Mixing
Space filling curve whose all level sets are finite (countable)
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
favorite
Is there a continuous surjective function $f:[0,1] to [0,1]^2$ such that
every level set $f^-1(y)$ is a finite set? If the answer is no, what about if we replace the finiteness of level sets by "countable level sets"?
gn.general-topology space-filling-curves
edited Aug 6 at 7:06
Martin Sleziak
2,52032026
asked Aug 6 at 6:54
Ali Taghavi
19151880
add a comment |Â
up vote
7
down vote
favorite
Is there a continuous surjective function $f:[0,1] to [0,1]^2$ such that
every level set $f^-1(y)$ is a finite set? If the answer is no, what about if we replace the finiteness of level sets by "countable level sets"?
gn.general-topology space-filling-curves
edited Aug 6 at 7:06
Martin Sleziak
2,52032026
asked Aug 6 at 6:54
Ali Taghavi
19151880
2
If I'm not wrong the Peano square filling-curve covers any point at most $4$ times; this follows easily from the analytic description of it
– Pietro Majer
Aug 6 at 7:27
2
You may look at ac.els-cdn.com/S0166864197002708/… They construct a map from $S^1$ to $S^2$, but you may restrict to a subinterval. More generally, the theory of Cannon-Thurston maps constructs Peano curves, where preimages of points typically seem to have cardinality $1$ or $2$.
– ThiKu
Aug 6 at 7:32
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Is there a continuous surjective function $f:[0,1] to [0,1]^2$ such that
every level set $f^-1(y)$ is a finite set? If the answer is no, what about if we replace the finiteness of level sets by "countable level sets"?
gn.general-topology space-filling-curves
edited Aug 6 at 7:06
Martin Sleziak
2,52032026
asked Aug 6 at 6:54
Ali Taghavi
19151880
Is there a continuous surjective function $f:[0,1] to [0,1]^2$ such that
every level set $f^-1(y)$ is a finite set? If the answer is no, what about if we replace the finiteness of level sets by "countable level sets"?
gn.general-topology space-filling-curves
edited Aug 6 at 7:06
Martin Sleziak
2,52032026
asked Aug 6 at 6:54
Ali Taghavi
19151880
edited Aug 6 at 7:06
Martin Sleziak
2,52032026
edited Aug 6 at 7:06
Martin Sleziak
2,52032026
edited Aug 6 at 7:06
Martin Sleziak
2,52032026
2,52032026
asked Aug 6 at 6:54
Ali Taghavi
19151880
asked Aug 6 at 6:54
Ali Taghavi
19151880
asked Aug 6 at 6:54
Ali Taghavi
19151880
19151880
2
If I'm not wrong the Peano square filling-curve covers any point at most $4$ times; this follows easily from the analytic description of it
– Pietro Majer
Aug 6 at 7:27
2
You may look at ac.els-cdn.com/S0166864197002708/… They construct a map from $S^1$ to $S^2$, but you may restrict to a subinterval. More generally, the theory of Cannon-Thurston maps constructs Peano curves, where preimages of points typically seem to have cardinality $1$ or $2$.
– ThiKu
Aug 6 at 7:32
add a comment |Â
2
If I'm not wrong the Peano square filling-curve covers any point at most $4$ times; this follows easily from the analytic description of it
– Pietro Majer
Aug 6 at 7:27
2
You may look at ac.els-cdn.com/S0166864197002708/… They construct a map from $S^1$ to $S^2$, but you may restrict to a subinterval. More generally, the theory of Cannon-Thurston maps constructs Peano curves, where preimages of points typically seem to have cardinality $1$ or $2$.
– ThiKu
Aug 6 at 7:32
2
2
If I'm not wrong the Peano square filling-curve covers any point at most $4$ times; this follows easily from the analytic description of it
– Pietro Majer
Aug 6 at 7:27
If I'm not wrong the Peano square filling-curve covers any point at most $4$ times; this follows easily from the analytic description of it
– Pietro Majer
Aug 6 at 7:27
2
2
You may look at ac.els-cdn.com/S0166864197002708/… They construct a map from $S^1$ to $S^2$, but you may restrict to a subinterval. More generally, the theory of Cannon-Thurston maps constructs Peano curves, where preimages of points typically seem to have cardinality $1$ or $2$.
– ThiKu
Aug 6 at 7:32
You may look at ac.els-cdn.com/S0166864197002708/… They construct a map from $S^1$ to $S^2$, but you may restrict to a subinterval. More generally, the theory of Cannon-Thurston maps constructs Peano curves, where preimages of points typically seem to have cardinality $1$ or $2$.
