Apostol Calculus Vol. 1, Theorem 1.28. The set of positive integers is unbounded above.
$begingroup$
Theorem 1.28: The set $P$ of positive integers $1,2,3,ldots$ is unbounded above.
Proof : Assume $P$ is bounded above. We shall show that this leads to a contradiction. Since $P$ is nonempty, $P$ has a least upper bound, say $b$. The number $b−1$, being less than $b$, cannot be an upper bound for $P$. Hence, there is at least one positive integer $n$ such that $n>b−1$. For this $n$ we have $n+1>b$. Since $n+1$ is in $P$, this contradicts the fact that $b$ is an upper bound for $P$.
In this proof, he says that
there is at least one positive integer $n$ such that $n>b−1$.
I can understand it intuitively, but can't deduce it from previous theorems and axioms.
calculus
edited Dec 25 '18 at 19:13
Hans Hüttel
3,2172921
asked Dec 25 '18 at 15:24
xydyxydy
11
$endgroup$
add a comment |
$begingroup$
Theorem 1.28: The set $P$ of positive integers $1,2,3,ldots$ is unbounded above.
Proof : Assume $P$ is bounded above. We shall show that this leads to a contradiction. Since $P$ is nonempty, $P$ has a least upper bound, say $b$. The number $b−1$, being less than $b$, cannot be an upper bound for $P$. Hence, there is at least one positive integer $n$ such that $n>b−1$. For this $n$ we have $n+1>b$. Since $n+1$ is in $P$, this contradicts the fact that $b$ is an upper bound for $P$.
In this proof, he says that
there is at least one positive integer $n$ such that $n>b−1$.
I can understand it intuitively, but can't deduce it from previous theorems and axioms.
calculus
edited Dec 25 '18 at 19:13
Hans Hüttel
3,2172921
asked Dec 25 '18 at 15:24
xydyxydy
11
$endgroup$
$begingroup$
As the previous sentence points out, $b$ is such an integer.
$endgroup$
– saulspatz
Dec 25 '18 at 15:27
2
$begingroup$
You left out the "Hence"! $b-1$ is not an upper bound for $P$, which means that there exists $nin P$ with $n>b-1$.
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:28
2
$begingroup$
@saulspatz I don't see how the previous sentence says what you say it says. In any case, how do you know $b$ is an integer?
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:30
$begingroup$
@David C. Ullrich I think that i get it. I've just realised I forgot how negation works. Thank you!
$endgroup$
– xydy
Dec 25 '18 at 16:00
add a comment |
$begingroup$
Theorem 1.28: The set $P$ of positive integers $1,2,3,ldots$ is unbounded above.
Proof : Assume $P$ is bounded above. We shall show that this leads to a contradiction. Since $P$ is nonempty, $P$ has a least upper bound, say $b$. The number $b−1$, being less than $b$, cannot be an upper bound for $P$. Hence, there is at least one positive integer $n$ such that $n>b−1$. For this $n$ we have $n+1>b$. Since $n+1$ is in $P$, this contradicts the fact that $b$ is an upper bound for $P$.
In this proof, he says that
there is at least one positive integer $n$ such that $n>b−1$.
I can understand it intuitively, but can't deduce it from previous theorems and axioms.
calculus
edited Dec 25 '18 at 19:13
Hans Hüttel
3,2172921
asked Dec 25 '18 at 15:24
xydyxydy
11
$endgroup$
Theorem 1.28: The set $P$ of positive integers $1,2,3,ldots$ is unbounded above.
Proof : Assume $P$ is bounded above. We shall show that this leads to a contradiction. Since $P$ is nonempty, $P$ has a least upper bound, say $b$. The number $b−1$, being less than $b$, cannot be an upper bound for $P$. Hence, there is at least one positive integer $n$ such that $n>b−1$. For this $n$ we have $n+1>b$. Since $n+1$ is in $P$, this contradicts the fact that $b$ is an upper bound for $P$.
In this proof, he says that
there is at least one positive integer $n$ such that $n>b−1$.
I can understand it intuitively, but can't deduce it from previous theorems and axioms.
calculus
calculus
edited Dec 25 '18 at 19:13
Hans Hüttel
3,2172921
asked Dec 25 '18 at 15:24
xydyxydy
11
edited Dec 25 '18 at 19:13
Hans Hüttel
3,2172921
asked Dec 25 '18 at 15:24
xydyxydy
11
edited Dec 25 '18 at 19:13
Hans Hüttel
3,2172921
edited Dec 25 '18 at 19:13
Hans Hüttel
3,2172921
edited Dec 25 '18 at 19:13
Hans Hüttel
3,2172921
3,2172921
asked Dec 25 '18 at 15:24
xydyxydy
11
asked Dec 25 '18 at 15:24
xydyxydy
11
asked Dec 25 '18 at 15:24
xydyxydy
11
11
$begingroup$
As the previous sentence points out, $b$ is such an integer.
