$U cap V $ is choose the correct option [closed]

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Under the usual topology in $mathbb{R}^3$ ,if $U ={(x,y,z) in mathbb{R}^3: x^2 +y^2 < 1} $and $V ={(x,y,z) in mathbb{R}^3 : z=0 }$ then , $U cap V $ is

choose the correct option

$a)$ both open and closed

$b)$ neither open nor closed

$c)$ open but not closed

$d)$ closed but not open

My attempt : i was trying to visualize the diagram,,,that inside circle V will contain, that is it will form a singleton set which is closed in $mathbb{R}$

so option d) will correct

Is it true ??

Any hints/solution will be appreciated

thanks u

asked Nov 24 at 21:51

jasmine

1,388416

closed as off-topic by amWhy, Andrés E. Caicedo, Leucippus, Shailesh, Rebellos Nov 25 at 0:27

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Andrés E. Caicedo, Leucippus, Shailesh, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.

2

Is it a homework?
– Michał Zapała
Nov 24 at 21:53

1

Seems like the intersection is ${(x, y, 0)mid x^2+y^2 < 1} subset mathbb R^3$.
– amWhy
Nov 24 at 21:55

1

$V$ doesn't restrict $x, y$. Only $z= 0$. $U$ doesn't restrict $z$; only requires that $x^2 + y^2 lt 1$.
– amWhy
Nov 24 at 21:58

add a comment |

up vote
-2
down vote

favorite

Under the usual topology in $mathbb{R}^3$ ,if $U ={(x,y,z) in mathbb{R}^3: x^2 +y^2 < 1} $and $V ={(x,y,z) in mathbb{R}^3 : z=0 }$ then , $U cap V $ is

choose the correct option

$a)$ both open and closed

$b)$ neither open nor closed

$c)$ open but not closed

$d)$ closed but not open

My attempt : i was trying to visualize the diagram,,,that inside circle V will contain, that is it will form a singleton set which is closed in $mathbb{R}$

so option d) will correct

Is it true ??

Any hints/solution will be appreciated

thanks u

asked Nov 24 at 21:51

jasmine

1,388416

closed as off-topic by amWhy, Andrés E. Caicedo, Leucippus, Shailesh, Rebellos Nov 25 at 0:27

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Andrés E. Caicedo, Leucippus, Shailesh, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.

2

Is it a homework?
– Michał Zapała
Nov 24 at 21:53

1

Seems like the intersection is ${(x, y, 0)mid x^2+y^2 < 1} subset mathbb R^3$.
– amWhy
Nov 24 at 21:55

1

$V$ doesn't restrict $x, y$. Only $z= 0$. $U$ doesn't restrict $z$; only requires that $x^2 + y^2 lt 1$.
– amWhy
Nov 24 at 21:58

add a comment |

up vote
-2
down vote

favorite

Under the usual topology in $mathbb{R}^3$ ,if $U ={(x,y,z) in mathbb{R}^3: x^2 +y^2 < 1} $and $V ={(x,y,z) in mathbb{R}^3 : z=0 }$ then , $U cap V $ is

choose the correct option

$a)$ both open and closed

$b)$ neither open nor closed

$c)$ open but not closed

$d)$ closed but not open

My attempt : i was trying to visualize the diagram,,,that inside circle V will contain, that is it will form a singleton set which is closed in $mathbb{R}$

so option d) will correct

Is it true ??

Any hints/solution will be appreciated

thanks u

asked Nov 24 at 21:51

jasmine

1,388416

Under the usual topology in $mathbb{R}^3$ ,if $U ={(x,y,z) in mathbb{R}^3: x^2 +y^2 < 1} $and $V ={(x,y,z) in mathbb{R}^3 : z=0 }$ then , $U cap V $ is

choose the correct option

$a)$ both open and closed

$b)$ neither open nor closed

$c)$ open but not closed

$d)$ closed but not open

My attempt : i was trying to visualize the diagram,,,that inside circle V will contain, that is it will form a singleton set which is closed in $mathbb{R}$

so option d) will correct

Is it true ??

