$f(mathbb{Q}) subset mathbb{R} - mathbb{Q}$ and $f(mathbb{R} - mathbb{Q}) subset mathbb{Q}$
$begingroup$
Continuous function $f$ such that $f(mathbb{Q}) subset mathbb{R} - mathbb{Q}$ and $f(mathbb{R} - mathbb{Q}) subset mathbb{Q}$
(I know this question has already been asked and answered but there the idea used was the cardinality of $mathbb{Q}$ and $mathbb{R}$ and $mathbb{Q}$. Here I want to try another idea- as described in my outline Reference: Continuous function that take irrationals to rationals and vice-versa..)
My attempt: I choose $m = 3$ and $c in mathbb{Q}$ and draw a family of lines $L_c: y = 3x + c$. If such a $f(x)$ exists and if $f(x) cap L_c neq emptyset$, then they intersect at $x$.
Two cases follow:
Case I: $x$ is rational.
Then $y = 3x + c in mathbb{Q}$ a contradiction to the definition of $f(x)$.
Case II: $x$ is irrational.
Then $y = 3x + c in mathbb{R} - mathbb{Q}$ again a contradiction to the definition of $f(x)$.
So my plan is draw the family $L_c$ on $mathbb{R^2}$ and show that $f(x)$ intersection atleast one of the line
Now if I don't want $f(x)$ to intersect $L_c$, then I have the following inequality $$3x + c < f(x) < 3x + c + frac{1}{n}.$$
I keep increasing $n$ and hence reach the conclusion that no such $f(x)$ exists. The problem with my proof is I want my conclusion to be $f(x) = 3x + c, cin mathbb{R} - mathbb{Q}$. Can someone fix my proof.
continuity
asked Dec 9 '18 at 14:10
henceprovedhenceproved
1628
$endgroup$
add a comment |
$begingroup$
Continuous function $f$ such that $f(mathbb{Q}) subset mathbb{R} - mathbb{Q}$ and $f(mathbb{R} - mathbb{Q}) subset mathbb{Q}$
(I know this question has already been asked and answered but there the idea used was the cardinality of $mathbb{Q}$ and $mathbb{R}$ and $mathbb{Q}$. Here I want to try another idea- as described in my outline Reference: Continuous function that take irrationals to rationals and vice-versa..)
My attempt: I choose $m = 3$ and $c in mathbb{Q}$ and draw a family of lines $L_c: y = 3x + c$. If such a $f(x)$ exists and if $f(x) cap L_c neq emptyset$, then they intersect at $x$.
Two cases follow:
Case I: $x$ is rational.
Then $y = 3x + c in mathbb{Q}$ a contradiction to the definition of $f(x)$.
Case II: $x$ is irrational.
Then $y = 3x + c in mathbb{R} - mathbb{Q}$ again a contradiction to the definition of $f(x)$.
So my plan is draw the family $L_c$ on $mathbb{R^2}$ and show that $f(x)$ intersection atleast one of the line
Now if I don't want $f(x)$ to intersect $L_c$, then I have the following inequality $$3x + c < f(x) < 3x + c + frac{1}{n}.$$
I keep increasing $n$ and hence reach the conclusion that no such $f(x)$ exists. The problem with my proof is I want my conclusion to be $f(x) = 3x + c, cin mathbb{R} - mathbb{Q}$. Can someone fix my proof.
continuity
asked Dec 9 '18 at 14:10
henceprovedhenceproved
1628
$endgroup$
$begingroup$
$f(x)cap L_c$ doesn't make sense. Are you thinking of $Gamma(f)cap L_c$ or something else?
$endgroup$
– user10354138
Dec 9 '18 at 14:15
$begingroup$
@user10354138 I mean the intersection points of $L_c$ and $f(x)$.
$endgroup$
– henceproved
Dec 9 '18 at 14:25
add a comment |
$begingroup$
Continuous function $f$ such that $f(mathbb{Q}) subset mathbb{R} - mathbb{Q}$ and $f(mathbb{R} - mathbb{Q}) subset mathbb{Q}$
(I know this question has already been asked and answered but there the idea used was the cardinality of $mathbb{Q}$ and $mathbb{R}$ and $mathbb{Q}$. Here I want to try another idea- as described in my outline Reference: Continuous function that take irrationals to rationals and vice-versa..)
My attempt: I choose $m = 3$ and $c in mathbb{Q}$ and draw a family of lines $L_c: y = 3x + c$. If such a $f(x)$ exists and if $f(x) cap L_c neq emptyset$, then they intersect at $x$.
Two cases follow:
Case I: $x$ is rational.
