Prove that $Graph (f)=bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}([frac{i}{2^n},frac{i+1}{2^n}))times...
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Let $f:mathbb Rlongrightarrow mathbb R$ a funcion. I try to prove that $$text{Graph}(f)=bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
but I have difficulty to prove the $supset$ inclusion. So let $$(x,y)in bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
then, for all $n$, there is $i_n$ s.t. $(x,y)in f^{-1}left(left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)right)times left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. In particular, $f(x)in left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$ for all $n$, i.e. $f(x)=bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. Now, I would like to prove that $$bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)={*},$$
is a singleton, but I can't.
measure-theory
asked Dec 10 '18 at 22:23
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb Rlongrightarrow mathbb R$ a funcion. I try to prove that $$text{Graph}(f)=bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
but I have difficulty to prove the $supset$ inclusion. So let $$(x,y)in bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
then, for all $n$, there is $i_n$ s.t. $(x,y)in f^{-1}left(left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)right)times left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. In particular, $f(x)in left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$ for all $n$, i.e. $f(x)=bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. Now, I would like to prove that $$bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)={*},$$
is a singleton, but I can't.
measure-theory
asked Dec 10 '18 at 22:23
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb Rlongrightarrow mathbb R$ a funcion. I try to prove that $$text{Graph}(f)=bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
but I have difficulty to prove the $supset$ inclusion. So let $$(x,y)in bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
then, for all $n$, there is $i_n$ s.t. $(x,y)in f^{-1}left(left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)right)times left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. In particular, $f(x)in left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$ for all $n$, i.e. $f(x)=bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. Now, I would like to prove that $$bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)={*},$$
is a singleton, but I can't.
measure-theory
asked Dec 10 '18 at 22:23
NewMathNewMath
4059
$endgroup$
Let $f:mathbb Rlongrightarrow mathbb R$ a funcion. I try to prove that $$text{Graph}(f)=bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
but I have difficulty to prove the $supset$ inclusion. So let $$(x,y)in bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
then, for all $n$, there is $i_n$ s.t. $(x,y)in f^{-1}left(left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)right)times left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. In particular, $f(x)in left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$ for all $n$, i.e. $f(x)=bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. Now, I would like to prove that $$bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)={*},$$
is a singleton, but I can't.
measure-theory
measure-theory
asked Dec 10 '18 at 22:23
NewMathNewMath
4059
asked Dec 10 '18 at 22:23
NewMathNewMath
4059
asked Dec 10 '18 at 22:23
NewMathNewMath
4059
asked Dec 10 '18 at 22:23
NewMathNewMath
4059
asked Dec 10 '18 at 22:23
NewMathNewMath
4059
4059
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1 Answer
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In one hand, $f(x)in bigcap_{ninmathbb N}I_n$. In the other hand, if $yin bigcap_{ninmathbb N}I_n$, then $|y-f(x)|leq frac{1}{2^n}to 0$ and thus $f(x)=y$. The claim follow.
answered Dec 10 '18 at 22:43
SurbSurb
37.7k94375
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1 Answer
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$begingroup$
In one hand, $f(x)in bigcap_{ninmathbb N}I_n$. In the other hand, if $yin bigcap_{ninmathbb N}I_n$, then $|y-f(x)|leq frac{1}{2^n}to 0$ and thus $f(x)=y$. The claim follow.
answered Dec 10 '18 at 22:43
SurbSurb
37.7k94375
$endgroup$
add a comment |
$begingroup$
In one hand, $f(x)in bigcap_{ninmathbb N}I_n$. In the other hand, if $yin bigcap_{ninmathbb N}I_n$, then $|y-f(x)|leq frac{1}{2^n}to 0$ and thus $f(x)=y$. The claim follow.
answered Dec 10 '18 at 22:43
SurbSurb
37.7k94375
$endgroup$
add a comment |
$begingroup$
In one hand, $f(x)in bigcap_{ninmathbb N}I_n$. In the other hand, if $yin bigcap_{ninmathbb N}I_n$, then $|y-f(x)|leq frac{1}{2^n}to 0$ and thus $f(x)=y$. The claim follow.
answered Dec 10 '18 at 22:43
SurbSurb
37.7k94375
$endgroup$
In one hand, $f(x)in bigcap_{ninmathbb N}I_n$. In the other hand, if $yin bigcap_{ninmathbb N}I_n$, then $|y-f(x)|leq frac{1}{2^n}to 0$ and thus $f(x)=y$. The claim follow.
answered Dec 10 '18 at 22:43
SurbSurb
37.7k94375
edited Dec 10 '18 at 22:51
answered Dec 10 '18 at 22:43
SurbSurb
37.7k94375
answered Dec 10 '18 at 22:43
SurbSurb
37.7k94375
answered Dec 10 '18 at 22:43
SurbSurb
37.7k94375
37.7k94375
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