How do I write $A # n$ in set builder notation? [closed]
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Let $A subset mathbb{R}$ be a set of real numbers, and let $ninmathbb{N}$ be a natural number. Define $A#n$ to be the set of all subsets of $A$ such that every pair of numbers has an absolute difference of at most $n$.
elementary-set-theory
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closed as off-topic by Andrés E. Caicedo, darij grinberg, Gregory J. Puleo, José Carlos Santos, max_zorn Feb 3 at 17:09
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$begingroup$
Let $A subset mathbb{R}$ be a set of real numbers, and let $ninmathbb{N}$ be a natural number. Define $A#n$ to be the set of all subsets of $A$ such that every pair of numbers has an absolute difference of at most $n$.
elementary-set-theory
$endgroup$
closed as off-topic by Andrés E. Caicedo, darij grinberg, Gregory J. Puleo, José Carlos Santos, max_zorn Feb 3 at 17:09
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." – Andrés E. Caicedo, darij grinberg, Gregory J. Puleo, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $A subset mathbb{R}$ be a set of real numbers, and let $ninmathbb{N}$ be a natural number. Define $A#n$ to be the set of all subsets of $A$ such that every pair of numbers has an absolute difference of at most $n$.
elementary-set-theory
$endgroup$
Let $A subset mathbb{R}$ be a set of real numbers, and let $ninmathbb{N}$ be a natural number. Define $A#n$ to be the set of all subsets of $A$ such that every pair of numbers has an absolute difference of at most $n$.
elementary-set-theory
elementary-set-theory
edited Feb 3 at 5:38


Tiago Emilio Siller
7581419
7581419
asked Feb 3 at 3:17


rohit jainrohit jain
43
43
closed as off-topic by Andrés E. Caicedo, darij grinberg, Gregory J. Puleo, José Carlos Santos, max_zorn Feb 3 at 17:09
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." – Andrés E. Caicedo, darij grinberg, Gregory J. Puleo, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Andrés E. Caicedo, darij grinberg, Gregory J. Puleo, José Carlos Santos, max_zorn Feb 3 at 17:09
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." – Andrés E. Caicedo, darij grinberg, Gregory J. Puleo, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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${ X : X subseteq A, forall x,y in X, |x - y| le n }$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
${ X : X subseteq A, forall x,y in X, |x - y| le n }$
$endgroup$
add a comment |
$begingroup$
${ X : X subseteq A, forall x,y in X, |x - y| le n }$
$endgroup$
add a comment |
$begingroup$
${ X : X subseteq A, forall x,y in X, |x - y| le n }$
$endgroup$
${ X : X subseteq A, forall x,y in X, |x - y| le n }$
edited Feb 3 at 5:51
bof
52.6k559121
52.6k559121
answered Feb 3 at 3:59
William ElliotWilliam Elliot
9,1812820
9,1812820
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