Let $A$ be bounded below, and $B = b in R : b$ is a lower bound for $A$. Show that $sup B = inf A$.
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This is Exercise 1.3.3 (a) from "Understanding Analysis" by Stephen Abbot, , page 17:
Let $A$ be bounded below, and define $B = b in R : b$ is a lower bound for $A$. Show that $sup B = inf A$.
Please check my proof:
Suppose $A$ be bound below, it exist number $Nleq n$ for every $nin mathbb R$ in the set and $N$ is $infA$. Then $B=b in mathbb R:b$ is a lower bound for $A$. It exist number $b$ in the set, since it contains only lower bound of $A$ then it has number $bleq M$ for every $b$ and $M$ is $sup$ of $B$. But $b$ is lower bound of $A$ and $A$ has $N$ as $sup$, $N$ is in $B$. $N$ is automatically is sup of $b$ therefore $M=N$ or $sup B=inf A$.
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 12:20
Martin Sleziak
43.5k6113260
asked Jan 16 '17 at 10:50
Lingnoi401
919419
 |Â
show 6 more comments
up vote
2
down vote
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This is Exercise 1.3.3 (a) from "Understanding Analysis" by Stephen Abbot, , page 17:
Let $A$ be bounded below, and define $B = b in R : b$ is a lower bound for $A$. Show that $sup B = inf A$.
Please check my proof:
Suppose $A$ be bound below, it exist number $Nleq n$ for every $nin mathbb R$ in the set and $N$ is $infA$. Then $B=b in mathbb R:b$ is a lower bound for $A$. It exist number $b$ in the set, since it contains only lower bound of $A$ then it has number $bleq M$ for every $b$ and $M$ is $sup$ of $B$. But $b$ is lower bound of $A$ and $A$ has $N$ as $sup$, $N$ is in $B$. $N$ is automatically is sup of $b$ therefore $M=N$ or $sup B=inf A$.
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 12:20
Martin Sleziak
43.5k6113260
asked Jan 16 '17 at 10:50
Lingnoi401
919419
1
wait ,why edited my question ?
– Lingnoi401
Jan 16 '17 at 11:00
1
@juniven exercise from Understanding Analysis by Stephen Abbot
– Lingnoi401
Jan 16 '17 at 11:08
1
it is in page 17
– Lingnoi401
Jan 16 '17 at 11:09
1
There I see Exercise 1.3.3 (a) Thank you.
– Î˜Î£Î¦GenSan
Jan 16 '17 at 11:13
2
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
– quid♦
Jan 16 '17 at 12:23
 |Â
show 6 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is Exercise 1.3.3 (a) from "Understanding Analysis" by Stephen Abbot, , page 17:
Let $A$ be bounded below, and define $B = b in R : b$ is a lower bound for $A$. Show that $sup B = inf A$.
Please check my proof:
Suppose $A$ be bound below, it exist number $Nleq n$ for every $nin mathbb R$ in the set and $N$ is $infA$. Then $B=b in mathbb R:b$ is a lower bound for $A$. It exist number $b$ in the set, since it contains only lower bound of $A$ then it has number $bleq M$ for every $b$ and $M$ is $sup$ of $B$. But $b$ is lower bound of $A$ and $A$ has $N$ as $sup$, $N$ is in $B$. $N$ is automatically is sup of $b$ therefore $M=N$ or $sup B=inf A$.
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 12:20
Martin Sleziak
43.5k6113260
asked Jan 16 '17 at 10:50
Lingnoi401
919419
This is Exercise 1.3.3 (a) from "Understanding Analysis" by Stephen Abbot, , page 17:
Let $A$ be bounded below, and define $B = b in R : b$ is a lower bound for $A$. Show that $sup B = inf A$.
Please check my proof:
Suppose $A$ be bound below, it exist number $Nleq n$ for every $nin mathbb R$ in the set and $N$ is $infA$. Then $B=b in mathbb R:b$ is a lower bound for $A$. It exist number $b$ in the set, since it contains only lower bound of $A$ then it has number $bleq M$ for every $b$ and $M$ is $sup$ of $B$. But $b$ is lower bound of $A$ and $A$ has $N$ as $sup$, $N$ is in $B$. $N$ is automatically is sup of $b$ therefore $M=N$ or $sup B=inf A$.
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 12:20
Martin Sleziak
43.5k6113260
asked Jan 16 '17 at 10:50
Lingnoi401
919419
edited Jan 16 '17 at 12:20
Martin Sleziak
43.5k6113260
edited Jan 16 '17 at 12:20
Martin Sleziak
43.5k6113260
edited Jan 16 '17 at 12:20
Martin Sleziak
43.5k6113260
43.5k6113260
asked Jan 16 '17 at 10:50
Lingnoi401
919419
asked Jan 16 '17 at 10:50
Lingnoi401
919419
asked Jan 16 '17 at 10:50
Lingnoi401
919419
919419
1
wait ,why edited my question ?
