Let $Asubseteq Bsubseteq mathbb{R}$. Suppose that $vert A vert = vert B vert$. [closed]
$begingroup$
Let $Asubseteq Bsubseteq mathbb{R}$. Suppose that $vert A vert = vert B vert$ (i.e., $A$ and $B$ have the same cardinality). It does not follow that
$A = B$.
1.) Give a counterexample to show why.
2.) What assumption on $B$ would force this to be true?
education
edited Jan 25 at 2:51
guchihe
1188
asked Jan 25 at 2:11
user638072user638072
61
$endgroup$
closed as off-topic by Shailesh, Leucippus, Trevor Gunn, metamorphy, Cesareo Jan 25 at 10:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Leucippus, Trevor Gunn, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $Asubseteq Bsubseteq mathbb{R}$. Suppose that $vert A vert = vert B vert$ (i.e., $A$ and $B$ have the same cardinality). It does not follow that
$A = B$.
1.) Give a counterexample to show why.
2.) What assumption on $B$ would force this to be true?
education
edited Jan 25 at 2:51
guchihe
1188
asked Jan 25 at 2:11
user638072user638072
61
$endgroup$
closed as off-topic by Shailesh, Leucippus, Trevor Gunn, metamorphy, Cesareo Jan 25 at 10:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Leucippus, Trevor Gunn, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
you should include your attempt when you ask a question.
$endgroup$
– Siong Thye Goh
Jan 25 at 2:17
add a comment |
$begingroup$
Let $Asubseteq Bsubseteq mathbb{R}$. Suppose that $vert A vert = vert B vert$ (i.e., $A$ and $B$ have the same cardinality). It does not follow that
$A = B$.
1.) Give a counterexample to show why.
2.) What assumption on $B$ would force this to be true?
education
edited Jan 25 at 2:51
guchihe
1188
asked Jan 25 at 2:11
user638072user638072
61
$endgroup$
Let $Asubseteq Bsubseteq mathbb{R}$. Suppose that $vert A vert = vert B vert$ (i.e., $A$ and $B$ have the same cardinality). It does not follow that
$A = B$.
1.) Give a counterexample to show why.
2.) What assumption on $B$ would force this to be true?
education
education
edited Jan 25 at 2:51
guchihe
1188
asked Jan 25 at 2:11
user638072user638072
61
edited Jan 25 at 2:51
guchihe
1188
asked Jan 25 at 2:11
user638072user638072
61
edited Jan 25 at 2:51
guchihe
1188
edited Jan 25 at 2:51
guchihe
1188
edited Jan 25 at 2:51
guchihe
1188
1188
asked Jan 25 at 2:11
user638072user638072
61
asked Jan 25 at 2:11
user638072user638072
61
asked Jan 25 at 2:11
user638072user638072
61
61
closed as off-topic by Shailesh, Leucippus, Trevor Gunn, metamorphy, Cesareo Jan 25 at 10:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Leucippus, Trevor Gunn, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shailesh, Leucippus, Trevor Gunn, metamorphy, Cesareo Jan 25 at 10:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Leucippus, Trevor Gunn, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
you should include your attempt when you ask a question.
$endgroup$
– Siong Thye Goh
Jan 25 at 2:17
add a comment |
1
$begingroup$
you should include your attempt when you ask a question.
$endgroup$
– Siong Thye Goh
Jan 25 at 2:17
1
1
$begingroup$
you should include your attempt when you ask a question.
$endgroup$
– Siong Thye Goh
Jan 25 at 2:17
$begingroup$
you should include your attempt when you ask a question.
$endgroup$
– Siong Thye Goh
Jan 25 at 2:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
1) B=$mathbb R$, A=(0,1)
2) B is finite
edited Jan 25 at 5:50
J. W. Tanner
3,0201320
answered Jan 25 at 2:17
Jagol95Jagol95
1637
$endgroup$
add a comment |
$begingroup$
For the former, you can set $A= mathbb{N}$ and $B = mathbb{Z}$, with the bijection $f:A to B$ such that $f(2k)= k$ and $f(2k+1) = -(k+1)$.
