$sup(Asetminus{z}) = sup(A)$ other direction and generalisation.
$begingroup$
Let A be a nonempty set, such that $zin A$ and $sup(A) not in A$
(a) $sup(A setminus {z}) = sup(A)$
(b) generalise this to $sup(A setminus {z_1 z_2 z_3 dots z_n})=sup(A)$
Let us denote $A$ with $z$ removed by $B$
For the first part I know that removing a point from $A$ can only really lower the supremum, because we might remove the right "end point", so the only possibility for $sup(B)$ is that $sup(B) leq sup(A)$, we can also see this via this formal argument:
Let $y in A $, but $y neq z$, then we have that $y leq sup (B)$, but certainly we have that $yin B$ so now this is an upper bound for $B$, since $sup(B)$ is the lowest of bounds for $B$, we have:
$$ sup(B) leq sup(A)$$
I fail so see the other direction, the problem is if we start out with an arbitrary element of $B$ is that we cannot say anything about $z$, at least I don't see this.
For (b) we simply solve this by induction on $n$, here is a rough sketch, I think this approach is fine, but correct me if I am wrong:
Our base case will hold by question (a), our inductive hypothesis generalises this for arbitrary $k in mathbb{N} $ to $A setminus {z_1,z_2,dots,z_k}$ and then for our inductive step we first consider $A setminus {z_1,z_2,dots,z_k }=C $ this is useful notation to make things shorter.
We can now write that: $sup(Csetminus{z_{k+1}})=sup(C)$ by question $(a)$,
We now use our inductive hypothesis ($sup(C)=sup(A)$) and state that: $$sup(Asetminus{z_1,z_2,dots,z_k,z_{k+1}})= sup(Csetminus{z_{k+1}})=sup(C)=sup(A)$$
By the principle of mathematical induction the desired relation holds. $square$
real-analysis supremum-and-infimum
edited Dec 1 '18 at 13:54
user10354138
7,4322925
asked Dec 1 '18 at 13:47
Wesley StrikWesley Strik
2,017423
$endgroup$
add a comment |
$begingroup$
Let A be a nonempty set, such that $zin A$ and $sup(A) not in A$
(a) $sup(A setminus {z}) = sup(A)$
(b) generalise this to $sup(A setminus {z_1 z_2 z_3 dots z_n})=sup(A)$
Let us denote $A$ with $z$ removed by $B$
For the first part I know that removing a point from $A$ can only really lower the supremum, because we might remove the right "end point", so the only possibility for $sup(B)$ is that $sup(B) leq sup(A)$, we can also see this via this formal argument:
Let $y in A $, but $y neq z$, then we have that $y leq sup (B)$, but certainly we have that $yin B$ so now this is an upper bound for $B$, since $sup(B)$ is the lowest of bounds for $B$, we have:
$$ sup(B) leq sup(A)$$
I fail so see the other direction, the problem is if we start out with an arbitrary element of $B$ is that we cannot say anything about $z$, at least I don't see this.
For (b) we simply solve this by induction on $n$, here is a rough sketch, I think this approach is fine, but correct me if I am wrong:
Our base case will hold by question (a), our inductive hypothesis generalises this for arbitrary $k in mathbb{N} $ to $A setminus {z_1,z_2,dots,z_k}$ and then for our inductive step we first consider $A setminus {z_1,z_2,dots,z_k }=C $ this is useful notation to make things shorter.
We can now write that: $sup(Csetminus{z_{k+1}})=sup(C)$ by question $(a)$,
We now use our inductive hypothesis ($sup(C)=sup(A)$) and state that: $$sup(Asetminus{z_1,z_2,dots,z_k,z_{k+1}})= sup(Csetminus{z_{k+1}})=sup(C)=sup(A)$$
By the principle of mathematical induction the desired relation holds. $square$
real-analysis supremum-and-infimum
edited Dec 1 '18 at 13:54
user10354138
7,4322925
asked Dec 1 '18 at 13:47
Wesley StrikWesley Strik
2,017423
$endgroup$
add a comment |
$begingroup$
Let A be a nonempty set, such that $zin A$ and $sup(A) not in A$
(a) $sup(A setminus {z}) = sup(A)$
(b) generalise this to $sup(A setminus {z_1 z_2 z_3 dots z_n})=sup(A)$
Let us denote $A$ with $z$ removed by $B$
For the first part I know that removing a point from $A$ can only really lower the supremum, because we might remove the right "end point", so the only possibility for $sup(B)$ is that $sup(B) leq sup(A)$, we can also see this via this formal argument:
Let $y in A $, but $y neq z$, then we have that $y leq sup (B)$, but certainly we have that $yin B$ so now this is an upper bound for $B$, since $sup(B)$ is the lowest of bounds for $B$, we have:
$$ sup(B) leq sup(A)$$
I fail so see the other direction, the problem is if we start out with an arbitrary element of $B$ is that we cannot say anything about $z$, at least I don't see this.
