Mixing
Show that if $2^kin Sspacespaceforall kinmathbb{N}$, and if $kin S$ and $kge2$, then $k-1in S$, then...
$begingroup$
Show that if $2^kin Sspacespaceforall kinmathbb{N}$, and if $kin S$ and $kge2$, then $k-1in S$, then $S=mathbb{N}$
My attempt:
Let $ninmathbb{N},$ hence $2^ninmathbb{N}$ by the first assumption. Now denote the set $A$ by the following: $$A:={ainmathbb{N}: a<2^nspacetext{and}space anot in S}$$
Because $Asubsetmathbb{N},$ it follows by the Well-Ordering Principle that if $A$ is bounded above it has a maximal element. Call that element $k$. By construction $k+1notin A$...
Here is where I am stuck. I would like to argue that $k+1in S$ since $k+1notin A$ ...
Any help from here would be appreciated!
real-analysis induction natural-numbers
asked Jan 22 at 6:16
livingtolearn-learningtolivelivingtolearn-learningtolive
34412
$endgroup$
add a comment |
$begingroup$
Show that if $2^kin Sspacespaceforall kinmathbb{N}$, and if $kin S$ and $kge2$, then $k-1in S$, then $S=mathbb{N}$
My attempt:
Let $ninmathbb{N},$ hence $2^ninmathbb{N}$ by the first assumption. Now denote the set $A$ by the following: $$A:={ainmathbb{N}: a<2^nspacetext{and}space anot in S}$$
Because $Asubsetmathbb{N},$ it follows by the Well-Ordering Principle that if $A$ is bounded above it has a maximal element. Call that element $k$. By construction $k+1notin A$...
Here is where I am stuck. I would like to argue that $k+1in S$ since $k+1notin A$ ...
Any help from here would be appreciated!
real-analysis induction natural-numbers
asked Jan 22 at 6:16
livingtolearn-learningtolivelivingtolearn-learningtolive
34412
$endgroup$
$begingroup$
Looks like your set doesn't have to contain $0$. $S$ is either $mathbb{N}$ or $mathbb{N}^*$
$endgroup$
– RcnSc
Jan 22 at 16:22
$begingroup$
(Just suggesting a simpler argument.) Suppose $S ne mathbb{N}$, i.e. there exists $m in mathbb{N} setminus S$. Assuming "$mathbb{N}$" does not include $0$, the second premise implies (i) $m + 1 in mathbb{N} setminus S$, and (ii) for all $n in mathbb{N}$, if $m + n in mathbb{N} setminus S$, then $m + n + 1 in mathbb{N} setminus S$. By induction on $n$, therefore, $m + n in mathbb{N} setminus S$ for all $n in mathbb{N}$. But $2^m > m$, so we can take $n = 2^m - m$. This implies $2^m in mathbb{N} setminus S$, contradicting the first premise. Therefore $S = mathbb{N}$.
$endgroup$
– Calum Gilhooley
Jan 22 at 18:48
add a comment |
$begingroup$
Show that if $2^kin Sspacespaceforall kinmathbb{N}$, and if $kin S$ and $kge2$, then $k-1in S$, then $S=mathbb{N}$
My attempt:
Let $ninmathbb{N},$ hence $2^ninmathbb{N}$ by the first assumption. Now denote the set $A$ by the following: $$A:={ainmathbb{N}: a<2^nspacetext{and}space anot in S}$$
Because $Asubsetmathbb{N},$ it follows by the Well-Ordering Principle that if $A$ is bounded above it has a maximal element. Call that element $k$. By construction $k+1notin A$...
Here is where I am stuck. I would like to argue that $k+1in S$ since $k+1notin A$ ...
Any help from here would be appreciated!
real-analysis induction natural-numbers
asked Jan 22 at 6:16
livingtolearn-learningtolivelivingtolearn-learningtolive
34412
$endgroup$
Show that if $2^kin Sspacespaceforall kinmathbb{N}$, and if $kin S$ and $kge2$, then $k-1in S$, then $S=mathbb{N}$
My attempt:
Let $ninmathbb{N},$ hence $2^ninmathbb{N}$ by the first assumption. Now denote the set $A$ by the following: $$A:={ainmathbb{N}: a<2^nspacetext{and}space anot in S}$$
Because $Asubsetmathbb{N},$ it follows by the Well-Ordering Principle that if $A$ is bounded above it has a maximal element. Call that element $k$. By construction $k+1notin A$...