For the second, if there is a bijection $f:Ato B$ and $B$ is finite, you can prove that $A=B$ viewing $Bsubseteq A$.
answered Jan 25 at 2:48
guchiheguchihe
1188
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
1) B=$mathbb R$, A=(0,1)
2) B is finite
edited Jan 25 at 5:50
J. W. Tanner
3,0201320
answered Jan 25 at 2:17
Jagol95Jagol95
1637
$endgroup$
add a comment |
$begingroup$
1) B=$mathbb R$, A=(0,1)
2) B is finite
edited Jan 25 at 5:50
J. W. Tanner
3,0201320
answered Jan 25 at 2:17
Jagol95Jagol95
1637
$endgroup$
add a comment |
$begingroup$
1) B=$mathbb R$, A=(0,1)
2) B is finite
edited Jan 25 at 5:50
J. W. Tanner
3,0201320
answered Jan 25 at 2:17
Jagol95Jagol95
1637
$endgroup$
1) B=$mathbb R$, A=(0,1)
2) B is finite
edited Jan 25 at 5:50
J. W. Tanner
3,0201320
answered Jan 25 at 2:17
Jagol95Jagol95
1637
edited Jan 25 at 5:50
J. W. Tanner
3,0201320
edited Jan 25 at 5:50
J. W. Tanner
3,0201320
edited Jan 25 at 5:50
J. W. Tanner
3,0201320
3,0201320
answered Jan 25 at 2:17
Jagol95Jagol95
1637
answered Jan 25 at 2:17
Jagol95Jagol95
1637
answered Jan 25 at 2:17
Jagol95Jagol95
1637
1637
add a comment |
add a comment |
$begingroup$
For the former, you can set $A= mathbb{N}$ and $B = mathbb{Z}$, with the bijection $f:A to B$ such that $f(2k)= k$ and $f(2k+1) = -(k+1)$.
For the second, if there is a bijection $f:Ato B$ and $B$ is finite, you can prove that $A=B$ viewing $Bsubseteq A$.
answered Jan 25 at 2:48
guchiheguchihe
1188
$endgroup$
add a comment |
$begingroup$
For the former, you can set $A= mathbb{N}$ and $B = mathbb{Z}$, with the bijection $f:A to B$ such that $f(2k)= k$ and $f(2k+1) = -(k+1)$.
For the second, if there is a bijection $f:Ato B$ and $B$ is finite, you can prove that $A=B$ viewing $Bsubseteq A$.
answered Jan 25 at 2:48
guchiheguchihe
1188
$endgroup$
add a comment |
$begingroup$
For the former, you can set $A= mathbb{N}$ and $B = mathbb{Z}$, with the bijection $f:A to B$ such that $f(2k)= k$ and $f(2k+1) = -(k+1)$.
For the second, if there is a bijection $f:Ato B$ and $B$ is finite, you can prove that $A=B$ viewing $Bsubseteq A$.
answered Jan 25 at 2:48
guchiheguchihe
1188
$endgroup$
For the former, you can set $A= mathbb{N}$ and $B = mathbb{Z}$, with the bijection $f:A to B$ such that $f(2k)= k$ and $f(2k+1) = -(k+1)$.
For the second, if there is a bijection $f:Ato B$ and $B$ is finite, you can prove that $A=B$ viewing $Bsubseteq A$.
answered Jan 25 at 2:48
guchiheguchihe
1188
answered Jan 25 at 2:48
guchiheguchihe
1188
answered Jan 25 at 2:48
guchiheguchihe
1188
answered Jan 25 at 2:48
guchiheguchihe
1188
1188
add a comment |
add a comment |
1
$begingroup$
you should include your attempt when you ask a question.
$endgroup$
– Siong Thye Goh
Jan 25 at 2:17