Is this proof that $mathbb{N}$ is not bounded above correct?
$begingroup$
The following proof is from https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch2.pdf
If $x in R$, then there exists $n in N$ such that $x < n$.
Suppose, for contradiction, that there exists $x in R$ such that $x > n$ for every
$n in N$. Then $x$ is an upper bound of $N$, so $N$ has a supremum $M = sup N in R$.
Since $n le M$ for every $n in N$, we have $n − 1 ≤ M − 1$ for every $n in N$, which implies that $n le M − 1$ for every $n in N$. But then $M − 1$ is an upper bound of $N$, which contradicts the assumption that $M$ is a least upper bound.
I'm not sure how exactly I'm supposed to use the fact that $n-1 le M-1$ for every $n in N$. I thought that I could show $n-1 le M-1 le M$ and go from there using a similar process shown below. I did the following:
$|n| le M$ implies that
$-M le n le M$
$-M + 1 le n + 1 le M +1$
so that
$-M le -M + 1 le n +1$
and
$-M le n+1 = |n+1|$
which implies that
$-M le n+1le M$
and from here I can show that $n le M-1$. I'm used to getting an interval from $|x| <M$, for example, which implies that $-M <x<M$. I'm not sure if my process is correct.
real-analysis
asked Dec 21 '18 at 4:54
K.MK.M
700412
$endgroup$
add a comment |
$begingroup$
The following proof is from https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch2.pdf
If $x in R$, then there exists $n in N$ such that $x < n$.
Suppose, for contradiction, that there exists $x in R$ such that $x > n$ for every
$n in N$. Then $x$ is an upper bound of $N$, so $N$ has a supremum $M = sup N in R$.
Since $n le M$ for every $n in N$, we have $n − 1 ≤ M − 1$ for every $n in N$, which implies that $n le M − 1$ for every $n in N$. But then $M − 1$ is an upper bound of $N$, which contradicts the assumption that $M$ is a least upper bound.
I'm not sure how exactly I'm supposed to use the fact that $n-1 le M-1$ for every $n in N$. I thought that I could show $n-1 le M-1 le M$ and go from there using a similar process shown below. I did the following:
$|n| le M$ implies that
$-M le n le M$
$-M + 1 le n + 1 le M +1$
so that
$-M le -M + 1 le n +1$
and
$-M le n+1 = |n+1|$
which implies that
$-M le n+1le M$
and from here I can show that $n le M-1$. I'm used to getting an interval from $|x| <M$, for example, which implies that $-M <x<M$. I'm not sure if my process is correct.
real-analysis
asked Dec 21 '18 at 4:54
K.MK.M
700412
$endgroup$
$begingroup$
I'm a bit confused. The cited paragraph in blocked quotes is completely done and nothing more needs doing.
$endgroup$
– fleablood
Dec 21 '18 at 5:17
$begingroup$
It's correct, I was just trying to see how we go from $n le M$ to $n le M -1$
$endgroup$
– K.M
Dec 21 '18 at 5:39
1
$begingroup$
If $n-1 le M-1$ for all natural numbers then it is true for $(n+1)-1le M-1$ for all natural numbers. $n$ is not a fixed value. That's the whole point.
$endgroup$
– fleablood
Dec 21 '18 at 6:04
add a comment |
$begingroup$
The following proof is from https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch2.pdf
If $x in R$, then there exists $n in N$ such that $x < n$.
Suppose, for contradiction, that there exists $x in R$ such that $x > n$ for every
$n in N$. Then $x$ is an upper bound of $N$, so $N$ has a supremum $M = sup N in R$.
Since $n le M$ for every $n in N$, we have $n − 1 ≤ M − 1$ for every $n in N$, which implies that $n le M − 1$ for every $n in N$. But then $M − 1$ is an upper bound of $N$, which contradicts the assumption that $M$ is a least upper bound.
I'm not sure how exactly I'm supposed to use the fact that $n-1 le M-1$ for every $n in N$. I thought that I could show $n-1 le M-1 le M$ and go from there using a similar process shown below. I did the following:
$|n| le M$ implies that
$-M le n le M$
$-M + 1 le n + 1 le M +1$
so that
$-M le -M + 1 le n +1$
and
$-M le n+1 = |n+1|$
which implies that
$-M le n+1le M$
and from here I can show that $n le M-1$. I'm used to getting an interval from $|x| <M$, for example, which implies that $-M <x<M$. I'm not sure if my process is correct.
real-analysis
asked Dec 21 '18 at 4:54
K.MK.M
700412
$endgroup$
The following proof is from https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch2.pdf
If $x in R$, then there exists $n in N$ such that $x < n$.
Suppose, for contradiction, that there exists $x in R$ such that $x > n$ for every
$n in N$. Then $x$ is an upper bound of $N$, so $N$ has a supremum $M = sup N in R$.
Since $n le M$ for every $n in N$, we have $n − 1 ≤ M − 1$ for every $n in N$, which implies that $n le M − 1$ for every $n in N$. But then $M − 1$ is an upper bound of $N$, which contradicts the assumption that $M$ is a least upper bound.