$endgroup$
– saulspatz
Dec 25 '18 at 15:27
2
$begingroup$
You left out the "Hence"! $b-1$ is not an upper bound for $P$, which means that there exists $nin P$ with $n>b-1$.
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:28
2
$begingroup$
@saulspatz I don't see how the previous sentence says what you say it says. In any case, how do you know $b$ is an integer?
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:30
$begingroup$
@David C. Ullrich I think that i get it. I've just realised I forgot how negation works. Thank you!
$endgroup$
– xydy
Dec 25 '18 at 16:00
add a comment |
$begingroup$
As the previous sentence points out, $b$ is such an integer.
$endgroup$
– saulspatz
Dec 25 '18 at 15:27
2
$begingroup$
You left out the "Hence"! $b-1$ is not an upper bound for $P$, which means that there exists $nin P$ with $n>b-1$.
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:28
2
$begingroup$
@saulspatz I don't see how the previous sentence says what you say it says. In any case, how do you know $b$ is an integer?
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:30
$begingroup$
@David C. Ullrich I think that i get it. I've just realised I forgot how negation works. Thank you!
$endgroup$
– xydy
Dec 25 '18 at 16:00
$begingroup$
As the previous sentence points out, $b$ is such an integer.
$endgroup$
– saulspatz
Dec 25 '18 at 15:27
$begingroup$
As the previous sentence points out, $b$ is such an integer.
$endgroup$
– saulspatz
Dec 25 '18 at 15:27
2
2
$begingroup$
You left out the "Hence"! $b-1$ is not an upper bound for $P$, which means that there exists $nin P$ with $n>b-1$.
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:28
$begingroup$
You left out the "Hence"! $b-1$ is not an upper bound for $P$, which means that there exists $nin P$ with $n>b-1$.
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:28
2
2
$begingroup$
@saulspatz I don't see how the previous sentence says what you say it says. In any case, how do you know $b$ is an integer?
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:30
$begingroup$
@saulspatz I don't see how the previous sentence says what you say it says. In any case, how do you know $b$ is an integer?
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:30
$begingroup$
@David C. Ullrich I think that i get it. I've just realised I forgot how negation works. Thank you!
$endgroup$
– xydy
Dec 25 '18 at 16:00
$begingroup$
@David C. Ullrich I think that i get it. I've just realised I forgot how negation works. Thank you!
$endgroup$
– xydy
Dec 25 '18 at 16:00
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I suspect it is a proof concerning $P$ as a subset of $mathbb R$ (please correct me if I am wrong).
Then the non-existence of such $n$ would imply that also $b-1$ is an upperbound of $P$.
This however contradicts that $b$ is the least upperbound of $P$.
answered Dec 25 '18 at 15:33
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
Asserting that $k$ is an upper bound of a set $Ssubsetmathbb R$ means that each element of $S$ is smaller than or equal to $k$. So, asserting that $k$ isn't an upper bound of $S$ means the, for some $sin S$, $s>k$. So, in particular, since $b-1$ is not an upper bound of $mathbb N$, there is a natural $n$ such that $n>b-1$.
answered Dec 25 '18 at 15:31
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
add a comment |
$begingroup$
In the previous sentence, $b$ is designated to be the least upper bound,
so that means for
all number $b'$ that is less than $b$, there must be an element $p$ in $P$ such that
$p>b'$. This can be reformulated as "For all $epsilon>0$, there is $pin P$ such that $p>b-epsilon$." Using this formulation to explain the made reasoning, the $epsilon$ is chosen to be $1$.
answered Dec 25 '18 at 17:12
Abdullah UYUAbdullah UYU
402415
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
I suspect it is a proof concerning $P$ as a subset of $mathbb R$ (please correct me if I am wrong).
Then the non-existence of such $n$ would imply that also $b-1$ is an upperbound of $P$.
This however contradicts that $b$ is the least upperbound of $P$.
answered Dec 25 '18 at 15:33
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
I suspect it is a proof concerning $P$ as a subset of $mathbb R$ (please correct me if I am wrong).
Then the non-existence of such $n$ would imply that also $b-1$ is an upperbound of $P$.
This however contradicts that $b$ is the least upperbound of $P$.
answered Dec 25 '18 at 15:33
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
I suspect it is a proof concerning $P$ as a subset of $mathbb R$ (please correct me if I am wrong).
Then the non-existence of such $n$ would imply that also $b-1$ is an upperbound of $P$.
This however contradicts that $b$ is the least upperbound of $P$.
answered Dec 25 '18 at 15:33
drhabdrhab
101k545136
$endgroup$
I suspect it is a proof concerning $P$ as a subset of $mathbb R$ (please correct me if I am wrong).
Then the non-existence of such $n$ would imply that also $b-1$ is an upperbound of $P$.
