Show that $X$ is the set of points at which the function $f$ is continuous.

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Question: Is there any function $f:mathbbRtomathbbR$ that is continuous precisely on the rational points of $mathbbR$?



Solution: Let $f:mathbbRto mathbbR$ be any arbitrary function. Give a positive integer $n$, let $U_n$ be the union of the elements of the following collection:

$$bigU:U ; text is open in ;mathbbR;&; text diam(f(U))<frac1nbig $$
Then $U_n$ is open in $mathbbR$ and let us consider the set $X=cap_n U_n$. Then $X$ is the set of points at which the function $f$ is continuous......



I can't able to understand the last line of the solution. That is, how $X$ becomes the set of points of continuity of $f$? Please help me to understand this.




continuity




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  • If $f$ is continuous at $y$, then for all $epsilon>0$ there is $delta_epsilon$ such that $x: subsetx: $. Since $x: $ are nested in each other when you decrease the $epsilon$, it is enough to check the definition for $epsilon=1/n$. Observe that $x: $ will be an element of $U: Utext is open and diam(f(U))<1/n$. Therefore, $yin U_n$ for all $n$. If $f$ is discontinuous at $z$, then for all neighborhood $U$ of $z$, $f(U)$ remains large. Therefore $znotin U_n$ for some $n$.
    – user577471
    Jul 19 at 17:31










  • Noticing density might help. And knowing an alternative formulation of the Baire category theorem will help as well.
    – Robert Wolfe
    Jul 19 at 17:32










  • @Robert. How can BCT will help here?
    – abcdmath
    Jul 19 at 17:56










  • Read the comments in here math.stackexchange.com/questions/67620/…
    – Robert Wolfe
    Jul 19 at 19:26














up vote
0
down vote

favorite












Question: Is there any function $f:mathbbRtomathbbR$ that is continuous precisely on the rational points of $mathbbR$?



Solution: Let $f:mathbbRto mathbbR$ be any arbitrary function. Give a positive integer $n$, let $U_n$ be the union of the elements of the following collection:

$$bigU:U ; text is open in ;mathbbR;&; text diam(f(U))<frac1nbig $$
Then $U_n$ is open in $mathbbR$ and let us consider the set $X=cap_n U_n$. Then $X$ is the set of points at which the function $f$ is continuous......



I can't able to understand the last line of the solution. That is, how $X$ becomes the set of points of continuity of $f$? Please help me to understand this.




continuity




share|cite|improve this question





















  • If $f$ is continuous at $y$, then for all $epsilon>0$ there is $delta_epsilon$ such that $x: subsetx: $. Since $x: $ are nested in each other when you decrease the $epsilon$, it is enough to check the definition for $epsilon=1/n$. Observe that $x: $ will be an element of $U: Utext is open and diam(f(U))<1/n$. Therefore, $yin U_n$ for all $n$. If $f$ is discontinuous at $z$, then for all neighborhood $U$ of $z$, $f(U)$ remains large. Therefore $znotin U_n$ for some $n$.
    – user577471
    Jul 19 at 17:31










  • Noticing density might help. And knowing an alternative formulation of the Baire category theorem will help as well.
    – Robert Wolfe
    Jul 19 at 17:32










  • @Robert. How can BCT will help here?
    – abcdmath
    Jul 19 at 17:56










  • Read the comments in here math.stackexchange.com/questions/67620/…
    – Robert Wolfe
    Jul 19 at 19:26












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Question: Is there any function $f:mathbbRtomathbbR$ that is continuous precisely on the rational points of $mathbbR$?



Solution: Let $f:mathbbRto mathbbR$ be any arbitrary function. Give a positive integer $n$, let $U_n$ be the union of the elements of the following collection:

$$bigU:U ; text is open in ;mathbbR;&; text diam(f(U))<frac1nbig $$
Then $U_n$ is open in $mathbbR$ and let us consider the set $X=cap_n U_n$. Then $X$ is the set of points at which the function $f$ is continuous......



I can't able to understand the last line of the solution. That is, how $X$ becomes the set of points of continuity of $f$? Please help me to understand this.




continuity




share|cite|improve this question













Question: Is there any function $f:mathbbRtomathbbR$ that is continuous precisely on the rational points of $mathbbR$?



