Mixing
Approve that if there is a lower bound for each $A subseteq R $ then there is $infA$
$begingroup$
I must approve that if there is a lower bound for each $ Asubseteq R $ then there is $infA$. ($A neq emptyset$)
I'm sure that there is a related post for this but I would like to validate my solution to better understand the topic of $sup$ and $inf$.
My solution:
Let us define a set A which is not empty set and there is a lower bound. Let us also define a set B such that
$B = { x in R : x$ is lower bound of A $}$
$B$ is not empty because there is a $m leqslant x $ for all $xin A$.
Is $infA =m $?
Two cases:
Let us consider that $ m neq infA$ then $exists e > 0$ such that
$forall x in A$, $m+e leqslant x$.
So, $m+e in B$ and $m+e>m$.
We wonder if there is $supB$. If there is $supB$ then $supB = infA$
Let us assume that there isn't $supB$. Then there is $e > 0$ for
each $x>0$ so that
$x leqslant M - e$ where $ x leqslant M $ for each $x in B$
Let us define $M = m + e $, then $x leqslant m$, which is
impossible because
$m+e >m$. Consequently, $ x > M -e $.
As a result, $supB=M=infA$
If $infA=m$ then there is $infA$
From (1) and (2), there is always $infA$
Could you validate my solution ?
EDIT:
- Can we assume that $supB = infA$ ?
- Can we select the $M = m + e$ ?
- Introducing the above two cases, can we assume that there is always $infA$ ?
calculus
asked Jan 27 at 10:28
Dimitris DimitriadisDimitris Dimitriadis
548
$endgroup$
add a comment |
$begingroup$
I must approve that if there is a lower bound for each $ Asubseteq R $ then there is $infA$. ($A neq emptyset$)
I'm sure that there is a related post for this but I would like to validate my solution to better understand the topic of $sup$ and $inf$.
My solution:
Let us define a set A which is not empty set and there is a lower bound. Let us also define a set B such that
$B = { x in R : x$ is lower bound of A $}$
$B$ is not empty because there is a $m leqslant x $ for all $xin A$.
Is $infA =m $?
Two cases:
Let us consider that $ m neq infA$ then $exists e > 0$ such that
$forall x in A$, $m+e leqslant x$.
So, $m+e in B$ and $m+e>m$.
We wonder if there is $supB$. If there is $supB$ then $supB = infA$
Let us assume that there isn't $supB$. Then there is $e > 0$ for
each $x>0$ so that
$x leqslant M - e$ where $ x leqslant M $ for each $x in B$
Let us define $M = m + e $, then $x leqslant m$, which is
impossible because
$m+e >m$. Consequently, $ x > M -e $.
As a result, $supB=M=infA$
If $infA=m$ then there is $infA$
From (1) and (2), there is always $infA$
Could you validate my solution ?
EDIT:
- Can we assume that $supB = infA$ ?
- Can we select the $M = m + e$ ?
- Introducing the above two cases, can we assume that there is always $infA$ ?
calculus
asked Jan 27 at 10:28
Dimitris DimitriadisDimitris Dimitriadis
548
$endgroup$
$begingroup$
Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
$endgroup$
– lzralbu
Jan 27 at 14:45
$begingroup$
Thanks, you are right. I will edit it.
$endgroup$
– Dimitris Dimitriadis
Jan 27 at 15:41
add a comment |
$begingroup$
I must approve that if there is a lower bound for each $ Asubseteq R $ then there is $infA$. ($A neq emptyset$)
I'm sure that there is a related post for this but I would like to validate my solution to better understand the topic of $sup$ and $inf$.
My solution:
Let us define a set A which is not empty set and there is a lower bound. Let us also define a set B such that
$B = { x in R : x$ is lower bound of A $}$
$B$ is not empty because there is a $m leqslant x $ for all $xin A$.
Is $infA =m $?
Two cases:
Let us consider that $ m neq infA$ then $exists e > 0$ such that
$forall x in A$, $m+e leqslant x$.