This however contradicts that $b$ is the least upperbound of $P$.
answered Dec 25 '18 at 15:33
drhabdrhab
101k545136
answered Dec 25 '18 at 15:33
drhabdrhab
101k545136
answered Dec 25 '18 at 15:33
drhabdrhab
101k545136
answered Dec 25 '18 at 15:33
drhabdrhab
101k545136
101k545136
add a comment |
add a comment |
$begingroup$
Asserting that $k$ is an upper bound of a set $Ssubsetmathbb R$ means that each element of $S$ is smaller than or equal to $k$. So, asserting that $k$ isn't an upper bound of $S$ means the, for some $sin S$, $s>k$. So, in particular, since $b-1$ is not an upper bound of $mathbb N$, there is a natural $n$ such that $n>b-1$.
answered Dec 25 '18 at 15:31
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
add a comment |
$begingroup$
Asserting that $k$ is an upper bound of a set $Ssubsetmathbb R$ means that each element of $S$ is smaller than or equal to $k$. So, asserting that $k$ isn't an upper bound of $S$ means the, for some $sin S$, $s>k$. So, in particular, since $b-1$ is not an upper bound of $mathbb N$, there is a natural $n$ such that $n>b-1$.
answered Dec 25 '18 at 15:31
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
add a comment |
$begingroup$
Asserting that $k$ is an upper bound of a set $Ssubsetmathbb R$ means that each element of $S$ is smaller than or equal to $k$. So, asserting that $k$ isn't an upper bound of $S$ means the, for some $sin S$, $s>k$. So, in particular, since $b-1$ is not an upper bound of $mathbb N$, there is a natural $n$ such that $n>b-1$.
answered Dec 25 '18 at 15:31
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
Asserting that $k$ is an upper bound of a set $Ssubsetmathbb R$ means that each element of $S$ is smaller than or equal to $k$. So, asserting that $k$ isn't an upper bound of $S$ means the, for some $sin S$, $s>k$. So, in particular, since $b-1$ is not an upper bound of $mathbb N$, there is a natural $n$ such that $n>b-1$.
answered Dec 25 '18 at 15:31
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 25 '18 at 15:31
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 25 '18 at 15:31
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 25 '18 at 15:31
José Carlos SantosJosé Carlos Santos
160k22127232
160k22127232
add a comment |
add a comment |
$begingroup$
In the previous sentence, $b$ is designated to be the least upper bound,
so that means for
all number $b'$ that is less than $b$, there must be an element $p$ in $P$ such that
$p>b'$. This can be reformulated as "For all $epsilon>0$, there is $pin P$ such that $p>b-epsilon$." Using this formulation to explain the made reasoning, the $epsilon$ is chosen to be $1$.
answered Dec 25 '18 at 17:12
Abdullah UYUAbdullah UYU
402415
$endgroup$
add a comment |
$begingroup$
In the previous sentence, $b$ is designated to be the least upper bound,
so that means for
all number $b'$ that is less than $b$, there must be an element $p$ in $P$ such that
$p>b'$. This can be reformulated as "For all $epsilon>0$, there is $pin P$ such that $p>b-epsilon$." Using this formulation to explain the made reasoning, the $epsilon$ is chosen to be $1$.
answered Dec 25 '18 at 17:12
Abdullah UYUAbdullah UYU
402415
$endgroup$
add a comment |
$begingroup$
In the previous sentence, $b$ is designated to be the least upper bound,
so that means for
all number $b'$ that is less than $b$, there must be an element $p$ in $P$ such that
$p>b'$. This can be reformulated as "For all $epsilon>0$, there is $pin P$ such that $p>b-epsilon$." Using this formulation to explain the made reasoning, the $epsilon$ is chosen to be $1$.
answered Dec 25 '18 at 17:12
Abdullah UYUAbdullah UYU
402415
$endgroup$
In the previous sentence, $b$ is designated to be the least upper bound,
so that means for
all number $b'$ that is less than $b$, there must be an element $p$ in $P$ such that
$p>b'$. This can be reformulated as "For all $epsilon>0$, there is $pin P$ such that $p>b-epsilon$." Using this formulation to explain the made reasoning, the $epsilon$ is chosen to be $1$.
answered Dec 25 '18 at 17:12
Abdullah UYUAbdullah UYU
402415
answered Dec 25 '18 at 17:12
Abdullah UYUAbdullah UYU
402415
answered Dec 25 '18 at 17:12
Abdullah UYUAbdullah UYU
402415
answered Dec 25 '18 at 17:12
Abdullah UYUAbdullah UYU
402415
402415
add a comment |
add a comment |
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$begingroup$
As the previous sentence points out, $b$ is such an integer.
$endgroup$
– saulspatz
Dec 25 '18 at 15:27
2
$begingroup$
You left out the "Hence"! $b-1$ is not an upper bound for $P$, which means that there exists $nin P$ with $n>b-1$.
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:28
2
$begingroup$
@saulspatz I don't see how the previous sentence says what you say it says. In any case, how do you know $b$ is an integer?
$endgroup$
– David C. Ullrich
Dec 25 '18 at 15:30
$begingroup$
@David C. Ullrich I think that i get it. I've just realised I forgot how negation works. Thank you!
$endgroup$
– xydy
Dec 25 '18 at 16:00