Function $f:[0,1] to [0,1]$ taking on each value in $[0,1]$ exactly twice
Clash Royale CLAN TAG#URR8PPP
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I want to find a function $f:[0,1] to [0,1]$ such that $f$ takes on each value in $[0,1]$ exactly twice. I think this means there are an infinite number of discontinuities. Can anyone help me figure this one out?
Anyone have any pointers?
real-analysis
edited Aug 22 at 8:37
Did
243k23208443
asked Nov 1 '12 at 22:57
Jackson Hart
4492626
 |Â
show 3 more comments
up vote
23
down vote
favorite
I want to find a function $f:[0,1] to [0,1]$ such that $f$ takes on each value in $[0,1]$ exactly twice. I think this means there are an infinite number of discontinuities. Can anyone help me figure this one out?
Anyone have any pointers?
real-analysis
edited Aug 22 at 8:37
Did
243k23208443
asked Nov 1 '12 at 22:57
Jackson Hart
4492626
7
That is actually a pretty cool problem.
– Michael Greinecker♦
Nov 1 '12 at 23:17
10
See Jo Heath, "Every exactly 2-to-1 function on the reals has an infinite number of discontinuities", Proceedings of the AMS, 98 (1986), 369-373 ams.org/journals/proc/1986-098-02/S0002-9939-1986-0854049-8/…
– Robert Israel
Nov 1 '12 at 23:23
5
Wouldn't a rather natural solution be to split the unit interval into a closed and a half-open interval and to construct bijections for each?
– EuYu
Nov 1 '12 at 23:28
@wj32 I am not sure exactly what function that could be though
– Jackson Hart
Nov 2 '12 at 2:30
6
Let $[0,1]$ be partitioned into the set of reciprocals of integers $A=1,1/2,1/3,...$ and the complement $B$ of $A$. Now map $A$ by a shift of the sequence (take 1 to 1/2, take 1/2 to 1/3, etc), and map $B$ by the identity. Then this will map $[0,1]$ bijectively to $[0,1)$.
– coffeemath
Nov 2 '12 at 4:07
 |Â
show 3 more comments
up vote
23
down vote
favorite
up vote
23
down vote
favorite
I want to find a function $f:[0,1] to [0,1]$ such that $f$ takes on each value in $[0,1]$ exactly twice. I think this means there are an infinite number of discontinuities. Can anyone help me figure this one out?
Anyone have any pointers?
real-analysis
edited Aug 22 at 8:37
Did
243k23208443
asked Nov 1 '12 at 22:57
Jackson Hart
4492626
I want to find a function $f:[0,1] to [0,1]$ such that $f$ takes on each value in $[0,1]$ exactly twice. I think this means there are an infinite number of discontinuities. Can anyone help me figure this one out?
Anyone have any pointers?
real-analysis
edited Aug 22 at 8:37
Did
243k23208443
asked Nov 1 '12 at 22:57
Jackson Hart
4492626
edited Aug 22 at 8:37
Did
243k23208443
edited Aug 22 at 8:37
Did
243k23208443
edited Aug 22 at 8:37
Did
243k23208443
243k23208443
asked Nov 1 '12 at 22:57
Jackson Hart
4492626
asked Nov 1 '12 at 22:57
Jackson Hart
4492626
asked Nov 1 '12 at 22:57
Jackson Hart
4492626
4492626
7
That is actually a pretty cool problem.
– Michael Greinecker♦
Nov 1 '12 at 23:17
10
See Jo Heath, "Every exactly 2-to-1 function on the reals has an infinite number of discontinuities", Proceedings of the AMS, 98 (1986), 369-373 ams.org/journals/proc/1986-098-02/S0002-9939-1986-0854049-8/…
– Robert Israel
Nov 1 '12 at 23:23
5
Wouldn't a rather natural solution be to split the unit interval into a closed and a half-open interval and to construct bijections for each?
– EuYu
Nov 1 '12 at 23:28
@wj32 I am not sure exactly what function that could be though
– Jackson Hart
Nov 2 '12 at 2:30
6
Let $[0,1]$ be partitioned into the set of reciprocals of integers $A=1,1/2,1/3,...$ and the complement $B$ of $A$. Now map $A$ by a shift of the sequence (take 1 to 1/2, take 1/2 to 1/3, etc), and map $B$ by the identity. Then this will map $[0,1]$ bijectively to $[0,1)$.
