If $x$ and $y$ are arbitrary real numbers with $x < y$, prove that there exists at least one real...
$begingroup$
This is a question from Apostol Calculus section I.3.12 q1
If $x$ and $y$ are arbitrary real numbers with $x < y$, prove that there exists at least one real satisfying $x < z < y$
I've seen a lot of answers such as this one question here that use the archimedean property of real numbers. tl;dr let $z = frac{y+x}{2}$
My question is the following: I attempted solving this using a different approach, relying on Theorems I.32 and 34. I'm wondering if the approach is valid, specifically the strategy. (I'm sure there are issues with the details, I'm new to proofs and working on smoothing out the details).
Theorem I.30 "Archimedean Property" If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive integer $n$ such that $nx > y$.
Theorem I.32: Let $h$ be a given positive integer and let $S$ be a set of real numbers.
(a) If $S$ has a supremum, then for some $x$ in $S$ we have$$x > sup S - h$$
(b) If S has an infimum, then for some $x text{ in } S$ we have $$x < inf S + h$$
Theorem I.34: Given two nonempty subsets $S$ and $T$ of $mathbb R$, such that $$s leq t$$ for every $s in S text{ and every } t in T$. Then $S$ has a supremum, and $T$ has an infimum, and they satisfy the inequality $$sup S leq inf T$$
My attempt
Let $S text{ and } T$ be nonempty sets of real numbers such that $x < y$ for all $x in S$ and all $y in T$.
Then, arbitrary $y$ in $T$ is an upper bound for $S$, so $S$ has $sup S$ such that $$sup S geq x text{ for all } x in S quad text{ (1) }$$ Moreover, from the definition of supremum, we have $$x leq sup S lt y quad text{ (2) }$$
Same reasoning for $T$. Arbitrary $x$ in $S$ is a lower bound for $T$, so $T$ has $inf T$ such that $$inf T leq y text{ for all } y in T quad text{ (3) }$$ Again from the definition of infimum we have, $$x lt inf T leq y quad text{ (4) }$$
Assume there exists a $z$ such that $x < z$ for all $x in S$ and $z < y$. for all $y in T$. Then $z$ is an upper bound of $S$ so $$sup S < z quad text{ (5) }$$ and, by definition of supremum, we have $$x leq sup S lt z quad text{ (6) }$$
Additionally, $z$ is a lower bound of $T$ so $$inf T > z quad text{ (7) }$$ and, by definition of infimum, we have $$z lt inf T leq y quad text{ (8) }$$
From (8) and (7) we have $$x leq sup S lt z lt inf T leq y$$ which becomes $$x < z < y$$
real-analysis calculus proof-verification proof-writing supremum-and-infimum
asked Jan 17 at 12:59
Jake KirschJake Kirsch
687
$endgroup$
|
show 5 more comments
$begingroup$
This is a question from Apostol Calculus section I.3.12 q1
If $x$ and $y$ are arbitrary real numbers with $x < y$, prove that there exists at least one real satisfying $x < z < y$
I've seen a lot of answers such as this one question here that use the archimedean property of real numbers. tl;dr let $z = frac{y+x}{2}$
My question is the following: I attempted solving this using a different approach, relying on Theorems I.32 and 34. I'm wondering if the approach is valid, specifically the strategy. (I'm sure there are issues with the details, I'm new to proofs and working on smoothing out the details).
Theorem I.30 "Archimedean Property" If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive integer $n$ such that $nx > y$.
Theorem I.32: Let $h$ be a given positive integer and let $S$ be a set of real numbers.
(a) If $S$ has a supremum, then for some $x$ in $S$ we have$$x > sup S - h$$
(b) If S has an infimum, then for some $x text{ in } S$ we have $$x < inf S + h$$
Theorem I.34: Given two nonempty subsets $S$ and $T$ of $mathbb R$, such that $$s leq t$$ for every $s in S text{ and every } t in T$. Then $S$ has a supremum, and $T$ has an infimum, and they satisfy the inequality $$sup S leq inf T$$
My attempt
Let $S text{ and } T$ be nonempty sets of real numbers such that $x < y$ for all $x in S$ and all $y in T$.
