usual top.)
Let A = Q = set of rationals,
Then (A, TA)=(Q, To) is a
Subspace. Let
S={PEQ: 0≤r≤1}
1/45+
I
Is S
open
in (A, TA)?
Is S closed in (A,TA)?
fact in (Rusual), C(x5)= (x-r, x+r)
fact
PES every base el. cont p
meets S
claim cl₂S = [0, 1]
S
117
Ę
P& ²1 (5)
(p<0)
C(pi-p) = (p-(-p), p+(-p)) = (2p, 0)
neet S=>
does not
→
PACI S.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9787c00-cfcc-43da-9235-b2564f9dd12e%2Fd4c994e2-ca7d-4117-b638-b9bafcbf1031%2Ftrkf9x_processed.jpeg&w=3840&q=75)
Transcribed Image Text:s.bbcollab.com/collab/ui/session/playback
ex Let (X,T) = (IR, usual top.)
Let A = Q = set of rationals,
Then (A, TA)=(Q, To) is a
Subspace. Let
S={PEQ: 0≤r≤1}
1/45+
I
Is S
open
in (A, TA)?
Is S closed in (A,TA)?
fact in (Rusual), C(x5)= (x-r, x+r)
fact
PES every base el. cont p
meets S
claim cl₂S = [0, 1]
S
117
Ę
P& ²1 (5)
(p<0)
C(pi-p) = (p-(-p), p+(-p)) = (2p, 0)
neet S=>
does not
→
PACI S.
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