Let $Asubseteq Bsubseteq mathbb{R}$. Suppose that $vert A vert = vert B vert$. [closed]

-1

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?

edited Jan 25 at 2:51

guchihe

1188

asked Jan 25 at 2:11

user638072

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

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?

edited Jan 25 at 2:51

guchihe

1188

asked Jan 25 at 2:11

user638072

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

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?

edited Jan 25 at 2:51

guchihe

1188

asked Jan 25 at 2:11

user638072

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

user638072

edited Jan 25 at 2:51

guchihe

1188

asked Jan 25 at 2:11

user638072

edited Jan 25 at 2:51

guchihe

1188

edited Jan 25 at 2:51

guchihe

1188

edited Jan 25 at 2:51

guchihe

1188

asked Jan 25 at 2:11

user638072

asked Jan 25 at 2:11

user638072

asked Jan 25 at 2:11

user638072

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

you should include your attempt when you ask a question.

– Siong Thye Goh
Jan 25 at 2:17

add a comment |

2 Answers
2

active

oldest

votes

1) B=$mathbb R$, A=(0,1)

2) B is finite

edited Jan 25 at 5:50

J. W. Tanner

3,0121320

answered Jan 25 at 2:17

Jagol95

1637

add a comment |

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

guchihe

1188

add a comment |

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

1) B=$mathbb R$, A=(0,1)

2) B is finite

edited Jan 25 at 5:50

J. W. Tanner

3,0121320

answered Jan 25 at 2:17

Jagol95

1637

add a comment |

1) B=$mathbb R$, A=(0,1)

2) B is finite

edited Jan 25 at 5:50

J. W. Tanner

3,0121320

answered Jan 25 at 2:17

Jagol95

1637

add a comment |

1) B=$mathbb R$, A=(0,1)

2) B is finite

edited Jan 25 at 5:50

J. W. Tanner

3,0121320

answered Jan 25 at 2:17

Jagol95

1637

1) B=$mathbb R$, A=(0,1)

2) B is finite

edited Jan 25 at 5:50

J. W. Tanner

3,0121320

answered Jan 25 at 2:17

Jagol95

1637

edited Jan 25 at 5:50

J. W. Tanner

3,0121320

edited Jan 25 at 5:50

J. W. Tanner

3,0121320

edited Jan 25 at 5:50

J. W. Tanner

3,0121320

answered Jan 25 at 2:17

Jagol95

1637

answered Jan 25 at 2:17

Jagol95

1637

answered Jan 25 at 2:17

Jagol95

1637

add a comment |

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

guchihe

1188

add a comment |

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

guchihe

1188

add a comment |

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

guchihe

1188

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

guchihe

1188

answered Jan 25 at 2:48

guchihe

1188

answered Jan 25 at 2:48

guchihe

1188

answered Jan 25 at 2:48

guchihe

1188

add a comment |

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