– ThiKu
Aug 6 at 7:32
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
11
down vote
accepted
Recall the definition of the Peano square-filling curve $f:[0,1]to[0,1]^2$, which is given in terms infinite ternary strings. If $ain [0,1]$ has a base $3$ representation of the form $0,a_1a_2a_3dots$, the point $f(a):=(b,c)$ has base $3$ digits resp. $$b_n:=bf k^a_2+a_4+dots a_2n-2a_2n-1$$
$$c_n:=bf k^a_1+a_3+dots a_2n-1a_2n$$
where $bf k$ is the involutory bijection of the set $0,1,2$ into itself given by $imapsto2-i$ (and exponents denote iterated composition, which in this case only depends on the parity of the exponent, since $bf k^2=bf id$). As Peano observes, this gives a bijection (actually a homeomorphism wrto product topologies of discrete spaces) $0,1,2^mathbbN to 0,1,2^mathbbNtimes 0,1,2^mathbbN$. In fact
$$a_2n-1:=bf k^c_1+c_2+dots +c_n-1b_n$$
$$a_2n:=bf k^b_1+b_2+dots +b_nc_n.$$
The point of the whole construction is that, thanks to the effect of the map $bf k$, the above bijection on ternary strings is compatible with the quotient map $operatornameval:0,1,2^mathbbN to[0,1]$, that takes a ternary string $(a_1,a_2,dots)$ to its value as a ternary expansion of a real number, $sum_n=1^infty 3^-na_n$. The latter map $operatornameval$ is surjective but of course not injective, due to points in $[0,1]$ with double representations of "ternary rationals", that is points in the set $T:= m/3^r: rinmathbbN, 0<m<3^r$. So passing to the quotient produces a (continuous, surjective, not injective) map $f:[0,1]to[0,1]^2$.
Now, going back to the question: it is immediate by the construction, that $f^-1(y)$ is a single point iff $yin T^ctimes T^c$; it has two elements iff $yin (T^c times T)cup (Ttimes T^c) $, four elements iff $yin Ttimes T$.
This is also mentioned in Peano original short paper.
answered Aug 6 at 8:35
Pietro Majer
37.9k277180
2
One can modify the construction to make $|f^-1y|le 3$; in general a space filling curve in $d$-dimensional space (for the covering dimension) has points of multiplicity at least $d+1$ and this bound is optimal.
– Anton Petrunin
Aug 6 at 16:26
@PietroMajer Thank you very much for your very great answer.
– Ali Taghavi
Aug 7 at 14:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
Recall the definition of the Peano square-filling curve $f:[0,1]to[0,1]^2$, which is given in terms infinite ternary strings. If $ain [0,1]$ has a base $3$ representation of the form $0,a_1a_2a_3dots$, the point $f(a):=(b,c)$ has base $3$ digits resp. $$b_n:=bf k^a_2+a_4+dots a_2n-2a_2n-1$$
$$c_n:=bf k^a_1+a_3+dots a_2n-1a_2n$$
where $bf k$ is the involutory bijection of the set $0,1,2$ into itself given by $imapsto2-i$ (and exponents denote iterated composition, which in this case only depends on the parity of the exponent, since $bf k^2=bf id$). As Peano observes, this gives a bijection (actually a homeomorphism wrto product topologies of discrete spaces) $0,1,2^mathbbN to 0,1,2^mathbbNtimes 0,1,2^mathbbN$. In fact
$$a_2n-1:=bf k^c_1+c_2+dots +c_n-1b_n$$
$$a_2n:=bf k^b_1+b_2+dots +b_nc_n.$$
The point of the whole construction is that, thanks to the effect of the map $bf k$, the above bijection on ternary strings is compatible with the quotient map $operatornameval:0,1,2^mathbbN to[0,1]$, that takes a ternary string $(a_1,a_2,dots)$ to its value as a ternary expansion of a real number, $sum_n=1^infty 3^-na_n$. The latter map $operatornameval$ is surjective but of course not injective, due to points in $[0,1]$ with double representations of "ternary rationals", that is points in the set $T:= m/3^r: rinmathbbN, 0<m<3^r$. So passing to the quotient produces a (continuous, surjective, not injective) map $f:[0,1]to[0,1]^2$.
Now, going back to the question: it is immediate by the construction, that $f^-1(y)$ is a single point iff $yin T^ctimes T^c$; it has two elements iff $yin (T^c times T)cup (Ttimes T^c) $, four elements iff $yin Ttimes T$.
This is also mentioned in Peano original short paper.
answered Aug 6 at 8:35
Pietro Majer
37.9k277180
2
One can modify the construction to make $|f^-1y|le 3$; in general a space filling curve in $d$-dimensional space (for the covering dimension) has points of multiplicity at least $d+1$ and this bound is optimal.
