Mixing
I heard that the empty set $emptyset$ is bounded. But I think this statement is not correct.
$begingroup$
In Rudin's "Principles of Mathematical Analysis", there is the following definition of bounded.
Definition: Suppose $S$ is an ordered set, and $E subset S$. If there exists a $beta in S$ such that $x leq beta$ for every $x in E$, we say that $E$ is bounded above, and call $beta$ an upper bound of $E$. Lower bounds are defined in the same way(with $geq$ in place of $leq$).
Let $S = emptyset$.
Let $E = emptyset$.
Then, there exists no $beta in S$ such that $x leq beta$ for every $x in E$.
So, in this case, I think $E (subset S)$ is not bounded above.
Am I correct or not?
Isn't $emptyset$ an orderd set?
elementary-set-theory order-theory upper-lower-bounds
asked Jan 19 at 8:41
tchappy hatchappy ha
767412
$endgroup$
add a comment |
$begingroup$
In Rudin's "Principles of Mathematical Analysis", there is the following definition of bounded.
Definition: Suppose $S$ is an ordered set, and $E subset S$. If there exists a $beta in S$ such that $x leq beta$ for every $x in E$, we say that $E$ is bounded above, and call $beta$ an upper bound of $E$. Lower bounds are defined in the same way(with $geq$ in place of $leq$).
Let $S = emptyset$.
Let $E = emptyset$.
Then, there exists no $beta in S$ such that $x leq beta$ for every $x in E$.
So, in this case, I think $E (subset S)$ is not bounded above.
Am I correct or not?
Isn't $emptyset$ an orderd set?
elementary-set-theory order-theory upper-lower-bounds
asked Jan 19 at 8:41
tchappy hatchappy ha
767412
$endgroup$
$begingroup$
If $S$ is empty then indeed it has no subsets that are bounded above, But IMV that is no reason to say that $S$ is not an ordered set. There are also non-empty sets endowed with an order that have no subsets that are bounded above.
$endgroup$
– drhab
Jan 19 at 10:20
add a comment |
$begingroup$
In Rudin's "Principles of Mathematical Analysis", there is the following definition of bounded.
Definition: Suppose $S$ is an ordered set, and $E subset S$. If there exists a $beta in S$ such that $x leq beta$ for every $x in E$, we say that $E$ is bounded above, and call $beta$ an upper bound of $E$. Lower bounds are defined in the same way(with $geq$ in place of $leq$).
Let $S = emptyset$.
Let $E = emptyset$.
Then, there exists no $beta in S$ such that $x leq beta$ for every $x in E$.
So, in this case, I think $E (subset S)$ is not bounded above.
Am I correct or not?
Isn't $emptyset$ an orderd set?
elementary-set-theory order-theory upper-lower-bounds
asked Jan 19 at 8:41
tchappy hatchappy ha
767412
$endgroup$
In Rudin's "Principles of Mathematical Analysis", there is the following definition of bounded.
Definition: Suppose $S$ is an ordered set, and $E subset S$. If there exists a $beta in S$ such that $x leq beta$ for every $x in E$, we say that $E$ is bounded above, and call $beta$ an upper bound of $E$. Lower bounds are defined in the same way(with $geq$ in place of $leq$).
Let $S = emptyset$.
Let $E = emptyset$.
Then, there exists no $beta in S$ such that $x leq beta$ for every $x in E$.
So, in this case, I think $E (subset S)$ is not bounded above.
Am I correct or not?
Isn't $emptyset$ an orderd set?
elementary-set-theory order-theory upper-lower-bounds
elementary-set-theory order-theory upper-lower-bounds
asked Jan 19 at 8:41
tchappy hatchappy ha
767412
asked Jan 19 at 8:41
tchappy hatchappy ha
767412
edited Jan 19 at 8:48
tchappy ha
asked Jan 19 at 8:41
tchappy hatchappy ha
767412
asked Jan 19 at 8:41
tchappy hatchappy ha
767412
asked Jan 19 at 8:41
tchappy hatchappy ha
767412
767412
$begingroup$
If $S$ is empty then indeed it has no subsets that are bounded above, But IMV that is no reason to say that $S$ is not an ordered set. There are also non-empty sets endowed with an order that have no subsets that are bounded above.
$endgroup$
– drhab
Jan 19 at 10:20
add a comment |
$begingroup$
If $S$ is empty then indeed it has no subsets that are bounded above, But IMV that is no reason to say that $S$ is not an ordered set. There are also non-empty sets endowed with an order that have no subsets that are bounded above.
$endgroup$
– drhab
Jan 19 at 10:20
$begingroup$
If $S$ is empty then indeed it has no subsets that are bounded above, But IMV that is no reason to say that $S$ is not an ordered set. There are also non-empty sets endowed with an order that have no subsets that are bounded above.