So, $m+e in B$ and $m+e>m$.
We wonder if there is $supB$. If there is $supB$ then $supB = infA$
Let us assume that there isn't $supB$. Then there is $e > 0$ for
each $x>0$ so that
$x leqslant M - e$ where $ x leqslant M $ for each $x in B$
Let us define $M = m + e $, then $x leqslant m$, which is
impossible because
$m+e >m$. Consequently, $ x > M -e $.
As a result, $supB=M=infA$
If $infA=m$ then there is $infA$
From (1) and (2), there is always $infA$
Could you validate my solution ?
EDIT:
- Can we assume that $supB = infA$ ?
- Can we select the $M = m + e$ ?
- Introducing the above two cases, can we assume that there is always $infA$ ?
calculus
asked Jan 27 at 10:28
Dimitris DimitriadisDimitris Dimitriadis
548
$endgroup$
I must approve that if there is a lower bound for each $ Asubseteq R $ then there is $infA$. ($A neq emptyset$)
I'm sure that there is a related post for this but I would like to validate my solution to better understand the topic of $sup$ and $inf$.
My solution:
Let us define a set A which is not empty set and there is a lower bound. Let us also define a set B such that
$B = { x in R : x$ is lower bound of A $}$
$B$ is not empty because there is a $m leqslant x $ for all $xin A$.
Is $infA =m $?
Two cases:
Let us consider that $ m neq infA$ then $exists e > 0$ such that
$forall x in A$, $m+e leqslant x$.
So, $m+e in B$ and $m+e>m$.
We wonder if there is $supB$. If there is $supB$ then $supB = infA$
Let us assume that there isn't $supB$. Then there is $e > 0$ for
each $x>0$ so that
$x leqslant M - e$ where $ x leqslant M $ for each $x in B$
Let us define $M = m + e $, then $x leqslant m$, which is
impossible because
$m+e >m$. Consequently, $ x > M -e $.
As a result, $supB=M=infA$
If $infA=m$ then there is $infA$
From (1) and (2), there is always $infA$
Could you validate my solution ?
EDIT:
- Can we assume that $supB = infA$ ?
- Can we select the $M = m + e$ ?
- Introducing the above two cases, can we assume that there is always $infA$ ?
calculus
calculus
asked Jan 27 at 10:28
Dimitris DimitriadisDimitris Dimitriadis
548
asked Jan 27 at 10:28
Dimitris DimitriadisDimitris Dimitriadis
548
edited Jan 27 at 15:42
Dimitris Dimitriadis
asked Jan 27 at 10:28
Dimitris DimitriadisDimitris Dimitriadis
548
asked Jan 27 at 10:28
Dimitris DimitriadisDimitris Dimitriadis
548
asked Jan 27 at 10:28
Dimitris DimitriadisDimitris Dimitriadis
548
548
$begingroup$
Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
$endgroup$
– lzralbu
Jan 27 at 14:45
$begingroup$
Thanks, you are right. I will edit it.
$endgroup$
– Dimitris Dimitriadis
Jan 27 at 15:41
add a comment |
$begingroup$
Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
$endgroup$
– lzralbu
Jan 27 at 14:45
$begingroup$
Thanks, you are right. I will edit it.
$endgroup$
– Dimitris Dimitriadis
Jan 27 at 15:41
$begingroup$
Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
$endgroup$
– lzralbu
Jan 27 at 14:45
$begingroup$
Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
$endgroup$
– lzralbu
Jan 27 at 14:45
$begingroup$
Thanks, you are right. I will edit it.
$endgroup$
– Dimitris Dimitriadis
Jan 27 at 15:41
$begingroup$
Thanks, you are right. I will edit it.
$endgroup$
– Dimitris Dimitriadis
Jan 27 at 15:41
add a comment |
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$begingroup$
Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
$endgroup$
– lzralbu
Jan 27 at 14:45
$begingroup$
Thanks, you are right. I will edit it.
$endgroup$
– Dimitris Dimitriadis
Jan 27 at 15:41