– Anton Petrunin
Aug 6 at 16:26
@PietroMajer Thank you very much for your very great answer.
– Ali Taghavi
Aug 7 at 14:10
add a comment |Â
up vote
11
down vote
accepted
Recall the definition of the Peano square-filling curve $f:[0,1]to[0,1]^2$, which is given in terms infinite ternary strings. If $ain [0,1]$ has a base $3$ representation of the form $0,a_1a_2a_3dots$, the point $f(a):=(b,c)$ has base $3$ digits resp. $$b_n:=bf k^a_2+a_4+dots a_2n-2a_2n-1$$
$$c_n:=bf k^a_1+a_3+dots a_2n-1a_2n$$
where $bf k$ is the involutory bijection of the set $0,1,2$ into itself given by $imapsto2-i$ (and exponents denote iterated composition, which in this case only depends on the parity of the exponent, since $bf k^2=bf id$). As Peano observes, this gives a bijection (actually a homeomorphism wrto product topologies of discrete spaces) $0,1,2^mathbbN to 0,1,2^mathbbNtimes 0,1,2^mathbbN$. In fact
$$a_2n-1:=bf k^c_1+c_2+dots +c_n-1b_n$$
$$a_2n:=bf k^b_1+b_2+dots +b_nc_n.$$
The point of the whole construction is that, thanks to the effect of the map $bf k$, the above bijection on ternary strings is compatible with the quotient map $operatornameval:0,1,2^mathbbN to[0,1]$, that takes a ternary string $(a_1,a_2,dots)$ to its value as a ternary expansion of a real number, $sum_n=1^infty 3^-na_n$. The latter map $operatornameval$ is surjective but of course not injective, due to points in $[0,1]$ with double representations of "ternary rationals", that is points in the set $T:= m/3^r: rinmathbbN, 0<m<3^r$. So passing to the quotient produces a (continuous, surjective, not injective) map $f:[0,1]to[0,1]^2$.
Now, going back to the question: it is immediate by the construction, that $f^-1(y)$ is a single point iff $yin T^ctimes T^c$; it has two elements iff $yin (T^c times T)cup (Ttimes T^c) $, four elements iff $yin Ttimes T$.
This is also mentioned in Peano original short paper.
answered Aug 6 at 8:35
Pietro Majer
37.9k277180
2
One can modify the construction to make $|f^-1y|le 3$; in general a space filling curve in $d$-dimensional space (for the covering dimension) has points of multiplicity at least $d+1$ and this bound is optimal.
– Anton Petrunin
Aug 6 at 16:26
@PietroMajer Thank you very much for your very great answer.
– Ali Taghavi
Aug 7 at 14:10
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
Recall the definition of the Peano square-filling curve $f:[0,1]to[0,1]^2$, which is given in terms infinite ternary strings. If $ain [0,1]$ has a base $3$ representation of the form $0,a_1a_2a_3dots$, the point $f(a):=(b,c)$ has base $3$ digits resp. $$b_n:=bf k^a_2+a_4+dots a_2n-2a_2n-1$$
$$c_n:=bf k^a_1+a_3+dots a_2n-1a_2n$$
where $bf k$ is the involutory bijection of the set $0,1,2$ into itself given by $imapsto2-i$ (and exponents denote iterated composition, which in this case only depends on the parity of the exponent, since $bf k^2=bf id$). As Peano observes, this gives a bijection (actually a homeomorphism wrto product topologies of discrete spaces) $0,1,2^mathbbN to 0,1,2^mathbbNtimes 0,1,2^mathbbN$. In fact
$$a_2n-1:=bf k^c_1+c_2+dots +c_n-1b_n$$
$$a_2n:=bf k^b_1+b_2+dots +b_nc_n.$$
The point of the whole construction is that, thanks to the effect of the map $bf k$, the above bijection on ternary strings is compatible with the quotient map $operatornameval:0,1,2^mathbbN to[0,1]$, that takes a ternary string $(a_1,a_2,dots)$ to its value as a ternary expansion of a real number, $sum_n=1^infty 3^-na_n$. The latter map $operatornameval$ is surjective but of course not injective, due to points in $[0,1]$ with double representations of "ternary rationals", that is points in the set $T:= m/3^r: rinmathbbN, 0<m<3^r$. So passing to the quotient produces a (continuous, surjective, not injective) map $f:[0,1]to[0,1]^2$.