– Lingnoi401
Jan 16 '17 at 11:00
1
@juniven exercise from Understanding Analysis by Stephen Abbot
– Lingnoi401
Jan 16 '17 at 11:08
1
it is in page 17
– Lingnoi401
Jan 16 '17 at 11:09
1
There I see Exercise 1.3.3 (a) Thank you.
– Î˜Î£Î¦GenSan
Jan 16 '17 at 11:13
2
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
– quid♦
Jan 16 '17 at 12:23
 |Â
show 6 more comments
1
wait ,why edited my question ?
– Lingnoi401
Jan 16 '17 at 11:00
1
@juniven exercise from Understanding Analysis by Stephen Abbot
– Lingnoi401
Jan 16 '17 at 11:08
1
it is in page 17
– Lingnoi401
Jan 16 '17 at 11:09
1
There I see Exercise 1.3.3 (a) Thank you.
– Î˜Î£Î¦GenSan
Jan 16 '17 at 11:13
2
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
– quid♦
Jan 16 '17 at 12:23
1
1
wait ,why edited my question ?
– Lingnoi401
Jan 16 '17 at 11:00
wait ,why edited my question ?
– Lingnoi401
Jan 16 '17 at 11:00
1
1
@juniven exercise from Understanding Analysis by Stephen Abbot
– Lingnoi401
Jan 16 '17 at 11:08
@juniven exercise from Understanding Analysis by Stephen Abbot
– Lingnoi401
Jan 16 '17 at 11:08
1
1
it is in page 17
– Lingnoi401
Jan 16 '17 at 11:09
it is in page 17
– Lingnoi401
Jan 16 '17 at 11:09
1
1
There I see Exercise 1.3.3 (a) Thank you.
– Î˜Î£Î¦GenSan
Jan 16 '17 at 11:13
There I see Exercise 1.3.3 (a) Thank you.
– Î˜Î£Î¦GenSan
Jan 16 '17 at 11:13
2
2
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
– quid♦
Jan 16 '17 at 12:23
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
– quid♦
Jan 16 '17 at 12:23
 |Â
show 6 more comments
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$inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $inf Ain B$. But it is also the biggest lower bound of $A$, so for every $bin B$ follows $bleqinf A$. This gives us $inf A=max B$. But if a set has a maximum, this maximum is also its supremum. Thus $inf A=max B=sup B$.
answered Aug 14 at 11:39
Bentapair
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$inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $inf Ain B$. But it is also the biggest lower bound of $A$, so for every $bin B$ follows $bleqinf A$. This gives us $inf A=max B$. But if a set has a maximum, this maximum is also its supremum. Thus $inf A=max B=sup B$.
answered Aug 14 at 11:39
Bentapair
1
add a comment |Â
up vote
0
down vote
$inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $inf Ain B$. But it is also the biggest lower bound of $A$, so for every $bin B$ follows $bleqinf A$. This gives us $inf A=max B$. But if a set has a maximum, this maximum is also its supremum. Thus $inf A=max B=sup B$.
answered Aug 14 at 11:39
Bentapair
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $inf Ain B$. But it is also the biggest lower bound of $A$, so for every $bin B$ follows $bleqinf A$. This gives us $inf A=max B$. But if a set has a maximum, this maximum is also its supremum. Thus $inf A=max B=sup B$.
answered Aug 14 at 11:39
Bentapair
1
$inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $inf Ain B$. But it is also the biggest lower bound of $A$, so for every $bin B$ follows $bleqinf A$. This gives us $inf A=max B$. But if a set has a maximum, this maximum is also its supremum. Thus $inf A=max B=sup B$.
answered Aug 14 at 11:39
Bentapair
1
answered Aug 14 at 11:39
Bentapair
1
answered Aug 14 at 11:39
Bentapair
1
answered Aug 14 at 11:39
Bentapair
1
1
add a comment |Â
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1
wait ,why edited my question ?
– Lingnoi401
Jan 16 '17 at 11:00
1
@juniven exercise from Understanding Analysis by Stephen Abbot
– Lingnoi401
Jan 16 '17 at 11:08
1
it is in page 17
– Lingnoi401
Jan 16 '17 at 11:09
1
There I see Exercise 1.3.3 (a) Thank you.
– Î˜Î£Î¦GenSan
Jan 16 '17 at 11:13
2
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
– quid♦
Jan 16 '17 at 12:23