– coffeemath
Nov 2 '12 at 4:07
 |Â
show 3 more comments
7
That is actually a pretty cool problem.
– Michael Greinecker♦
Nov 1 '12 at 23:17
10
See Jo Heath, "Every exactly 2-to-1 function on the reals has an infinite number of discontinuities", Proceedings of the AMS, 98 (1986), 369-373 ams.org/journals/proc/1986-098-02/S0002-9939-1986-0854049-8/…
– Robert Israel
Nov 1 '12 at 23:23
5
Wouldn't a rather natural solution be to split the unit interval into a closed and a half-open interval and to construct bijections for each?
– EuYu
Nov 1 '12 at 23:28
@wj32 I am not sure exactly what function that could be though
– Jackson Hart
Nov 2 '12 at 2:30
6
Let $[0,1]$ be partitioned into the set of reciprocals of integers $A=1,1/2,1/3,...$ and the complement $B$ of $A$. Now map $A$ by a shift of the sequence (take 1 to 1/2, take 1/2 to 1/3, etc), and map $B$ by the identity. Then this will map $[0,1]$ bijectively to $[0,1)$.
– coffeemath
Nov 2 '12 at 4:07
7
7
That is actually a pretty cool problem.
– Michael Greinecker♦
Nov 1 '12 at 23:17
That is actually a pretty cool problem.
– Michael Greinecker♦
Nov 1 '12 at 23:17
10
10
See Jo Heath, "Every exactly 2-to-1 function on the reals has an infinite number of discontinuities", Proceedings of the AMS, 98 (1986), 369-373 ams.org/journals/proc/1986-098-02/S0002-9939-1986-0854049-8/…
– Robert Israel
Nov 1 '12 at 23:23
See Jo Heath, "Every exactly 2-to-1 function on the reals has an infinite number of discontinuities", Proceedings of the AMS, 98 (1986), 369-373 ams.org/journals/proc/1986-098-02/S0002-9939-1986-0854049-8/…
– Robert Israel
Nov 1 '12 at 23:23
5
5
Wouldn't a rather natural solution be to split the unit interval into a closed and a half-open interval and to construct bijections for each?
– EuYu
Nov 1 '12 at 23:28
Wouldn't a rather natural solution be to split the unit interval into a closed and a half-open interval and to construct bijections for each?
– EuYu
Nov 1 '12 at 23:28
@wj32 I am not sure exactly what function that could be though
– Jackson Hart
Nov 2 '12 at 2:30
@wj32 I am not sure exactly what function that could be though
– Jackson Hart
Nov 2 '12 at 2:30
6
6
Let $[0,1]$ be partitioned into the set of reciprocals of integers $A=1,1/2,1/3,...$ and the complement $B$ of $A$. Now map $A$ by a shift of the sequence (take 1 to 1/2, take 1/2 to 1/3, etc), and map $B$ by the identity. Then this will map $[0,1]$ bijectively to $[0,1)$.
– coffeemath
Nov 2 '12 at 4:07
Let $[0,1]$ be partitioned into the set of reciprocals of integers $A=1,1/2,1/3,...$ and the complement $B$ of $A$. Now map $A$ by a shift of the sequence (take 1 to 1/2, take 1/2 to 1/3, etc), and map $B$ by the identity. Then this will map $[0,1]$ bijectively to $[0,1)$.
– coffeemath
Nov 2 '12 at 4:07
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Let $x_alpha$ be a well-ordering of $[0,1]$.
For any ordinal $alpha = theta + n < frakc$ where $theta$ is a limit ordinal or $0$ and $n$ is a finite ordinal, let $F(theta + n cdot 2) = F(theta + n cdot 2 + 1) = x_alpha$.
Now define $f(x_alpha) = F(alpha)$ for all $alpha lt frakc$ and it is clear that $f$ has the required property.
answered Nov 2 '12 at 4:55
Dan Brumleve
11.8k53586
3
This is a overblown considering @EuYu's comment. I do need practice with this kind of proof but it's not appropriate for this problem.
– Dan Brumleve
Nov 2 '12 at 5:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let $x_alpha$ be a well-ordering of $[0,1]$.
For any ordinal $alpha = theta + n < frakc$ where $theta$ is a limit ordinal or $0$ and $n$ is a finite ordinal, let $F(theta + n cdot 2) = F(theta + n cdot 2 + 1) = x_alpha$.