Now, going back to the question: it is immediate by the construction, that $f^-1(y)$ is a single point iff $yin T^ctimes T^c$; it has two elements iff $yin (T^c times T)cup (Ttimes T^c) $, four elements iff $yin Ttimes T$.
This is also mentioned in Peano original short paper.
answered Aug 6 at 8:35
Pietro Majer
37.9k277180
Recall the definition of the Peano square-filling curve $f:[0,1]to[0,1]^2$, which is given in terms infinite ternary strings. If $ain [0,1]$ has a base $3$ representation of the form $0,a_1a_2a_3dots$, the point $f(a):=(b,c)$ has base $3$ digits resp. $$b_n:=bf k^a_2+a_4+dots a_2n-2a_2n-1$$
$$c_n:=bf k^a_1+a_3+dots a_2n-1a_2n$$
where $bf k$ is the involutory bijection of the set $0,1,2$ into itself given by $imapsto2-i$ (and exponents denote iterated composition, which in this case only depends on the parity of the exponent, since $bf k^2=bf id$). As Peano observes, this gives a bijection (actually a homeomorphism wrto product topologies of discrete spaces) $0,1,2^mathbbN to 0,1,2^mathbbNtimes 0,1,2^mathbbN$. In fact
$$a_2n-1:=bf k^c_1+c_2+dots +c_n-1b_n$$
$$a_2n:=bf k^b_1+b_2+dots +b_nc_n.$$
The point of the whole construction is that, thanks to the effect of the map $bf k$, the above bijection on ternary strings is compatible with the quotient map $operatornameval:0,1,2^mathbbN to[0,1]$, that takes a ternary string $(a_1,a_2,dots)$ to its value as a ternary expansion of a real number, $sum_n=1^infty 3^-na_n$. The latter map $operatornameval$ is surjective but of course not injective, due to points in $[0,1]$ with double representations of "ternary rationals", that is points in the set $T:= m/3^r: rinmathbbN, 0<m<3^r$. So passing to the quotient produces a (continuous, surjective, not injective) map $f:[0,1]to[0,1]^2$.
Now, going back to the question: it is immediate by the construction, that $f^-1(y)$ is a single point iff $yin T^ctimes T^c$; it has two elements iff $yin (T^c times T)cup (Ttimes T^c) $, four elements iff $yin Ttimes T$.
This is also mentioned in Peano original short paper.
answered Aug 6 at 8:35
Pietro Majer
37.9k277180
edited Aug 6 at 8:44
answered Aug 6 at 8:35
Pietro Majer
37.9k277180
answered Aug 6 at 8:35
Pietro Majer
37.9k277180
answered Aug 6 at 8:35
Pietro Majer
37.9k277180
37.9k277180
2
One can modify the construction to make $|f^-1y|le 3$; in general a space filling curve in $d$-dimensional space (for the covering dimension) has points of multiplicity at least $d+1$ and this bound is optimal.
– Anton Petrunin
Aug 6 at 16:26
@PietroMajer Thank you very much for your very great answer.
– Ali Taghavi
Aug 7 at 14:10
add a comment |Â
2
One can modify the construction to make $|f^-1y|le 3$; in general a space filling curve in $d$-dimensional space (for the covering dimension) has points of multiplicity at least $d+1$ and this bound is optimal.
– Anton Petrunin
Aug 6 at 16:26
@PietroMajer Thank you very much for your very great answer.
– Ali Taghavi
Aug 7 at 14:10
2
2
One can modify the construction to make $|f^-1y|le 3$; in general a space filling curve in $d$-dimensional space (for the covering dimension) has points of multiplicity at least $d+1$ and this bound is optimal.
– Anton Petrunin
Aug 6 at 16:26
One can modify the construction to make $|f^-1y|le 3$; in general a space filling curve in $d$-dimensional space (for the covering dimension) has points of multiplicity at least $d+1$ and this bound is optimal.
– Anton Petrunin
Aug 6 at 16:26
@PietroMajer Thank you very much for your very great answer.
– Ali Taghavi
Aug 7 at 14:10
@PietroMajer Thank you very much for your very great answer.
– Ali Taghavi
Aug 7 at 14:10
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f307640%2fspace-filling-curve-whose-all-level-sets-are-finite-countable%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
If I'm not wrong the Peano square filling-curve covers any point at most $4$ times; this follows easily from the analytic description of it
– Pietro Majer
Aug 6 at 7:27
2
You may look at ac.els-cdn.com/S0166864197002708/… They construct a map from $S^1$ to $S^2$, but you may restrict to a subinterval. More generally, the theory of Cannon-Thurston maps constructs Peano curves, where preimages of points typically seem to have cardinality $1$ or $2$.
– ThiKu
Aug 6 at 7:32