Here is where I am stuck. I would like to argue that $k+1in S$ since $k+1notin A$ ...
Any help from here would be appreciated!
real-analysis induction natural-numbers
real-analysis induction natural-numbers
asked Jan 22 at 6:16
livingtolearn-learningtolivelivingtolearn-learningtolive
34412
asked Jan 22 at 6:16
livingtolearn-learningtolivelivingtolearn-learningtolive
34412
edited Jan 22 at 11:07
livingtolearn-learningtolive
asked Jan 22 at 6:16
livingtolearn-learningtolivelivingtolearn-learningtolive
34412
asked Jan 22 at 6:16
livingtolearn-learningtolivelivingtolearn-learningtolive
34412
asked Jan 22 at 6:16
livingtolearn-learningtolivelivingtolearn-learningtolive
34412
34412
$begingroup$
Looks like your set doesn't have to contain $0$. $S$ is either $mathbb{N}$ or $mathbb{N}^*$
$endgroup$
– RcnSc
Jan 22 at 16:22
$begingroup$
(Just suggesting a simpler argument.) Suppose $S ne mathbb{N}$, i.e. there exists $m in mathbb{N} setminus S$. Assuming "$mathbb{N}$" does not include $0$, the second premise implies (i) $m + 1 in mathbb{N} setminus S$, and (ii) for all $n in mathbb{N}$, if $m + n in mathbb{N} setminus S$, then $m + n + 1 in mathbb{N} setminus S$. By induction on $n$, therefore, $m + n in mathbb{N} setminus S$ for all $n in mathbb{N}$. But $2^m > m$, so we can take $n = 2^m - m$. This implies $2^m in mathbb{N} setminus S$, contradicting the first premise. Therefore $S = mathbb{N}$.
$endgroup$
– Calum Gilhooley
Jan 22 at 18:48
add a comment |
$begingroup$
Looks like your set doesn't have to contain $0$. $S$ is either $mathbb{N}$ or $mathbb{N}^*$
$endgroup$
– RcnSc
Jan 22 at 16:22
$begingroup$
(Just suggesting a simpler argument.) Suppose $S ne mathbb{N}$, i.e. there exists $m in mathbb{N} setminus S$. Assuming "$mathbb{N}$" does not include $0$, the second premise implies (i) $m + 1 in mathbb{N} setminus S$, and (ii) for all $n in mathbb{N}$, if $m + n in mathbb{N} setminus S$, then $m + n + 1 in mathbb{N} setminus S$. By induction on $n$, therefore, $m + n in mathbb{N} setminus S$ for all $n in mathbb{N}$. But $2^m > m$, so we can take $n = 2^m - m$. This implies $2^m in mathbb{N} setminus S$, contradicting the first premise. Therefore $S = mathbb{N}$.
$endgroup$
– Calum Gilhooley
Jan 22 at 18:48
$begingroup$
Looks like your set doesn't have to contain $0$. $S$ is either $mathbb{N}$ or $mathbb{N}^*$
$endgroup$
– RcnSc
Jan 22 at 16:22
$begingroup$
Looks like your set doesn't have to contain $0$. $S$ is either $mathbb{N}$ or $mathbb{N}^*$
$endgroup$
– RcnSc
Jan 22 at 16:22
$begingroup$
(Just suggesting a simpler argument.) Suppose $S ne mathbb{N}$, i.e. there exists $m in mathbb{N} setminus S$. Assuming "$mathbb{N}$" does not include $0$, the second premise implies (i) $m + 1 in mathbb{N} setminus S$, and (ii) for all $n in mathbb{N}$, if $m + n in mathbb{N} setminus S$, then $m + n + 1 in mathbb{N} setminus S$. By induction on $n$, therefore, $m + n in mathbb{N} setminus S$ for all $n in mathbb{N}$. But $2^m > m$, so we can take $n = 2^m - m$. This implies $2^m in mathbb{N} setminus S$, contradicting the first premise. Therefore $S = mathbb{N}$.