Now define $f(x_alpha) = F(alpha)$ for all $alpha lt frakc$ and it is clear that $f$ has the required property.
answered Nov 2 '12 at 4:55
Dan Brumleve
11.8k53586
3
This is a overblown considering @EuYu's comment. I do need practice with this kind of proof but it's not appropriate for this problem.
– Dan Brumleve
Nov 2 '12 at 5:53
add a comment |Â
up vote
4
down vote
accepted
Let $x_alpha$ be a well-ordering of $[0,1]$.
For any ordinal $alpha = theta + n < frakc$ where $theta$ is a limit ordinal or $0$ and $n$ is a finite ordinal, let $F(theta + n cdot 2) = F(theta + n cdot 2 + 1) = x_alpha$.
Now define $f(x_alpha) = F(alpha)$ for all $alpha lt frakc$ and it is clear that $f$ has the required property.
answered Nov 2 '12 at 4:55
Dan Brumleve
11.8k53586
3
This is a overblown considering @EuYu's comment. I do need practice with this kind of proof but it's not appropriate for this problem.
– Dan Brumleve
Nov 2 '12 at 5:53
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let $x_alpha$ be a well-ordering of $[0,1]$.
For any ordinal $alpha = theta + n < frakc$ where $theta$ is a limit ordinal or $0$ and $n$ is a finite ordinal, let $F(theta + n cdot 2) = F(theta + n cdot 2 + 1) = x_alpha$.
Now define $f(x_alpha) = F(alpha)$ for all $alpha lt frakc$ and it is clear that $f$ has the required property.
answered Nov 2 '12 at 4:55
Dan Brumleve
11.8k53586
Let $x_alpha$ be a well-ordering of $[0,1]$.
For any ordinal $alpha = theta + n < frakc$ where $theta$ is a limit ordinal or $0$ and $n$ is a finite ordinal, let $F(theta + n cdot 2) = F(theta + n cdot 2 + 1) = x_alpha$.
Now define $f(x_alpha) = F(alpha)$ for all $alpha lt frakc$ and it is clear that $f$ has the required property.
answered Nov 2 '12 at 4:55
Dan Brumleve
11.8k53586
edited Nov 2 '12 at 10:41
answered Nov 2 '12 at 4:55
Dan Brumleve
11.8k53586
answered Nov 2 '12 at 4:55
Dan Brumleve
11.8k53586
answered Nov 2 '12 at 4:55
Dan Brumleve
11.8k53586
11.8k53586
3
This is a overblown considering @EuYu's comment. I do need practice with this kind of proof but it's not appropriate for this problem.
– Dan Brumleve
Nov 2 '12 at 5:53
add a comment |Â
3
This is a overblown considering @EuYu's comment. I do need practice with this kind of proof but it's not appropriate for this problem.
– Dan Brumleve
Nov 2 '12 at 5:53
3
3
This is a overblown considering @EuYu's comment. I do need practice with this kind of proof but it's not appropriate for this problem.
– Dan Brumleve
Nov 2 '12 at 5:53
This is a overblown considering @EuYu's comment. I do need practice with this kind of proof but it's not appropriate for this problem.
– Dan Brumleve
Nov 2 '12 at 5:53
add a comment |Â
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7
That is actually a pretty cool problem.
– Michael Greinecker♦
Nov 1 '12 at 23:17
10
See Jo Heath, "Every exactly 2-to-1 function on the reals has an infinite number of discontinuities", Proceedings of the AMS, 98 (1986), 369-373 ams.org/journals/proc/1986-098-02/S0002-9939-1986-0854049-8/…
– Robert Israel
Nov 1 '12 at 23:23
5
Wouldn't a rather natural solution be to split the unit interval into a closed and a half-open interval and to construct bijections for each?
– EuYu
Nov 1 '12 at 23:28
@wj32 I am not sure exactly what function that could be though
– Jackson Hart
Nov 2 '12 at 2:30
6
Let $[0,1]$ be partitioned into the set of reciprocals of integers $A=1,1/2,1/3,...$ and the complement $B$ of $A$. Now map $A$ by a shift of the sequence (take 1 to 1/2, take 1/2 to 1/3, etc), and map $B$ by the identity. Then this will map $[0,1]$ bijectively to $[0,1)$.
– coffeemath
Nov 2 '12 at 4:07