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What does this Cantor space in $left[frac13,1right]$ look like?
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What does this Cantor space in $left[frac13,1right]$ look like? Does it look anything like $xinBbb Z[frac16]cap[frac13,1]:lvert xrvert_2geq1$ and if so, does it imply some p-adic type completion of $xinBbb Z[frac16]:lvert xrvert_2geq1}$ by $lvertcdotrvert_4$?
I'm looking at the numbers expressible in base $4$ using only the digits drawn from $1,3$
By the usual bijection to the unit interval it seems obvious it will be in the subset $left[frac13,1right]$ because the smallest number is $.overline1_4$ and the largest $.overline3_4$
It can be written $displaystylesum_n=0^infty fraca_n4^n+1$ with $a_nin1,3$
Or it could equally be written:
$displaystylesum_n=0^infty frac2a_n+14^n+1$ with $a_nin0,1$
I'm struggling to visualise whether the usual removing a centre third structure is preserved or whether it's something degenerate. My instinct says its construction follows the usual diagram of the Cantor set with the centre third removed rather than a half or quarter, but perhaps with the left hand edge tapering in towards the bottom-right at each step.
Or perhaps it's just $frac23cdot C+frac13$
A bit of a random shot in the dark but does this look anything like $xinBbb Z[frac16]cap[frac13,1]:lvert xrvert_2geq1$ and if so, does it imply some p-adic type completion of $Bbb Z[frac16]$ by a "4-adic" valuation $lvertcdotrvert_4$? (please note I'm aware $C$ is an uncountable set including limit points).
elementary-set-theory metric-spaces p-adic-number-theory cantor-set
asked Jul 28 at 9:20
Robert Frost
3,884936
add a comment |Â
up vote
-1
down vote
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What does this Cantor space in $left[frac13,1right]$ look like? Does it look anything like $xinBbb Z[frac16]cap[frac13,1]:lvert xrvert_2geq1$ and if so, does it imply some p-adic type completion of $xinBbb Z[frac16]:lvert xrvert_2geq1}$ by $lvertcdotrvert_4$?
I'm looking at the numbers expressible in base $4$ using only the digits drawn from $1,3$
By the usual bijection to the unit interval it seems obvious it will be in the subset $left[frac13,1right]$ because the smallest number is $.overline1_4$ and the largest $.overline3_4$
It can be written $displaystylesum_n=0^infty fraca_n4^n+1$ with $a_nin1,3$
Or it could equally be written:
$displaystylesum_n=0^infty frac2a_n+14^n+1$ with $a_nin0,1$
I'm struggling to visualise whether the usual removing a centre third structure is preserved or whether it's something degenerate. My instinct says its construction follows the usual diagram of the Cantor set with the centre third removed rather than a half or quarter, but perhaps with the left hand edge tapering in towards the bottom-right at each step.
Or perhaps it's just $frac23cdot C+frac13$
A bit of a random shot in the dark but does this look anything like $xinBbb Z[frac16]cap[frac13,1]:lvert xrvert_2geq1$ and if so, does it imply some p-adic type completion of $Bbb Z[frac16]$ by a "4-adic" valuation $lvertcdotrvert_4$? (please note I'm aware $C$ is an uncountable set including limit points).
elementary-set-theory metric-spaces p-adic-number-theory cantor-set
asked Jul 28 at 9:20
Robert Frost
3,884936
1
You start your question by saying "this Cantor space", but there's nothing for the word "this" to point back to. Which Cantor space are you talking about?
– Henning Makholm
Jul 28 at 11:09
@HenningMakholm sorry "I'm looking at the numbers expressible in base $4$ using only the digits drawn from $1,3$" I've given it more prominence now, it was a bit lost.
– Robert Frost
Jul 28 at 11:10
Why are you only defining what you're talking about halfway down the question, instead of before you begin to ask questions about it?
– Henning Makholm
Jul 28 at 11:11
1
@HenningMakholm because I'm an idiot
– Robert Frost
Jul 28 at 11:11
@HenningMakholm ...it's starting to look like it's $[frac13,1]$ with the centre half progressively removed.
– Robert Frost
Jul 28 at 11:12
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
What does this Cantor space in $left[frac13,1right]$ look like? Does it look anything like $xinBbb Z[frac16]cap[frac13,1]:lvert xrvert_2geq1$ and if so, does it imply some p-adic type completion of $xinBbb Z[frac16]:lvert xrvert_2geq1}$ by $lvertcdotrvert_4$?
I'm looking at the numbers expressible in base $4$ using only the digits drawn from $1,3$
By the usual bijection to the unit interval it seems obvious it will be in the subset $left[frac13,1right]$ because the smallest number is $.overline1_4$ and the largest $.overline3_4$
It can be written $displaystylesum_n=0^infty fraca_n4^n+1$ with $a_nin1,3$
Or it could equally be written:
$displaystylesum_n=0^infty frac2a_n+14^n+1$ with $a_nin0,1$
I'm struggling to visualise whether the usual removing a centre third structure is preserved or whether it's something degenerate. My instinct says its construction follows the usual diagram of the Cantor set with the centre third removed rather than a half or quarter, but perhaps with the left hand edge tapering in towards the bottom-right at each step.