$endgroup$
– Calum Gilhooley
Jan 22 at 18:48
$begingroup$
(Just suggesting a simpler argument.) Suppose $S ne mathbb{N}$, i.e. there exists $m in mathbb{N} setminus S$. Assuming "$mathbb{N}$" does not include $0$, the second premise implies (i) $m + 1 in mathbb{N} setminus S$, and (ii) for all $n in mathbb{N}$, if $m + n in mathbb{N} setminus S$, then $m + n + 1 in mathbb{N} setminus S$. By induction on $n$, therefore, $m + n in mathbb{N} setminus S$ for all $n in mathbb{N}$. But $2^m > m$, so we can take $n = 2^m - m$. This implies $2^m in mathbb{N} setminus S$, contradicting the first premise. Therefore $S = mathbb{N}$.
$endgroup$
– Calum Gilhooley
Jan 22 at 18:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For every $n in mathbb{N}$, if there is a $k in mathbb{N}$ such that $2^k = n$, then we know it's in $S$. Otherwise, there exists a $k in mathbb{N}$ such that $2^k > n$ as the set of values of $2^k$ has no upper bound. Choose the smallest such $k$ (although you can choose any larger $k$) and call $2^k = m$, where $m$ is in $S$. As $k in mathbb{N}$, we have that $m ge 2$, so we know that $m - 1$ is also in $S$. You repeat this step $m - n$ times, with the $m$ value of each next step being replaced by the $m - 1$ of the previous step, at which time you will get down to $n$, with this being assured to occur as $n ge 1$ and the steps stop at $m = 2$ giving that $m - 1 = 1$ is in $S$. This shows that all values $n in mathbb{N}$ have that $n in S$, thus giving that $S = mathbb{N}$.
There is a likely a simpler & more succinct way to show this, but this is not my area of expertise. Also, I hope I'm not misinterpreting anything, nor assuming or using anything I shouldn't be.
edited Jan 22 at 15:57
Jordan Green
1,133410
answered Jan 22 at 6:47
John OmielanJohn Omielan
3,7801215
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For every $n in mathbb{N}$, if there is a $k in mathbb{N}$ such that $2^k = n$, then we know it's in $S$. Otherwise, there exists a $k in mathbb{N}$ such that $2^k > n$ as the set of values of $2^k$ has no upper bound. Choose the smallest such $k$ (although you can choose any larger $k$) and call $2^k = m$, where $m$ is in $S$. As $k in mathbb{N}$, we have that $m ge 2$, so we know that $m - 1$ is also in $S$. You repeat this step $m - n$ times, with the $m$ value of each next step being replaced by the $m - 1$ of the previous step, at which time you will get down to $n$, with this being assured to occur as $n ge 1$ and the steps stop at $m = 2$ giving that $m - 1 = 1$ is in $S$. This shows that all values $n in mathbb{N}$ have that $n in S$, thus giving that $S = mathbb{N}$.
There is a likely a simpler & more succinct way to show this, but this is not my area of expertise. Also, I hope I'm not misinterpreting anything, nor assuming or using anything I shouldn't be.
edited Jan 22 at 15:57
Jordan Green
1,133410
answered Jan 22 at 6:47
John OmielanJohn Omielan
3,7801215
$endgroup$
add a comment |
$begingroup$
For every $n in mathbb{N}$, if there is a $k in mathbb{N}$ such that $2^k = n$, then we know it's in $S$. Otherwise, there exists a $k in mathbb{N}$ such that $2^k > n$ as the set of values of $2^k$ has no upper bound. Choose the smallest such $k$ (although you can choose any larger $k$) and call $2^k = m$, where $m$ is in $S$. As $k in mathbb{N}$, we have that $m ge 2$, so we know that $m - 1$ is also in $S$. You repeat this step $m - n$ times, with the $m$ value of each next step being replaced by the $m - 1$ of the previous step, at which time you will get down to $n$, with this being assured to occur as $n ge 1$ and the steps stop at $m = 2$ giving that $m - 1 = 1$ is in $S$. This shows that all values $n in mathbb{N}$ have that $n in S$, thus giving that $S = mathbb{N}$.