For (b) we simply solve this by induction on $n$, here is a rough sketch, I think this approach is fine, but correct me if I am wrong:
Our base case will hold by question (a), our inductive hypothesis generalises this for arbitrary $k in mathbb{N} $ to $A setminus {z_1,z_2,dots,z_k}$ and then for our inductive step we first consider $A setminus {z_1,z_2,dots,z_k }=C $ this is useful notation to make things shorter.
We can now write that: $sup(Csetminus{z_{k+1}})=sup(C)$ by question $(a)$,
We now use our inductive hypothesis ($sup(C)=sup(A)$) and state that: $$sup(Asetminus{z_1,z_2,dots,z_k,z_{k+1}})= sup(Csetminus{z_{k+1}})=sup(C)=sup(A)$$
By the principle of mathematical induction the desired relation holds. $square$
real-analysis supremum-and-infimum
edited Dec 1 '18 at 13:54
user10354138
7,4322925
asked Dec 1 '18 at 13:47
Wesley StrikWesley Strik
2,017423
$endgroup$
Let A be a nonempty set, such that $zin A$ and $sup(A) not in A$
(a) $sup(A setminus {z}) = sup(A)$
(b) generalise this to $sup(A setminus {z_1 z_2 z_3 dots z_n})=sup(A)$
Let us denote $A$ with $z$ removed by $B$
For the first part I know that removing a point from $A$ can only really lower the supremum, because we might remove the right "end point", so the only possibility for $sup(B)$ is that $sup(B) leq sup(A)$, we can also see this via this formal argument:
Let $y in A $, but $y neq z$, then we have that $y leq sup (B)$, but certainly we have that $yin B$ so now this is an upper bound for $B$, since $sup(B)$ is the lowest of bounds for $B$, we have:
$$ sup(B) leq sup(A)$$
I fail so see the other direction, the problem is if we start out with an arbitrary element of $B$ is that we cannot say anything about $z$, at least I don't see this.
For (b) we simply solve this by induction on $n$, here is a rough sketch, I think this approach is fine, but correct me if I am wrong:
Our base case will hold by question (a), our inductive hypothesis generalises this for arbitrary $k in mathbb{N} $ to $A setminus {z_1,z_2,dots,z_k}$ and then for our inductive step we first consider $A setminus {z_1,z_2,dots,z_k }=C $ this is useful notation to make things shorter.
We can now write that: $sup(Csetminus{z_{k+1}})=sup(C)$ by question $(a)$,
We now use our inductive hypothesis ($sup(C)=sup(A)$) and state that: $$sup(Asetminus{z_1,z_2,dots,z_k,z_{k+1}})= sup(Csetminus{z_{k+1}})=sup(C)=sup(A)$$
By the principle of mathematical induction the desired relation holds. $square$
real-analysis supremum-and-infimum
real-analysis supremum-and-infimum
edited Dec 1 '18 at 13:54
user10354138
7,4322925
asked Dec 1 '18 at 13:47
Wesley StrikWesley Strik
2,017423
edited Dec 1 '18 at 13:54
user10354138
7,4322925
asked Dec 1 '18 at 13:47
Wesley StrikWesley Strik
2,017423
edited Dec 1 '18 at 13:54
user10354138
7,4322925
edited Dec 1 '18 at 13:54
user10354138
7,4322925
edited Dec 1 '18 at 13:54
user10354138
7,4322925
7,4322925
asked Dec 1 '18 at 13:47
Wesley StrikWesley Strik
2,017423
asked Dec 1 '18 at 13:47
Wesley StrikWesley Strik
2,017423
asked Dec 1 '18 at 13:47
Wesley StrikWesley Strik
2,017423
2,017423
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $sup(B)<sup(A)$. Then, take $xin A$ such that $sup(B)<x<sup(A)$ (We can do it by definition of $sup$ and because $sup(A)notin A$). Then, we have that $xnotin B$ hence $x$ must be $z$. But it is impossible because it would imply that $z=sup(A)$. Then, we have that $sup(A)=sup(B)$.
answered Dec 1 '18 at 14:07
GödelGödel
1,418319
$endgroup$
$begingroup$
clever, just suppose it does not and we get a contradiction. I often forget contradiction as an approach - thanks for reminding me.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
I feel with every bit of steering in the right direction I'm becoming better at analysis, thank you :)
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
@WesleyGroupshaveFeelingsToo You are welcome.
