Let $S$ be the set of all the numbers of the form $sum_{k=0}^{infty} frac{a_k}{3^k}$ where $a_k=0text{, }2$.
$begingroup$
Let $S$ be the set of all the numbers of the form $sum_{k=0}^{infty} frac{a_k}{3^k}$ where $a_k=0text{, }2$.
I want to answer the following question about $S$?
Is $S$ uncountable?
Is $S$ closed?
Attempt:
If I consider the ternary representation of reals, then $S=[0,2]-$ {all those expression where $a_k=1$}
When $a_k=1$ then we have $$sum_{k=0}^{infty}3^{-k}$$ which is a basic geometric series with common ratio $1/3$ and initial term 1. So it adds up to $3/2$
Then $S=[0,2]-{3/2}$ so it is uncountable.
Am I going the right way?
real-analysis
asked Nov 26 '18 at 13:27
StammeringMathematicianStammeringMathematician
2,3441322
$endgroup$
|
show 1 more comment
$begingroup$
Let $S$ be the set of all the numbers of the form $sum_{k=0}^{infty} frac{a_k}{3^k}$ where $a_k=0text{, }2$.
I want to answer the following question about $S$?
Is $S$ uncountable?
Is $S$ closed?
Attempt:
If I consider the ternary representation of reals, then $S=[0,2]-$ {all those expression where $a_k=1$}
When $a_k=1$ then we have $$sum_{k=0}^{infty}3^{-k}$$ which is a basic geometric series with common ratio $1/3$ and initial term 1. So it adds up to $3/2$
Then $S=[0,2]-{3/2}$ so it is uncountable.
Am I going the right way?
real-analysis
asked Nov 26 '18 at 13:27
StammeringMathematicianStammeringMathematician
2,3441322
$endgroup$
$begingroup$
Hint: argue that it is in bijection with the usual binary representations of numbers on $[0,1]$
$endgroup$
– lulu
Nov 26 '18 at 13:32
$begingroup$
@lulu Suppose I take the expression where a_k=2 for all $k$ then we have $sum 23^{-k}$ and this sums to $4/3$ which is not in $[0,1]$ ??
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:36
1
$begingroup$
So, change the bijection. Your string is just a string of $0's$ and $2's$ and such strings are clearly in bijection with strings of $0's$ and $1's$.
$endgroup$
– lulu
Nov 26 '18 at 13:42
$begingroup$
@lulu I have understood what you are trying to say. Thanks for that. Can you please check my approach and whatever I have written in the OP. I will be thankful.
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:50
$begingroup$
I don't understand what you wrote. Amongst the ternary numbers, the complement of those for which no $a_i=1$ is not just the single element for which all the $a_i=1$.
$endgroup$
– lulu
Nov 26 '18 at 13:52
|
show 1 more comment
$begingroup$
Let $S$ be the set of all the numbers of the form $sum_{k=0}^{infty} frac{a_k}{3^k}$ where $a_k=0text{, }2$.
I want to answer the following question about $S$?
Is $S$ uncountable?
Is $S$ closed?
Attempt:
If I consider the ternary representation of reals, then $S=[0,2]-$ {all those expression where $a_k=1$}
When $a_k=1$ then we have $$sum_{k=0}^{infty}3^{-k}$$ which is a basic geometric series with common ratio $1/3$ and initial term 1. So it adds up to $3/2$
Then $S=[0,2]-{3/2}$ so it is uncountable.
Am I going the right way?
real-analysis
asked Nov 26 '18 at 13:27
StammeringMathematicianStammeringMathematician
2,3441322
$endgroup$
Let $S$ be the set of all the numbers of the form $sum_{k=0}^{infty} frac{a_k}{3^k}$ where $a_k=0text{, }2$.
I want to answer the following question about $S$?
Is $S$ uncountable?
Is $S$ closed?
Attempt:
If I consider the ternary representation of reals, then $S=[0,2]-$ {all those expression where $a_k=1$}
When $a_k=1$ then we have $$sum_{k=0}^{infty}3^{-k}$$ which is a basic geometric series with common ratio $1/3$ and initial term 1. So it adds up to $3/2$
Then $S=[0,2]-{3/2}$ so it is uncountable.
Am I going the right way?
real-analysis
real-analysis
asked Nov 26 '18 at 13:27
StammeringMathematicianStammeringMathematician
2,3441322
asked Nov 26 '18 at 13:27
StammeringMathematicianStammeringMathematician
2,3441322
asked Nov 26 '18 at 13:27
StammeringMathematicianStammeringMathematician
2,3441322
asked Nov 26 '18 at 13:27
StammeringMathematicianStammeringMathematician
2,3441322
asked Nov 26 '18 at 13:27
StammeringMathematicianStammeringMathematician
2,3441322
2,3441322
$begingroup$
Hint: argue that it is in bijection with the usual binary representations of numbers on $[0,1]$
$endgroup$
– lulu
Nov 26 '18 at 13:32
$begingroup$
@lulu Suppose I take the expression where a_k=2 for all $k$ then we have $sum 23^{-k}$ and this sums to $4/3$ which is not in $[0,1]$ ??