Or perhaps it's just $frac23cdot C+frac13$
A bit of a random shot in the dark but does this look anything like $xinBbb Z[frac16]cap[frac13,1]:lvert xrvert_2geq1$ and if so, does it imply some p-adic type completion of $Bbb Z[frac16]$ by a "4-adic" valuation $lvertcdotrvert_4$? (please note I'm aware $C$ is an uncountable set including limit points).
elementary-set-theory metric-spaces p-adic-number-theory cantor-set
asked Jul 28 at 9:20
Robert Frost
3,884936
What does this Cantor space in $left[frac13,1right]$ look like? Does it look anything like $xinBbb Z[frac16]cap[frac13,1]:lvert xrvert_2geq1$ and if so, does it imply some p-adic type completion of $xinBbb Z[frac16]:lvert xrvert_2geq1}$ by $lvertcdotrvert_4$?
I'm looking at the numbers expressible in base $4$ using only the digits drawn from $1,3$
By the usual bijection to the unit interval it seems obvious it will be in the subset $left[frac13,1right]$ because the smallest number is $.overline1_4$ and the largest $.overline3_4$
It can be written $displaystylesum_n=0^infty fraca_n4^n+1$ with $a_nin1,3$
Or it could equally be written:
$displaystylesum_n=0^infty frac2a_n+14^n+1$ with $a_nin0,1$
I'm struggling to visualise whether the usual removing a centre third structure is preserved or whether it's something degenerate. My instinct says its construction follows the usual diagram of the Cantor set with the centre third removed rather than a half or quarter, but perhaps with the left hand edge tapering in towards the bottom-right at each step.
Or perhaps it's just $frac23cdot C+frac13$
A bit of a random shot in the dark but does this look anything like $xinBbb Z[frac16]cap[frac13,1]:lvert xrvert_2geq1$ and if so, does it imply some p-adic type completion of $Bbb Z[frac16]$ by a "4-adic" valuation $lvertcdotrvert_4$? (please note I'm aware $C$ is an uncountable set including limit points).
elementary-set-theory metric-spaces p-adic-number-theory cantor-set
asked Jul 28 at 9:20
Robert Frost
3,884936
edited Jul 28 at 11:10
asked Jul 28 at 9:20
Robert Frost
3,884936
asked Jul 28 at 9:20
Robert Frost
3,884936
asked Jul 28 at 9:20
Robert Frost
3,884936
3,884936
1
You start your question by saying "this Cantor space", but there's nothing for the word "this" to point back to. Which Cantor space are you talking about?
– Henning Makholm
Jul 28 at 11:09
@HenningMakholm sorry "I'm looking at the numbers expressible in base $4$ using only the digits drawn from $1,3$" I've given it more prominence now, it was a bit lost.
– Robert Frost
Jul 28 at 11:10
Why are you only defining what you're talking about halfway down the question, instead of before you begin to ask questions about it?
– Henning Makholm
Jul 28 at 11:11
1
@HenningMakholm because I'm an idiot
– Robert Frost
Jul 28 at 11:11
@HenningMakholm ...it's starting to look like it's $[frac13,1]$ with the centre half progressively removed.
– Robert Frost
Jul 28 at 11:12
add a comment |Â
1
You start your question by saying "this Cantor space", but there's nothing for the word "this" to point back to. Which Cantor space are you talking about?
– Henning Makholm
Jul 28 at 11:09
@HenningMakholm sorry "I'm looking at the numbers expressible in base $4$ using only the digits drawn from $1,3$" I've given it more prominence now, it was a bit lost.
– Robert Frost
Jul 28 at 11:10
Why are you only defining what you're talking about halfway down the question, instead of before you begin to ask questions about it?
– Henning Makholm
Jul 28 at 11:11
1
@HenningMakholm because I'm an idiot
– Robert Frost
Jul 28 at 11:11
@HenningMakholm ...it's starting to look like it's $[frac13,1]$ with the centre half progressively removed.
– Robert Frost
Jul 28 at 11:12
1
1
You start your question by saying "this Cantor space", but there's nothing for the word "this" to point back to. Which Cantor space are you talking about?
– Henning Makholm
Jul 28 at 11:09
You start your question by saying "this Cantor space", but there's nothing for the word "this" to point back to. Which Cantor space are you talking about?
– Henning Makholm
Jul 28 at 11:09
@HenningMakholm sorry "I'm looking at the numbers expressible in base $4$ using only the digits drawn from $1,3$" I've given it more prominence now, it was a bit lost.