Then, arbitrary $y$ in $T$ is an upper bound for $S$, so $S$ has $sup S$ such that $$sup S geq x text{ for all } x in S quad text{ (1) }$$ Moreover, from the definition of supremum, we have $$x leq sup S lt y quad text{ (2) }$$
Same reasoning for $T$. Arbitrary $x$ in $S$ is a lower bound for $T$, so $T$ has $inf T$ such that $$inf T leq y text{ for all } y in T quad text{ (3) }$$ Again from the definition of infimum we have, $$x lt inf T leq y quad text{ (4) }$$
Assume there exists a $z$ such that $x < z$ for all $x in S$ and $z < y$. for all $y in T$. Then $z$ is an upper bound of $S$ so $$sup S < z quad text{ (5) }$$ and, by definition of supremum, we have $$x leq sup S lt z quad text{ (6) }$$
Additionally, $z$ is a lower bound of $T$ so $$inf T > z quad text{ (7) }$$ and, by definition of infimum, we have $$z lt inf T leq y quad text{ (8) }$$
From (8) and (7) we have $$x leq sup S lt z lt inf T leq y$$ which becomes $$x < z < y$$
real-analysis calculus proof-verification proof-writing supremum-and-infimum
asked Jan 17 at 12:59
Jake KirschJake Kirsch
687
$endgroup$
$begingroup$
Did you really mean $x<y<z$ in the title?
$endgroup$
– José Carlos Santos
Jan 17 at 13:01
$begingroup$
Do you want to obtain $x<z<y$ as in your last line (in which case take $z=(x+y)/2$), or $x<y<z$ (in which case take $z=2y-x$) as in your title (and a problem duplication underneath it)?
$endgroup$
– J.G.
Jan 17 at 13:02
$begingroup$
@JoséCarlosSantos apologies, fixed the title. I understand that the approach of taking $z = frac{x +y}{2}$, my question is about if you can solve this not using that theorem. I'll fix the intro paragraph to make that more clear
$endgroup$
– Jake Kirsch
Jan 17 at 13:04
$begingroup$
@JakeKirsch The statement duplication still contains the typo.
$endgroup$
– J.G.
Jan 17 at 13:05
$begingroup$
@J.G. thanks, fixed that part as well
$endgroup$
– Jake Kirsch
Jan 17 at 13:11
|
show 5 more comments
$begingroup$
This is a question from Apostol Calculus section I.3.12 q1
If $x$ and $y$ are arbitrary real numbers with $x < y$, prove that there exists at least one real satisfying $x < z < y$
I've seen a lot of answers such as this one question here that use the archimedean property of real numbers. tl;dr let $z = frac{y+x}{2}$
My question is the following: I attempted solving this using a different approach, relying on Theorems I.32 and 34. I'm wondering if the approach is valid, specifically the strategy. (I'm sure there are issues with the details, I'm new to proofs and working on smoothing out the details).
Theorem I.30 "Archimedean Property" If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive integer $n$ such that $nx > y$.
Theorem I.32: Let $h$ be a given positive integer and let $S$ be a set of real numbers.
(a) If $S$ has a supremum, then for some $x$ in $S$ we have$$x > sup S - h$$
(b) If S has an infimum, then for some $x text{ in } S$ we have $$x < inf S + h$$
Theorem I.34: Given two nonempty subsets $S$ and $T$ of $mathbb R$, such that $$s leq t$$ for every $s in S text{ and every } t in T$. Then $S$ has a supremum, and $T$ has an infimum, and they satisfy the inequality $$sup S leq inf T$$
My attempt
Let $S text{ and } T$ be nonempty sets of real numbers such that $x < y$ for all $x in S$ and all $y in T$.