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:36
1
$begingroup$
So, change the bijection. Your string is just a string of $0's$ and $2's$ and such strings are clearly in bijection with strings of $0's$ and $1's$.
$endgroup$
– lulu
Nov 26 '18 at 13:42
$begingroup$
@lulu I have understood what you are trying to say. Thanks for that. Can you please check my approach and whatever I have written in the OP. I will be thankful.
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:50
$begingroup$
I don't understand what you wrote. Amongst the ternary numbers, the complement of those for which no $a_i=1$ is not just the single element for which all the $a_i=1$.
$endgroup$
– lulu
Nov 26 '18 at 13:52
|
show 1 more comment
$begingroup$
Hint: argue that it is in bijection with the usual binary representations of numbers on $[0,1]$
$endgroup$
– lulu
Nov 26 '18 at 13:32
$begingroup$
@lulu Suppose I take the expression where a_k=2 for all $k$ then we have $sum 23^{-k}$ and this sums to $4/3$ which is not in $[0,1]$ ??
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:36
1
$begingroup$
So, change the bijection. Your string is just a string of $0's$ and $2's$ and such strings are clearly in bijection with strings of $0's$ and $1's$.
$endgroup$
– lulu
Nov 26 '18 at 13:42
$begingroup$
@lulu I have understood what you are trying to say. Thanks for that. Can you please check my approach and whatever I have written in the OP. I will be thankful.
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:50
$begingroup$
I don't understand what you wrote. Amongst the ternary numbers, the complement of those for which no $a_i=1$ is not just the single element for which all the $a_i=1$.
$endgroup$
– lulu
Nov 26 '18 at 13:52
$begingroup$
Hint: argue that it is in bijection with the usual binary representations of numbers on $[0,1]$
$endgroup$
– lulu
Nov 26 '18 at 13:32
$begingroup$
Hint: argue that it is in bijection with the usual binary representations of numbers on $[0,1]$
$endgroup$
– lulu
Nov 26 '18 at 13:32
$begingroup$
@lulu Suppose I take the expression where a_k=2 for all $k$ then we have $sum 23^{-k}$ and this sums to $4/3$ which is not in $[0,1]$ ??
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:36
$begingroup$
@lulu Suppose I take the expression where a_k=2 for all $k$ then we have $sum 23^{-k}$ and this sums to $4/3$ which is not in $[0,1]$ ??
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:36
1
1
$begingroup$
So, change the bijection. Your string is just a string of $0's$ and $2's$ and such strings are clearly in bijection with strings of $0's$ and $1's$.
$endgroup$
– lulu
Nov 26 '18 at 13:42
$begingroup$
So, change the bijection. Your string is just a string of $0's$ and $2's$ and such strings are clearly in bijection with strings of $0's$ and $1's$.
$endgroup$
– lulu
Nov 26 '18 at 13:42
$begingroup$
@lulu I have understood what you are trying to say. Thanks for that. Can you please check my approach and whatever I have written in the OP. I will be thankful.
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:50
$begingroup$
@lulu I have understood what you are trying to say. Thanks for that. Can you please check my approach and whatever I have written in the OP. I will be thankful.
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:50
$begingroup$
I don't understand what you wrote. Amongst the ternary numbers, the complement of those for which no $a_i=1$ is not just the single element for which all the $a_i=1$.
$endgroup$
– lulu
Nov 26 '18 at 13:52
$begingroup$
I don't understand what you wrote. Amongst the ternary numbers, the complement of those for which no $a_i=1$ is not just the single element for which all the $a_i=1$.
$endgroup$
– lulu
Nov 26 '18 at 13:52
|
show 1 more comment
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$begingroup$
Hint: argue that it is in bijection with the usual binary representations of numbers on $[0,1]$
$endgroup$
– lulu
Nov 26 '18 at 13:32
$begingroup$
@lulu Suppose I take the expression where a_k=2 for all $k$ then we have $sum 23^{-k}$ and this sums to $4/3$ which is not in $[0,1]$ ??
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:36
1
$begingroup$
So, change the bijection. Your string is just a string of $0's$ and $2's$ and such strings are clearly in bijection with strings of $0's$ and $1's$.
$endgroup$
– lulu
Nov 26 '18 at 13:42
$begingroup$
@lulu I have understood what you are trying to say. Thanks for that. Can you please check my approach and whatever I have written in the OP. I will be thankful.
$endgroup$
– StammeringMathematician
Nov 26 '18 at 13:50
$begingroup$
I don't understand what you wrote. Amongst the ternary numbers, the complement of those for which no $a_i=1$ is not just the single element for which all the $a_i=1$.
$endgroup$
– lulu
Nov 26 '18 at 13:52