Any hints/solution will be appreciated

thanks u

general-topology

asked Nov 24 at 21:51

jasmine

1,388416

asked Nov 24 at 21:51

jasmine

1,388416

asked Nov 24 at 21:51

jasmine

1,388416

asked Nov 24 at 21:51

jasmine

1,388416

asked Nov 24 at 21:51

jasmine

1,388416

closed as off-topic by amWhy, Andrés E. Caicedo, Leucippus, Shailesh, Rebellos Nov 25 at 0:27

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Andrés E. Caicedo, Leucippus, Shailesh, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Andrés E. Caicedo, Leucippus, Shailesh, Rebellos Nov 25 at 0:27

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Andrés E. Caicedo, Leucippus, Shailesh, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.

2

Is it a homework?
– Michał Zapała
Nov 24 at 21:53

1

Seems like the intersection is ${(x, y, 0)mid x^2+y^2 < 1} subset mathbb R^3$.
– amWhy
Nov 24 at 21:55

1

$V$ doesn't restrict $x, y$. Only $z= 0$. $U$ doesn't restrict $z$; only requires that $x^2 + y^2 lt 1$.
– amWhy
Nov 24 at 21:58

add a comment |

2

Is it a homework?
– Michał Zapała
Nov 24 at 21:53

1

Seems like the intersection is ${(x, y, 0)mid x^2+y^2 < 1} subset mathbb R^3$.
– amWhy
Nov 24 at 21:55

1

$V$ doesn't restrict $x, y$. Only $z= 0$. $U$ doesn't restrict $z$; only requires that $x^2 + y^2 lt 1$.
– amWhy
Nov 24 at 21:58

Is it a homework?
– Michał Zapała
Nov 24 at 21:53

Seems like the intersection is ${(x, y, 0)mid x^2+y^2 < 1} subset mathbb R^3$.
– amWhy
Nov 24 at 21:55

$V$ doesn't restrict $x, y$. Only $z= 0$. $U$ doesn't restrict $z$; only requires that $x^2 + y^2 lt 1$.
– amWhy
Nov 24 at 21:58

add a comment |

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

The intersection is ${(x,y,0): x^2 + y^2 < 1}$ which is neither open nor closed.

Not open as all balls stick out in 3 dimensions: any ball around $(x,y,0)$ has points with $z neq 0$. Not closed as $(1,0,0)$ is not in the set but in its closure, e.g.

answered Nov 24 at 22:02

Henno Brandsma

102k345109

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

The intersection is ${(x,y,0): x^2 + y^2 < 1}$ which is neither open nor closed.

Not open as all balls stick out in 3 dimensions: any ball around $(x,y,0)$ has points with $z neq 0$. Not closed as $(1,0,0)$ is not in the set but in its closure, e.g.

answered Nov 24 at 22:02

Henno Brandsma

102k345109

add a comment |

up vote
1
down vote

accepted

The intersection is ${(x,y,0): x^2 + y^2 < 1}$ which is neither open nor closed.

Not open as all balls stick out in 3 dimensions: any ball around $(x,y,0)$ has points with $z neq 0$. Not closed as $(1,0,0)$ is not in the set but in its closure, e.g.

answered Nov 24 at 22:02

Henno Brandsma

102k345109

add a comment |

up vote
1
down vote

accepted

The intersection is ${(x,y,0): x^2 + y^2 < 1}$ which is neither open nor closed.

Not open as all balls stick out in 3 dimensions: any ball around $(x,y,0)$ has points with $z neq 0$. Not closed as $(1,0,0)$ is not in the set but in its closure, e.g.

answered Nov 24 at 22:02

Henno Brandsma

102k345109

The intersection is ${(x,y,0): x^2 + y^2 < 1}$ which is neither open nor closed.

Not open as all balls stick out in 3 dimensions: any ball around $(x,y,0)$ has points with $z neq 0$. Not closed as $(1,0,0)$ is not in the set but in its closure, e.g.

answered Nov 24 at 22:02

Henno Brandsma

102k345109

answered Nov 24 at 22:02

Henno Brandsma

102k345109

answered Nov 24 at 22:02

Henno Brandsma

102k345109

answered Nov 24 at 22:02

Henno Brandsma

102k345109

add a comment |

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