$endgroup$
– drhab
Jan 19 at 10:20
$begingroup$
If $S$ is empty then indeed it has no subsets that are bounded above, But IMV that is no reason to say that $S$ is not an ordered set. There are also non-empty sets endowed with an order that have no subsets that are bounded above.
$endgroup$
– drhab
Jan 19 at 10:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Being bounded is relative to a given ordered set. If $S$ is empty, then indeed it is not bounded, which is just one more reason to agree that in general the empty set shouldn't be counted as a partial order. But if $S$ is non-empty, then the empty set is bounded, and every element is an upper and lower bound.
There is an analogy here with $(0,1)$. As a subset of itself, it is not bounded. There is no maximal element. But as a subset of $[0,1]$ or $Bbb R$ it is certainly bounded by $0$ and $1$.
edited Feb 15 at 9:05
José Carlos Santos
164k22131234
answered Jan 19 at 8:49
Asaf Karagila♦Asaf Karagila
305k33435766
$endgroup$
$begingroup$
Thank you very much, Asaf Karagila. Thank you for the nice example too.
$endgroup$
– tchappy ha
Jan 19 at 8:53
1
$begingroup$
There also good reasons to maintain the empty set as partial order. It serves e.g. as initial object of the category of posets and makes a forgetful functor from sets to posets possible.
$endgroup$
– drhab
Jan 19 at 9:57
$begingroup$
@drhab: Yes, or when you consider ordinals. But we are "adult", so we know how to distinguish these situations.
$endgroup$
– Asaf Karagila♦
Jan 19 at 12:35
$begingroup$
Adults do not argue about that.
$endgroup$
– drhab
Jan 19 at 16:58
add a comment |
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1 Answer
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$begingroup$
Being bounded is relative to a given ordered set. If $S$ is empty, then indeed it is not bounded, which is just one more reason to agree that in general the empty set shouldn't be counted as a partial order. But if $S$ is non-empty, then the empty set is bounded, and every element is an upper and lower bound.
There is an analogy here with $(0,1)$. As a subset of itself, it is not bounded. There is no maximal element. But as a subset of $[0,1]$ or $Bbb R$ it is certainly bounded by $0$ and $1$.
edited Feb 15 at 9:05
José Carlos Santos
164k22131234
answered Jan 19 at 8:49
Asaf Karagila♦Asaf Karagila
305k33435766
$endgroup$
$begingroup$
Thank you very much, Asaf Karagila. Thank you for the nice example too.
$endgroup$
– tchappy ha
Jan 19 at 8:53
1
$begingroup$
There also good reasons to maintain the empty set as partial order. It serves e.g. as initial object of the category of posets and makes a forgetful functor from sets to posets possible.
$endgroup$
– drhab
Jan 19 at 9:57
$begingroup$
@drhab: Yes, or when you consider ordinals. But we are "adult", so we know how to distinguish these situations.
$endgroup$
– Asaf Karagila♦
Jan 19 at 12:35
$begingroup$
Adults do not argue about that.
$endgroup$
– drhab
Jan 19 at 16:58
add a comment |
$begingroup$
Being bounded is relative to a given ordered set. If $S$ is empty, then indeed it is not bounded, which is just one more reason to agree that in general the empty set shouldn't be counted as a partial order. But if $S$ is non-empty, then the empty set is bounded, and every element is an upper and lower bound.
There is an analogy here with $(0,1)$. As a subset of itself, it is not bounded. There is no maximal element. But as a subset of $[0,1]$ or $Bbb R$ it is certainly bounded by $0$ and $1$.
edited Feb 15 at 9:05
José Carlos Santos
164k22131234
answered Jan 19 at 8:49
Asaf Karagila♦Asaf Karagila
305k33435766
$endgroup$
$begingroup$
Thank you very much, Asaf Karagila. Thank you for the nice example too.
$endgroup$
– tchappy ha
Jan 19 at 8:53
1
$begingroup$
There also good reasons to maintain the empty set as partial order. It serves e.g. as initial object of the category of posets and makes a forgetful functor from sets to posets possible.
$endgroup$
– drhab
Jan 19 at 9:57
$begingroup$
@drhab: Yes, or when you consider ordinals. But we are "adult", so we know how to distinguish these situations.
$endgroup$
– Asaf Karagila♦
Jan 19 at 12:35
$begingroup$
Adults do not argue about that.
$endgroup$
– drhab
Jan 19 at 16:58
add a comment |
$begingroup$
Being bounded is relative to a given ordered set. If $S$ is empty, then indeed it is not bounded, which is just one more reason to agree that in general the empty set shouldn't be counted as a partial order. But if $S$ is non-empty, then the empty set is bounded, and every element is an upper and lower bound.