Then, arbitrary $y$ in $T$ is an upper bound for $S$, so $S$ has $sup S$ such that $$sup S geq x text{ for all } x in S quad text{ (1) }$$ Moreover, from the definition of supremum, we have $$x leq sup S lt y quad text{ (2) }$$
Same reasoning for $T$. Arbitrary $x$ in $S$ is a lower bound for $T$, so $T$ has $inf T$ such that $$inf T leq y text{ for all } y in T quad text{ (3) }$$ Again from the definition of infimum we have, $$x lt inf T leq y quad text{ (4) }$$
Assume there exists a $z$ such that $x < z$ for all $x in S$ and $z < y$. for all $y in T$. Then $z$ is an upper bound of $S$ so $$sup S < z quad text{ (5) }$$ and, by definition of supremum, we have $$x leq sup S lt z quad text{ (6) }$$
Additionally, $z$ is a lower bound of $T$ so $$inf T > z quad text{ (7) }$$ and, by definition of infimum, we have $$z lt inf T leq y quad text{ (8) }$$
From (8) and (7) we have $$x leq sup S lt z lt inf T leq y$$ which becomes $$x < z < y$$
real-analysis calculus proof-verification proof-writing supremum-and-infimum
asked Jan 17 at 12:59
Jake KirschJake Kirsch
687
$endgroup$
This is a question from Apostol Calculus section I.3.12 q1
If $x$ and $y$ are arbitrary real numbers with $x < y$, prove that there exists at least one real satisfying $x < z < y$
I've seen a lot of answers such as this one question here that use the archimedean property of real numbers. tl;dr let $z = frac{y+x}{2}$
My question is the following: I attempted solving this using a different approach, relying on Theorems I.32 and 34. I'm wondering if the approach is valid, specifically the strategy. (I'm sure there are issues with the details, I'm new to proofs and working on smoothing out the details).
Theorem I.30 "Archimedean Property" If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive integer $n$ such that $nx > y$.
Theorem I.32: Let $h$ be a given positive integer and let $S$ be a set of real numbers.
(a) If $S$ has a supremum, then for some $x$ in $S$ we have$$x > sup S - h$$
(b) If S has an infimum, then for some $x text{ in } S$ we have $$x < inf S + h$$
Theorem I.34: Given two nonempty subsets $S$ and $T$ of $mathbb R$, such that $$s leq t$$ for every $s in S text{ and every } t in T$. Then $S$ has a supremum, and $T$ has an infimum, and they satisfy the inequality $$sup S leq inf T$$
My attempt
Let $S text{ and } T$ be nonempty sets of real numbers such that $x < y$ for all $x in S$ and all $y in T$.
Then, arbitrary $y$ in $T$ is an upper bound for $S$, so $S$ has $sup S$ such that $$sup S geq x text{ for all } x in S quad text{ (1) }$$ Moreover, from the definition of supremum, we have $$x leq sup S lt y quad text{ (2) }$$
Same reasoning for $T$. Arbitrary $x$ in $S$ is a lower bound for $T$, so $T$ has $inf T$ such that $$inf T leq y text{ for all } y in T quad text{ (3) }$$ Again from the definition of infimum we have, $$x lt inf T leq y quad text{ (4) }$$
Assume there exists a $z$ such that $x < z$ for all $x in S$ and $z < y$. for all $y in T$. Then $z$ is an upper bound of $S$ so $$sup S < z quad text{ (5) }$$ and, by definition of supremum, we have $$x leq sup S lt z quad text{ (6) }$$
Additionally, $z$ is a lower bound of $T$ so $$inf T > z quad text{ (7) }$$ and, by definition of infimum, we have $$z lt inf T leq y quad text{ (8) }$$
From (8) and (7) we have $$x leq sup S lt z lt inf T leq y$$ which becomes $$x < z < y$$
real-analysis calculus proof-verification proof-writing supremum-and-infimum
real-analysis calculus proof-verification proof-writing supremum-and-infimum
asked Jan 17 at 12:59
Jake KirschJake Kirsch
687
asked Jan 17 at 12:59
Jake KirschJake Kirsch
687
edited Jan 17 at 13:06
Jake Kirsch
asked Jan 17 at 12:59
Jake KirschJake Kirsch
687
asked Jan 17 at 12:59
Jake KirschJake Kirsch
687
asked Jan 17 at 12:59
Jake KirschJake Kirsch
687
687
$begingroup$
Did you really mean $x<y<z$ in the title?
$endgroup$
– José Carlos Santos
Jan 17 at 13:01
$begingroup$
Do you want to obtain $x<z<y$ as in your last line (in which case take $z=(x+y)/2$), or $x<y<z$ (in which case take $z=2y-x$) as in your title (and a problem duplication underneath it)?
$endgroup$
– J.G.
Jan 17 at 13:02
$begingroup$
@JoséCarlosSantos apologies, fixed the title. I understand that the approach of taking $z = frac{x +y}{2}$, my question is about if you can solve this not using that theorem. I'll fix the intro paragraph to make that more clear
$endgroup$
– Jake Kirsch
Jan 17 at 13:04
$begingroup$
@JakeKirsch The statement duplication still contains the typo.