Then $y = 3x + c in mathbb{Q}$ a contradiction to the definition of $f(x)$.
Case II: $x$ is irrational.
Then $y = 3x + c in mathbb{R} - mathbb{Q}$ again a contradiction to the definition of $f(x)$.
So my plan is draw the family $L_c$ on $mathbb{R^2}$ and show that $f(x)$ intersection atleast one of the line
Now if I don't want $f(x)$ to intersect $L_c$, then I have the following inequality $$3x + c < f(x) < 3x + c + frac{1}{n}.$$
I keep increasing $n$ and hence reach the conclusion that no such $f(x)$ exists. The problem with my proof is I want my conclusion to be $f(x) = 3x + c, cin mathbb{R} - mathbb{Q}$. Can someone fix my proof.
continuity
asked Dec 9 '18 at 14:10
henceprovedhenceproved
1628
$endgroup$
Continuous function $f$ such that $f(mathbb{Q}) subset mathbb{R} - mathbb{Q}$ and $f(mathbb{R} - mathbb{Q}) subset mathbb{Q}$
(I know this question has already been asked and answered but there the idea used was the cardinality of $mathbb{Q}$ and $mathbb{R}$ and $mathbb{Q}$. Here I want to try another idea- as described in my outline Reference: Continuous function that take irrationals to rationals and vice-versa..)
My attempt: I choose $m = 3$ and $c in mathbb{Q}$ and draw a family of lines $L_c: y = 3x + c$. If such a $f(x)$ exists and if $f(x) cap L_c neq emptyset$, then they intersect at $x$.
Two cases follow:
Case I: $x$ is rational.
Then $y = 3x + c in mathbb{Q}$ a contradiction to the definition of $f(x)$.
Case II: $x$ is irrational.
Then $y = 3x + c in mathbb{R} - mathbb{Q}$ again a contradiction to the definition of $f(x)$.
So my plan is draw the family $L_c$ on $mathbb{R^2}$ and show that $f(x)$ intersection atleast one of the line
Now if I don't want $f(x)$ to intersect $L_c$, then I have the following inequality $$3x + c < f(x) < 3x + c + frac{1}{n}.$$
I keep increasing $n$ and hence reach the conclusion that no such $f(x)$ exists. The problem with my proof is I want my conclusion to be $f(x) = 3x + c, cin mathbb{R} - mathbb{Q}$. Can someone fix my proof.
continuity
continuity
asked Dec 9 '18 at 14:10
henceprovedhenceproved
1628
asked Dec 9 '18 at 14:10
henceprovedhenceproved
1628
asked Dec 9 '18 at 14:10
henceprovedhenceproved
1628
asked Dec 9 '18 at 14:10
henceprovedhenceproved
1628
asked Dec 9 '18 at 14:10
henceprovedhenceproved
1628
1628
$begingroup$
$f(x)cap L_c$ doesn't make sense. Are you thinking of $Gamma(f)cap L_c$ or something else?
$endgroup$
– user10354138
Dec 9 '18 at 14:15
$begingroup$
@user10354138 I mean the intersection points of $L_c$ and $f(x)$.
$endgroup$
– henceproved
Dec 9 '18 at 14:25
add a comment |
$begingroup$
$f(x)cap L_c$ doesn't make sense. Are you thinking of $Gamma(f)cap L_c$ or something else?
$endgroup$
– user10354138
Dec 9 '18 at 14:15
$begingroup$
@user10354138 I mean the intersection points of $L_c$ and $f(x)$.
$endgroup$
– henceproved
Dec 9 '18 at 14:25
$begingroup$
$f(x)cap L_c$ doesn't make sense. Are you thinking of $Gamma(f)cap L_c$ or something else?
$endgroup$
– user10354138
Dec 9 '18 at 14:15
$begingroup$
$f(x)cap L_c$ doesn't make sense. Are you thinking of $Gamma(f)cap L_c$ or something else?
$endgroup$
– user10354138
Dec 9 '18 at 14:15
$begingroup$
@user10354138 I mean the intersection points of $L_c$ and $f(x)$.
$endgroup$
– henceproved
Dec 9 '18 at 14:25
$begingroup$
@user10354138 I mean the intersection points of $L_c$ and $f(x)$.
$endgroup$
– henceproved
Dec 9 '18 at 14:25
add a comment |
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$begingroup$
$f(x)cap L_c$ doesn't make sense. Are you thinking of $Gamma(f)cap L_c$ or something else?
$endgroup$
– user10354138
Dec 9 '18 at 14:15
$begingroup$
@user10354138 I mean the intersection points of $L_c$ and $f(x)$.
$endgroup$
– henceproved
Dec 9 '18 at 14:25