There is an analogy here with $(0,1)$. As a subset of itself, it is not bounded. There is no maximal element. But as a subset of $[0,1]$ or $Bbb R$ it is certainly bounded by $0$ and $1$.
edited Feb 15 at 9:05
José Carlos Santos
164k22131234
answered Jan 19 at 8:49
Asaf Karagila♦Asaf Karagila
305k33435766
$endgroup$
Being bounded is relative to a given ordered set. If $S$ is empty, then indeed it is not bounded, which is just one more reason to agree that in general the empty set shouldn't be counted as a partial order. But if $S$ is non-empty, then the empty set is bounded, and every element is an upper and lower bound.
There is an analogy here with $(0,1)$. As a subset of itself, it is not bounded. There is no maximal element. But as a subset of $[0,1]$ or $Bbb R$ it is certainly bounded by $0$ and $1$.
edited Feb 15 at 9:05
José Carlos Santos
164k22131234
answered Jan 19 at 8:49
Asaf Karagila♦Asaf Karagila
305k33435766
edited Feb 15 at 9:05
José Carlos Santos
164k22131234
edited Feb 15 at 9:05
José Carlos Santos
164k22131234
edited Feb 15 at 9:05
José Carlos Santos
164k22131234
164k22131234
answered Jan 19 at 8:49
Asaf Karagila♦Asaf Karagila
305k33435766
answered Jan 19 at 8:49
Asaf Karagila♦Asaf Karagila
305k33435766
answered Jan 19 at 8:49
Asaf Karagila♦Asaf Karagila
305k33435766
305k33435766
$begingroup$
Thank you very much, Asaf Karagila. Thank you for the nice example too.
$endgroup$
– tchappy ha
Jan 19 at 8:53
1
$begingroup$
There also good reasons to maintain the empty set as partial order. It serves e.g. as initial object of the category of posets and makes a forgetful functor from sets to posets possible.
$endgroup$
– drhab
Jan 19 at 9:57
$begingroup$
@drhab: Yes, or when you consider ordinals. But we are "adult", so we know how to distinguish these situations.
$endgroup$
– Asaf Karagila♦
Jan 19 at 12:35
$begingroup$
Adults do not argue about that.
$endgroup$
– drhab
Jan 19 at 16:58
add a comment |
$begingroup$
Thank you very much, Asaf Karagila. Thank you for the nice example too.
$endgroup$
– tchappy ha
Jan 19 at 8:53
1
$begingroup$
There also good reasons to maintain the empty set as partial order. It serves e.g. as initial object of the category of posets and makes a forgetful functor from sets to posets possible.
$endgroup$
– drhab
Jan 19 at 9:57
$begingroup$
@drhab: Yes, or when you consider ordinals. But we are "adult", so we know how to distinguish these situations.
$endgroup$
– Asaf Karagila♦
Jan 19 at 12:35
$begingroup$
Adults do not argue about that.
$endgroup$
– drhab
Jan 19 at 16:58
$begingroup$
Thank you very much, Asaf Karagila. Thank you for the nice example too.
$endgroup$
– tchappy ha
Jan 19 at 8:53
$begingroup$
Thank you very much, Asaf Karagila. Thank you for the nice example too.
$endgroup$
– tchappy ha
Jan 19 at 8:53
1
1
$begingroup$
There also good reasons to maintain the empty set as partial order. It serves e.g. as initial object of the category of posets and makes a forgetful functor from sets to posets possible.
$endgroup$
– drhab
Jan 19 at 9:57
$begingroup$
There also good reasons to maintain the empty set as partial order. It serves e.g. as initial object of the category of posets and makes a forgetful functor from sets to posets possible.
$endgroup$
– drhab
Jan 19 at 9:57
$begingroup$
@drhab: Yes, or when you consider ordinals. But we are "adult", so we know how to distinguish these situations.
$endgroup$
– Asaf Karagila♦
Jan 19 at 12:35
$begingroup$
@drhab: Yes, or when you consider ordinals. But we are "adult", so we know how to distinguish these situations.
$endgroup$
– Asaf Karagila♦
Jan 19 at 12:35
$begingroup$
Adults do not argue about that.
$endgroup$
– drhab
Jan 19 at 16:58
$begingroup$
Adults do not argue about that.
$endgroup$
– drhab
Jan 19 at 16:58
add a comment |
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$begingroup$
If $S$ is empty then indeed it has no subsets that are bounded above, But IMV that is no reason to say that $S$ is not an ordered set. There are also non-empty sets endowed with an order that have no subsets that are bounded above.
$endgroup$
– drhab
Jan 19 at 10:20