$endgroup$
– J.G.
Jan 17 at 13:05
$begingroup$
@J.G. thanks, fixed that part as well
$endgroup$
– Jake Kirsch
Jan 17 at 13:11
|
show 5 more comments
$begingroup$
Did you really mean $x<y<z$ in the title?
$endgroup$
– José Carlos Santos
Jan 17 at 13:01
$begingroup$
Do you want to obtain $x<z<y$ as in your last line (in which case take $z=(x+y)/2$), or $x<y<z$ (in which case take $z=2y-x$) as in your title (and a problem duplication underneath it)?
$endgroup$
– J.G.
Jan 17 at 13:02
$begingroup$
@JoséCarlosSantos apologies, fixed the title. I understand that the approach of taking $z = frac{x +y}{2}$, my question is about if you can solve this not using that theorem. I'll fix the intro paragraph to make that more clear
$endgroup$
– Jake Kirsch
Jan 17 at 13:04
$begingroup$
@JakeKirsch The statement duplication still contains the typo.
$endgroup$
– J.G.
Jan 17 at 13:05
$begingroup$
@J.G. thanks, fixed that part as well
$endgroup$
– Jake Kirsch
Jan 17 at 13:11
$begingroup$
Did you really mean $x<y<z$ in the title?
$endgroup$
– José Carlos Santos
Jan 17 at 13:01
$begingroup$
Did you really mean $x<y<z$ in the title?
$endgroup$
– José Carlos Santos
Jan 17 at 13:01
$begingroup$
Do you want to obtain $x<z<y$ as in your last line (in which case take $z=(x+y)/2$), or $x<y<z$ (in which case take $z=2y-x$) as in your title (and a problem duplication underneath it)?
$endgroup$
– J.G.
Jan 17 at 13:02
$begingroup$
Do you want to obtain $x<z<y$ as in your last line (in which case take $z=(x+y)/2$), or $x<y<z$ (in which case take $z=2y-x$) as in your title (and a problem duplication underneath it)?
$endgroup$
– J.G.
Jan 17 at 13:02
$begingroup$
@JoséCarlosSantos apologies, fixed the title. I understand that the approach of taking $z = frac{x +y}{2}$, my question is about if you can solve this not using that theorem. I'll fix the intro paragraph to make that more clear
$endgroup$
– Jake Kirsch
Jan 17 at 13:04
$begingroup$
@JoséCarlosSantos apologies, fixed the title. I understand that the approach of taking $z = frac{x +y}{2}$, my question is about if you can solve this not using that theorem. I'll fix the intro paragraph to make that more clear
$endgroup$
– Jake Kirsch
Jan 17 at 13:04
$begingroup$
@JakeKirsch The statement duplication still contains the typo.
$endgroup$
– J.G.
Jan 17 at 13:05
$begingroup$
@JakeKirsch The statement duplication still contains the typo.
$endgroup$
– J.G.
Jan 17 at 13:05
$begingroup$
@J.G. thanks, fixed that part as well
$endgroup$
– Jake Kirsch
Jan 17 at 13:11
$begingroup$
@J.G. thanks, fixed that part as well
$endgroup$
– Jake Kirsch
Jan 17 at 13:11
|
show 5 more comments
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$begingroup$
Did you really mean $x<y<z$ in the title?
$endgroup$
– José Carlos Santos
Jan 17 at 13:01
$begingroup$
Do you want to obtain $x<z<y$ as in your last line (in which case take $z=(x+y)/2$), or $x<y<z$ (in which case take $z=2y-x$) as in your title (and a problem duplication underneath it)?
$endgroup$
– J.G.
Jan 17 at 13:02
$begingroup$
@JoséCarlosSantos apologies, fixed the title. I understand that the approach of taking $z = frac{x +y}{2}$, my question is about if you can solve this not using that theorem. I'll fix the intro paragraph to make that more clear
$endgroup$
– Jake Kirsch
Jan 17 at 13:04
$begingroup$
@JakeKirsch The statement duplication still contains the typo.
$endgroup$
– J.G.
Jan 17 at 13:05
$begingroup$
@J.G. thanks, fixed that part as well
$endgroup$
– Jake Kirsch
Jan 17 at 13:11