I'm not sure how exactly I'm supposed to use the fact that $n-1 le M-1$ for every $n in N$. I thought that I could show $n-1 le M-1 le M$ and go from there using a similar process shown below. I did the following:
$|n| le M$ implies that
$-M le n le M$
$-M + 1 le n + 1 le M +1$
so that
$-M le -M + 1 le n +1$
and
$-M le n+1 = |n+1|$
which implies that
$-M le n+1le M$
and from here I can show that $n le M-1$. I'm used to getting an interval from $|x| <M$, for example, which implies that $-M <x<M$. I'm not sure if my process is correct.
real-analysis
real-analysis
asked Dec 21 '18 at 4:54
K.MK.M
700412
asked Dec 21 '18 at 4:54
K.MK.M
700412
edited Dec 21 '18 at 5:37
K.M
asked Dec 21 '18 at 4:54
K.MK.M
700412
asked Dec 21 '18 at 4:54
K.MK.M
700412
asked Dec 21 '18 at 4:54
K.MK.M
700412
700412
$begingroup$
I'm a bit confused. The cited paragraph in blocked quotes is completely done and nothing more needs doing.
$endgroup$
– fleablood
Dec 21 '18 at 5:17
$begingroup$
It's correct, I was just trying to see how we go from $n le M$ to $n le M -1$
$endgroup$
– K.M
Dec 21 '18 at 5:39
1
$begingroup$
If $n-1 le M-1$ for all natural numbers then it is true for $(n+1)-1le M-1$ for all natural numbers. $n$ is not a fixed value. That's the whole point.
$endgroup$
– fleablood
Dec 21 '18 at 6:04
add a comment |
$begingroup$
I'm a bit confused. The cited paragraph in blocked quotes is completely done and nothing more needs doing.
$endgroup$
– fleablood
Dec 21 '18 at 5:17
$begingroup$
It's correct, I was just trying to see how we go from $n le M$ to $n le M -1$
$endgroup$
– K.M
Dec 21 '18 at 5:39
1
$begingroup$
If $n-1 le M-1$ for all natural numbers then it is true for $(n+1)-1le M-1$ for all natural numbers. $n$ is not a fixed value. That's the whole point.
$endgroup$
– fleablood
Dec 21 '18 at 6:04
$begingroup$
I'm a bit confused. The cited paragraph in blocked quotes is completely done and nothing more needs doing.
$endgroup$
– fleablood
Dec 21 '18 at 5:17
$begingroup$
I'm a bit confused. The cited paragraph in blocked quotes is completely done and nothing more needs doing.
$endgroup$
– fleablood
Dec 21 '18 at 5:17
$begingroup$
It's correct, I was just trying to see how we go from $n le M$ to $n le M -1$
$endgroup$
– K.M
Dec 21 '18 at 5:39
$begingroup$
It's correct, I was just trying to see how we go from $n le M$ to $n le M -1$
$endgroup$
– K.M
Dec 21 '18 at 5:39
1
1
$begingroup$
If $n-1 le M-1$ for all natural numbers then it is true for $(n+1)-1le M-1$ for all natural numbers. $n$ is not a fixed value. That's the whole point.
$endgroup$
– fleablood
Dec 21 '18 at 6:04
$begingroup$
If $n-1 le M-1$ for all natural numbers then it is true for $(n+1)-1le M-1$ for all natural numbers. $n$ is not a fixed value. That's the whole point.
$endgroup$
– fleablood
Dec 21 '18 at 6:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I'm not sure how exactly I'm supposed to use the fact that n−1≤M−1 for every n∈N.
You are thinking way too hard. If $n-1le M-1$ for all $n$ then it is true for $(n+1)-1 = n le M-1$. So $nle M-1$ for all $n in mathbb N$.
So $M-1$ is an upper bound of $N$. But $M$ is the least upper bound. And that's a contradiction.
answered Dec 21 '18 at 5:21
fleabloodfleablood
71.6k22686
$endgroup$
$begingroup$
Even if I did more than necessary, is it correct?
$endgroup$
– K.M
Dec 21 '18 at 5:40
add a comment |
$begingroup$
I'm not sure you need to play tricks in order to prove this; in the axiomatic treatment of the reals you get the least upper bound property, from which you can derive the Archimedean property of the reals. Once you know that you cannot have infinitely large reals, and that the naturals are contained within the reals, you should be done.
answered Dec 21 '18 at 5:09
DanielJackDanielJack
661
$endgroup$
add a comment |
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2 Answers
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$begingroup$
I'm not sure how exactly I'm supposed to use the fact that n−1≤M−1 for every n∈N.
You are thinking way too hard. If $n-1le M-1$ for all $n$ then it is true for $(n+1)-1 = n le M-1$. So $nle M-1$ for all $n in mathbb N$.