There is a likely a simpler & more succinct way to show this, but this is not my area of expertise. Also, I hope I'm not misinterpreting anything, nor assuming or using anything I shouldn't be.
edited Jan 22 at 15:57
Jordan Green
1,133410
answered Jan 22 at 6:47
John OmielanJohn Omielan
3,7801215
$endgroup$
add a comment |
$begingroup$
For every $n in mathbb{N}$, if there is a $k in mathbb{N}$ such that $2^k = n$, then we know it's in $S$. Otherwise, there exists a $k in mathbb{N}$ such that $2^k > n$ as the set of values of $2^k$ has no upper bound. Choose the smallest such $k$ (although you can choose any larger $k$) and call $2^k = m$, where $m$ is in $S$. As $k in mathbb{N}$, we have that $m ge 2$, so we know that $m - 1$ is also in $S$. You repeat this step $m - n$ times, with the $m$ value of each next step being replaced by the $m - 1$ of the previous step, at which time you will get down to $n$, with this being assured to occur as $n ge 1$ and the steps stop at $m = 2$ giving that $m - 1 = 1$ is in $S$. This shows that all values $n in mathbb{N}$ have that $n in S$, thus giving that $S = mathbb{N}$.
There is a likely a simpler & more succinct way to show this, but this is not my area of expertise. Also, I hope I'm not misinterpreting anything, nor assuming or using anything I shouldn't be.
edited Jan 22 at 15:57
Jordan Green
1,133410
answered Jan 22 at 6:47
John OmielanJohn Omielan
3,7801215
$endgroup$
For every $n in mathbb{N}$, if there is a $k in mathbb{N}$ such that $2^k = n$, then we know it's in $S$. Otherwise, there exists a $k in mathbb{N}$ such that $2^k > n$ as the set of values of $2^k$ has no upper bound. Choose the smallest such $k$ (although you can choose any larger $k$) and call $2^k = m$, where $m$ is in $S$. As $k in mathbb{N}$, we have that $m ge 2$, so we know that $m - 1$ is also in $S$. You repeat this step $m - n$ times, with the $m$ value of each next step being replaced by the $m - 1$ of the previous step, at which time you will get down to $n$, with this being assured to occur as $n ge 1$ and the steps stop at $m = 2$ giving that $m - 1 = 1$ is in $S$. This shows that all values $n in mathbb{N}$ have that $n in S$, thus giving that $S = mathbb{N}$.
There is a likely a simpler & more succinct way to show this, but this is not my area of expertise. Also, I hope I'm not misinterpreting anything, nor assuming or using anything I shouldn't be.
edited Jan 22 at 15:57
Jordan Green
1,133410
answered Jan 22 at 6:47
John OmielanJohn Omielan
3,7801215
edited Jan 22 at 15:57
Jordan Green
1,133410
edited Jan 22 at 15:57
Jordan Green
1,133410
edited Jan 22 at 15:57
Jordan Green
1,133410
1,133410
answered Jan 22 at 6:47
John OmielanJohn Omielan
3,7801215
answered Jan 22 at 6:47
John OmielanJohn Omielan
3,7801215
answered Jan 22 at 6:47
John OmielanJohn Omielan
3,7801215
3,7801215
add a comment |
add a comment |
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$begingroup$
Looks like your set doesn't have to contain $0$. $S$ is either $mathbb{N}$ or $mathbb{N}^*$
$endgroup$
– RcnSc
Jan 22 at 16:22
$begingroup$
(Just suggesting a simpler argument.) Suppose $S ne mathbb{N}$, i.e. there exists $m in mathbb{N} setminus S$. Assuming "$mathbb{N}$" does not include $0$, the second premise implies (i) $m + 1 in mathbb{N} setminus S$, and (ii) for all $n in mathbb{N}$, if $m + n in mathbb{N} setminus S$, then $m + n + 1 in mathbb{N} setminus S$. By induction on $n$, therefore, $m + n in mathbb{N} setminus S$ for all $n in mathbb{N}$. But $2^m > m$, so we can take $n = 2^m - m$. This implies $2^m in mathbb{N} setminus S$, contradicting the first premise. Therefore $S = mathbb{N}$.
$endgroup$
– Calum Gilhooley
Jan 22 at 18:48