$endgroup$
– Gödel
Dec 1 '18 at 14:30
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Suppose that $sup(B)<sup(A)$. Then, take $xin A$ such that $sup(B)<x<sup(A)$ (We can do it by definition of $sup$ and because $sup(A)notin A$). Then, we have that $xnotin B$ hence $x$ must be $z$. But it is impossible because it would imply that $z=sup(A)$. Then, we have that $sup(A)=sup(B)$.
answered Dec 1 '18 at 14:07
GödelGödel
1,418319
$endgroup$
$begingroup$
clever, just suppose it does not and we get a contradiction. I often forget contradiction as an approach - thanks for reminding me.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
I feel with every bit of steering in the right direction I'm becoming better at analysis, thank you :)
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
@WesleyGroupshaveFeelingsToo You are welcome.
$endgroup$
– Gödel
Dec 1 '18 at 14:30
add a comment |
$begingroup$
Suppose that $sup(B)<sup(A)$. Then, take $xin A$ such that $sup(B)<x<sup(A)$ (We can do it by definition of $sup$ and because $sup(A)notin A$). Then, we have that $xnotin B$ hence $x$ must be $z$. But it is impossible because it would imply that $z=sup(A)$. Then, we have that $sup(A)=sup(B)$.
answered Dec 1 '18 at 14:07
GödelGödel
1,418319
$endgroup$
$begingroup$
clever, just suppose it does not and we get a contradiction. I often forget contradiction as an approach - thanks for reminding me.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
I feel with every bit of steering in the right direction I'm becoming better at analysis, thank you :)
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
@WesleyGroupshaveFeelingsToo You are welcome.
$endgroup$
– Gödel
Dec 1 '18 at 14:30
add a comment |
$begingroup$
Suppose that $sup(B)<sup(A)$. Then, take $xin A$ such that $sup(B)<x<sup(A)$ (We can do it by definition of $sup$ and because $sup(A)notin A$). Then, we have that $xnotin B$ hence $x$ must be $z$. But it is impossible because it would imply that $z=sup(A)$. Then, we have that $sup(A)=sup(B)$.
answered Dec 1 '18 at 14:07
GödelGödel
1,418319
$endgroup$
Suppose that $sup(B)<sup(A)$. Then, take $xin A$ such that $sup(B)<x<sup(A)$ (We can do it by definition of $sup$ and because $sup(A)notin A$). Then, we have that $xnotin B$ hence $x$ must be $z$. But it is impossible because it would imply that $z=sup(A)$. Then, we have that $sup(A)=sup(B)$.
answered Dec 1 '18 at 14:07
GödelGödel
1,418319
answered Dec 1 '18 at 14:07
GödelGödel
1,418319
answered Dec 1 '18 at 14:07
GödelGödel
1,418319
answered Dec 1 '18 at 14:07
GödelGödel
1,418319
1,418319
$begingroup$
clever, just suppose it does not and we get a contradiction. I often forget contradiction as an approach - thanks for reminding me.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
I feel with every bit of steering in the right direction I'm becoming better at analysis, thank you :)
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
@WesleyGroupshaveFeelingsToo You are welcome.
$endgroup$
– Gödel
Dec 1 '18 at 14:30
add a comment |
$begingroup$
clever, just suppose it does not and we get a contradiction. I often forget contradiction as an approach - thanks for reminding me.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
I feel with every bit of steering in the right direction I'm becoming better at analysis, thank you :)
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
@WesleyGroupshaveFeelingsToo You are welcome.
$endgroup$
– Gödel
Dec 1 '18 at 14:30
$begingroup$
clever, just suppose it does not and we get a contradiction. I often forget contradiction as an approach - thanks for reminding me.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
clever, just suppose it does not and we get a contradiction. I often forget contradiction as an approach - thanks for reminding me.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
I feel with every bit of steering in the right direction I'm becoming better at analysis, thank you :)
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
I feel with every bit of steering in the right direction I'm becoming better at analysis, thank you :)
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:27
$begingroup$
@WesleyGroupshaveFeelingsToo You are welcome.
$endgroup$
– Gödel
Dec 1 '18 at 14:30
$begingroup$
@WesleyGroupshaveFeelingsToo You are welcome.
$endgroup$
– Gödel
Dec 1 '18 at 14:30
add a comment |
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