So $M-1$ is an upper bound of $N$. But $M$ is the least upper bound. And that's a contradiction.
answered Dec 21 '18 at 5:21
fleabloodfleablood
71.6k22686
$endgroup$
$begingroup$
Even if I did more than necessary, is it correct?
$endgroup$
– K.M
Dec 21 '18 at 5:40
add a comment |
$begingroup$
I'm not sure how exactly I'm supposed to use the fact that n−1≤M−1 for every n∈N.
You are thinking way too hard. If $n-1le M-1$ for all $n$ then it is true for $(n+1)-1 = n le M-1$. So $nle M-1$ for all $n in mathbb N$.
So $M-1$ is an upper bound of $N$. But $M$ is the least upper bound. And that's a contradiction.
answered Dec 21 '18 at 5:21
fleabloodfleablood
71.6k22686
$endgroup$
$begingroup$
Even if I did more than necessary, is it correct?
$endgroup$
– K.M
Dec 21 '18 at 5:40
add a comment |
$begingroup$
I'm not sure how exactly I'm supposed to use the fact that n−1≤M−1 for every n∈N.
You are thinking way too hard. If $n-1le M-1$ for all $n$ then it is true for $(n+1)-1 = n le M-1$. So $nle M-1$ for all $n in mathbb N$.
So $M-1$ is an upper bound of $N$. But $M$ is the least upper bound. And that's a contradiction.
answered Dec 21 '18 at 5:21
fleabloodfleablood
71.6k22686
$endgroup$
I'm not sure how exactly I'm supposed to use the fact that n−1≤M−1 for every n∈N.
You are thinking way too hard. If $n-1le M-1$ for all $n$ then it is true for $(n+1)-1 = n le M-1$. So $nle M-1$ for all $n in mathbb N$.
So $M-1$ is an upper bound of $N$. But $M$ is the least upper bound. And that's a contradiction.
answered Dec 21 '18 at 5:21
fleabloodfleablood
71.6k22686
answered Dec 21 '18 at 5:21
fleabloodfleablood
71.6k22686
answered Dec 21 '18 at 5:21
fleabloodfleablood
71.6k22686
answered Dec 21 '18 at 5:21
fleabloodfleablood
71.6k22686
71.6k22686
$begingroup$
Even if I did more than necessary, is it correct?
$endgroup$
– K.M
Dec 21 '18 at 5:40
add a comment |
$begingroup$
Even if I did more than necessary, is it correct?
$endgroup$
– K.M
Dec 21 '18 at 5:40
$begingroup$
Even if I did more than necessary, is it correct?
$endgroup$
– K.M
Dec 21 '18 at 5:40
$begingroup$
Even if I did more than necessary, is it correct?
$endgroup$
– K.M
Dec 21 '18 at 5:40
add a comment |
$begingroup$
I'm not sure you need to play tricks in order to prove this; in the axiomatic treatment of the reals you get the least upper bound property, from which you can derive the Archimedean property of the reals. Once you know that you cannot have infinitely large reals, and that the naturals are contained within the reals, you should be done.
answered Dec 21 '18 at 5:09
DanielJackDanielJack
661
$endgroup$
add a comment |
$begingroup$
I'm not sure you need to play tricks in order to prove this; in the axiomatic treatment of the reals you get the least upper bound property, from which you can derive the Archimedean property of the reals. Once you know that you cannot have infinitely large reals, and that the naturals are contained within the reals, you should be done.
answered Dec 21 '18 at 5:09
DanielJackDanielJack
661
$endgroup$
add a comment |
$begingroup$
I'm not sure you need to play tricks in order to prove this; in the axiomatic treatment of the reals you get the least upper bound property, from which you can derive the Archimedean property of the reals. Once you know that you cannot have infinitely large reals, and that the naturals are contained within the reals, you should be done.
answered Dec 21 '18 at 5:09
DanielJackDanielJack
661
$endgroup$
I'm not sure you need to play tricks in order to prove this; in the axiomatic treatment of the reals you get the least upper bound property, from which you can derive the Archimedean property of the reals. Once you know that you cannot have infinitely large reals, and that the naturals are contained within the reals, you should be done.
answered Dec 21 '18 at 5:09
DanielJackDanielJack
661
answered Dec 21 '18 at 5:09
DanielJackDanielJack
661
answered Dec 21 '18 at 5:09
DanielJackDanielJack
661
answered Dec 21 '18 at 5:09
DanielJackDanielJack
661
661
add a comment |
add a comment |
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$begingroup$
I'm a bit confused. The cited paragraph in blocked quotes is completely done and nothing more needs doing.
$endgroup$
– fleablood
Dec 21 '18 at 5:17
$begingroup$
It's correct, I was just trying to see how we go from $n le M$ to $n le M -1$
$endgroup$
– K.M
Dec 21 '18 at 5:39
1
$begingroup$
If $n-1 le M-1$ for all natural numbers then it is true for $(n+1)-1le M-1$ for all natural numbers. $n$ is not a fixed value. That's the whole point.
$endgroup$
– fleablood
Dec 21 '18 at 6:04