– Robert Frost
Jul 28 at 11:10
@HenningMakholm sorry "I'm looking at the numbers expressible in base $4$ using only the digits drawn from $1,3$" I've given it more prominence now, it was a bit lost.
– Robert Frost
Jul 28 at 11:10
Why are you only defining what you're talking about halfway down the question, instead of before you begin to ask questions about it?
– Henning Makholm
Jul 28 at 11:11
Why are you only defining what you're talking about halfway down the question, instead of before you begin to ask questions about it?
– Henning Makholm
Jul 28 at 11:11
1
1
@HenningMakholm because I'm an idiot
– Robert Frost
Jul 28 at 11:11
@HenningMakholm because I'm an idiot
– Robert Frost
Jul 28 at 11:11
@HenningMakholm ...it's starting to look like it's $[frac13,1]$ with the centre half progressively removed.
– Robert Frost
Jul 28 at 11:12
@HenningMakholm ...it's starting to look like it's $[frac13,1]$ with the centre half progressively removed.
– Robert Frost
Jul 28 at 11:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your construction is closely analogous to the usual middle-thirds Cantor set, except at each step you remove the middle half, leaving a fourth at each end.
To wit, if you take your set, scale it by $frac14$ and add either $frac 14$ or $frac34$, you get the two self-similar components of the set.
The inner endpoints of those components are $0.1333ldots_4=0.2_4=frac 12$ and $0.3111ldots_4 = frac56$, so each has length $frac16$ and there is $frac26$ of removed interval between them.
answered Jul 28 at 11:20
Henning Makholm
225k16290516
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your construction is closely analogous to the usual middle-thirds Cantor set, except at each step you remove the middle half, leaving a fourth at each end.
To wit, if you take your set, scale it by $frac14$ and add either $frac 14$ or $frac34$, you get the two self-similar components of the set.
The inner endpoints of those components are $0.1333ldots_4=0.2_4=frac 12$ and $0.3111ldots_4 = frac56$, so each has length $frac16$ and there is $frac26$ of removed interval between them.
answered Jul 28 at 11:20
Henning Makholm
225k16290516
add a comment |Â
up vote
1
down vote
accepted
Your construction is closely analogous to the usual middle-thirds Cantor set, except at each step you remove the middle half, leaving a fourth at each end.
To wit, if you take your set, scale it by $frac14$ and add either $frac 14$ or $frac34$, you get the two self-similar components of the set.
The inner endpoints of those components are $0.1333ldots_4=0.2_4=frac 12$ and $0.3111ldots_4 = frac56$, so each has length $frac16$ and there is $frac26$ of removed interval between them.
answered Jul 28 at 11:20
Henning Makholm
225k16290516
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your construction is closely analogous to the usual middle-thirds Cantor set, except at each step you remove the middle half, leaving a fourth at each end.
To wit, if you take your set, scale it by $frac14$ and add either $frac 14$ or $frac34$, you get the two self-similar components of the set.
The inner endpoints of those components are $0.1333ldots_4=0.2_4=frac 12$ and $0.3111ldots_4 = frac56$, so each has length $frac16$ and there is $frac26$ of removed interval between them.
answered Jul 28 at 11:20
Henning Makholm
225k16290516
Your construction is closely analogous to the usual middle-thirds Cantor set, except at each step you remove the middle half, leaving a fourth at each end.
To wit, if you take your set, scale it by $frac14$ and add either $frac 14$ or $frac34$, you get the two self-similar components of the set.
The inner endpoints of those components are $0.1333ldots_4=0.2_4=frac 12$ and $0.3111ldots_4 = frac56$, so each has length $frac16$ and there is $frac26$ of removed interval between them.
answered Jul 28 at 11:20
Henning Makholm
225k16290516
edited Jul 28 at 14:21
answered Jul 28 at 11:20
Henning Makholm
225k16290516
answered Jul 28 at 11:20
Henning Makholm
225k16290516
answered Jul 28 at 11:20
Henning Makholm
225k16290516
225k16290516
add a comment |Â
add a comment |Â
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1
You start your question by saying "this Cantor space", but there's nothing for the word "this" to point back to. Which Cantor space are you talking about?
– Henning Makholm
Jul 28 at 11:09
@HenningMakholm sorry "I'm looking at the numbers expressible in base $4$ using only the digits drawn from $1,3$" I've given it more prominence now, it was a bit lost.
– Robert Frost
Jul 28 at 11:10
Why are you only defining what you're talking about halfway down the question, instead of before you begin to ask questions about it?
– Henning Makholm
Jul 28 at 11:11
1
@HenningMakholm because I'm an idiot
– Robert Frost
Jul 28 at 11:11
@HenningMakholm ...it's starting to look like it's $[frac13,1]$ with the centre half progressively removed.
– Robert Frost
Jul 28 at 11:12