Supremum and infimum of a set
up vote
2
down vote
favorite
I wanted to check my reasoning for the following question:
Determine minimum, maximum, supremum and infimum of the set:
$$B=left{ -frac{1}{n} in mathbb{Q}: n in mathbb{N}_+ right}$$
Whenever $n=1$ we have that $-frac{1}{1}=-1$, this is the minimum element of the set, since for all $n$ we know that $ngeq 1$ and thus $frac{1}{n} leq 1$, this gives $- frac{1}{n} geq -1 $. As the value of $n$ gets bigger (we will later define limits), we notice that $-frac{1}{n}$ will approach zero, but zero is not contained in the set. We realise that $sup(B)=0$ and summarise:
$$min(B)=inf(B)=-1$$
$$sup(B)=0 $$ $max(B) $ does not exist.
proof-verification elementary-set-theory supremum-and-infimum
edited 1 hour ago
Chinnapparaj R
4,3551725
asked 1 hour ago
WesleyGroupshaveFeelingsToo
743217
add a comment |
up vote
2
down vote
favorite
I wanted to check my reasoning for the following question:
Determine minimum, maximum, supremum and infimum of the set:
$$B=left{ -frac{1}{n} in mathbb{Q}: n in mathbb{N}_+ right}$$
Whenever $n=1$ we have that $-frac{1}{1}=-1$, this is the minimum element of the set, since for all $n$ we know that $ngeq 1$ and thus $frac{1}{n} leq 1$, this gives $- frac{1}{n} geq -1 $. As the value of $n$ gets bigger (we will later define limits), we notice that $-frac{1}{n}$ will approach zero, but zero is not contained in the set. We realise that $sup(B)=0$ and summarise:
$$min(B)=inf(B)=-1$$
$$sup(B)=0 $$ $max(B) $ does not exist.
proof-verification elementary-set-theory supremum-and-infimum
edited 1 hour ago
Chinnapparaj R
4,3551725
asked 1 hour ago
WesleyGroupshaveFeelingsToo
743217
4
Your answers are correct.
– Kavi Rama Murthy
1 hour ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I wanted to check my reasoning for the following question:
Determine minimum, maximum, supremum and infimum of the set:
$$B=left{ -frac{1}{n} in mathbb{Q}: n in mathbb{N}_+ right}$$
Whenever $n=1$ we have that $-frac{1}{1}=-1$, this is the minimum element of the set, since for all $n$ we know that $ngeq 1$ and thus $frac{1}{n} leq 1$, this gives $- frac{1}{n} geq -1 $. As the value of $n$ gets bigger (we will later define limits), we notice that $-frac{1}{n}$ will approach zero, but zero is not contained in the set. We realise that $sup(B)=0$ and summarise:
$$min(B)=inf(B)=-1$$
$$sup(B)=0 $$ $max(B) $ does not exist.
proof-verification elementary-set-theory supremum-and-infimum
edited 1 hour ago
Chinnapparaj R
4,3551725
asked 1 hour ago
WesleyGroupshaveFeelingsToo
743217
I wanted to check my reasoning for the following question:
Determine minimum, maximum, supremum and infimum of the set:
$$B=left{ -frac{1}{n} in mathbb{Q}: n in mathbb{N}_+ right}$$
Whenever $n=1$ we have that $-frac{1}{1}=-1$, this is the minimum element of the set, since for all $n$ we know that $ngeq 1$ and thus $frac{1}{n} leq 1$, this gives $- frac{1}{n} geq -1 $. As the value of $n$ gets bigger (we will later define limits), we notice that $-frac{1}{n}$ will approach zero, but zero is not contained in the set. We realise that $sup(B)=0$ and summarise:
$$min(B)=inf(B)=-1$$
$$sup(B)=0 $$ $max(B) $ does not exist.
proof-verification elementary-set-theory supremum-and-infimum
proof-verification elementary-set-theory supremum-and-infimum
edited 1 hour ago
Chinnapparaj R
4,3551725
asked 1 hour ago
WesleyGroupshaveFeelingsToo
743217
edited 1 hour ago
Chinnapparaj R
4,3551725
asked 1 hour ago
WesleyGroupshaveFeelingsToo
743217
edited 1 hour ago
Chinnapparaj R
4,3551725
edited 1 hour ago
Chinnapparaj R
4,3551725
edited 1 hour ago
Chinnapparaj R
4,3551725
4,3551725
asked 1 hour ago
WesleyGroupshaveFeelingsToo
743217
asked 1 hour ago
WesleyGroupshaveFeelingsToo
743217
asked 1 hour ago
WesleyGroupshaveFeelingsToo
743217
743217
4
Your answers are correct.
– Kavi Rama Murthy
1 hour ago
add a comment |
4
Your answers are correct.
– Kavi Rama Murthy
1 hour ago
4
4
Your answers are correct.
– Kavi Rama Murthy
1 hour ago
Your answers are correct.
– Kavi Rama Murthy
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You are correct!
Indeed, if you feel uncomforatble to work with negative signs, just compute $inf A$ and $sup A$ where $A={1/n:n in Bbb N}$ and then use $$sup(-A)=-inf A\inf(-A)=-sup A$$
answered 1 hour ago
Chinnapparaj R
4,3551725
1
Good one! This is actually a later exercise.
– WesleyGroupshaveFeelingsToo
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You are correct!
Indeed, if you feel uncomforatble to work with negative signs, just compute $inf A$ and $sup A$ where $A={1/n:n in Bbb N}$ and then use $$sup(-A)=-inf A\inf(-A)=-sup A$$
answered 1 hour ago
Chinnapparaj R
4,3551725
1
Good one! This is actually a later exercise.
– WesleyGroupshaveFeelingsToo
1 hour ago
add a comment |
up vote
1
down vote
accepted
You are correct!
Indeed, if you feel uncomforatble to work with negative signs, just compute $inf A$ and $sup A$ where $A={1/n:n in Bbb N}$ and then use $$sup(-A)=-inf A\inf(-A)=-sup A$$
answered 1 hour ago
Chinnapparaj R
4,3551725
1
Good one! This is actually a later exercise.
– WesleyGroupshaveFeelingsToo
1 hour ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are correct!
Indeed, if you feel uncomforatble to work with negative signs, just compute $inf A$ and $sup A$ where $A={1/n:n in Bbb N}$ and then use $$sup(-A)=-inf A\inf(-A)=-sup A$$
answered 1 hour ago
Chinnapparaj R
4,3551725
You are correct!
Indeed, if you feel uncomforatble to work with negative signs, just compute $inf A$ and $sup A$ where $A={1/n:n in Bbb N}$ and then use $$sup(-A)=-inf A\inf(-A)=-sup A$$
answered 1 hour ago
Chinnapparaj R
4,3551725
answered 1 hour ago
Chinnapparaj R
4,3551725
answered 1 hour ago
Chinnapparaj R
4,3551725
answered 1 hour ago
Chinnapparaj R
4,3551725
4,3551725
1
Good one! This is actually a later exercise.
– WesleyGroupshaveFeelingsToo
1 hour ago
add a comment |
1
Good one! This is actually a later exercise.
– WesleyGroupshaveFeelingsToo
1 hour ago
1
1
Good one! This is actually a later exercise.
– WesleyGroupshaveFeelingsToo
1 hour ago
Good one! This is actually a later exercise.
– WesleyGroupshaveFeelingsToo
1 hour ago
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997927%2fsupremum-and-infimum-of-a-set%23new-answer', 'question_page');
}
);
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
4
Your answers are correct.
– Kavi Rama Murthy
1 hour ago