Solution: Let $f:mathbbRto mathbbR$ be any arbitrary function. Give a positive integer $n$, let $U_n$ be the union of the elements of the following collection:

$$bigU:U ; text is open in ;mathbbR;&; text diam(f(U))<frac1nbig $$
Then $U_n$ is open in $mathbbR$ and let us consider the set $X=cap_n U_n$. Then $X$ is the set of points at which the function $f$ is continuous......



I can't able to understand the last line of the solution. That is, how $X$ becomes the set of points of continuity of $f$? Please help me to understand this.





continuity





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 19 at 17:28









lulu

35.2k14172




35.2k14172









asked Jul 19 at 17:21









abcdmath

29310




29310











  • If $f$ is continuous at $y$, then for all $epsilon>0$ there is $delta_epsilon$ such that $x: subsetx: $. Since $x: $ are nested in each other when you decrease the $epsilon$, it is enough to check the definition for $epsilon=1/n$. Observe that $x: $ will be an element of $U: Utext is open and diam(f(U))<1/n$. Therefore, $yin U_n$ for all $n$. If $f$ is discontinuous at $z$, then for all neighborhood $U$ of $z$, $f(U)$ remains large. Therefore $znotin U_n$ for some $n$.
    – user577471
    Jul 19 at 17:31










  • Noticing density might help. And knowing an alternative formulation of the Baire category theorem will help as well.
    – Robert Wolfe
    Jul 19 at 17:32










  • @Robert. How can BCT will help here?
    – abcdmath
    Jul 19 at 17:56










  • Read the comments in here math.stackexchange.com/questions/67620/…
    – Robert Wolfe
    Jul 19 at 19:26
















  • If $f$ is continuous at $y$, then for all $epsilon>0$ there is $delta_epsilon$ such that $x: subsetx: $. Since $x: $ are nested in each other when you decrease the $epsilon$, it is enough to check the definition for $epsilon=1/n$. Observe that $x: $ will be an element of $U: Utext is open and diam(f(U))<1/n$. Therefore, $yin U_n$ for all $n$. If $f$ is discontinuous at $z$, then for all neighborhood $U$ of $z$, $f(U)$ remains large. Therefore $znotin U_n$ for some $n$.
    – user577471
    Jul 19 at 17:31










  • Noticing density might help. And knowing an alternative formulation of the Baire category theorem will help as well.
    – Robert Wolfe
    Jul 19 at 17:32










  • @Robert. How can BCT will help here?
    – abcdmath
    Jul 19 at 17:56










  • Read the comments in here math.stackexchange.com/questions/67620/…
    – Robert Wolfe
    Jul 19 at 19:26















If $f$ is continuous at $y$, then for all $epsilon>0$ there is $delta_epsilon$ such that $x: subsetx: $. Since $x: $ are nested in each other when you decrease the $epsilon$, it is enough to check the definition for $epsilon=1/n$. Observe that $x: $ will be an element of $U: Utext is open and diam(f(U))<1/n$. Therefore, $yin U_n$ for all $n$. If $f$ is discontinuous at $z$, then for all neighborhood $U$ of $z$, $f(U)$ remains large. Therefore $znotin U_n$ for some $n$.
– user577471
Jul 19 at 17:31




If $f$ is continuous at $y$, then for all $epsilon>0$ there is $delta_epsilon$ such that $x: subsetx: $. Since $x: $ are nested in each other when you decrease the $epsilon$, it is enough to check the definition for $epsilon=1/n$. Observe that $x: $ will be an element of $U: Utext is open and diam(f(U))<1/n$. Therefore, $yin U_n$ for all $n$. If $f$ is discontinuous at $z$, then for all neighborhood $U$ of $z$, $f(U)$ remains large. Therefore $znotin U_n$ for some $n$.
– user577471
Jul 19 at 17:31












Noticing density might help. And knowing an alternative formulation of the Baire category theorem will help as well.
– Robert Wolfe
Jul 19 at 17:32




Noticing density might help. And knowing an alternative formulation of the Baire category theorem will help as well.
– Robert Wolfe
Jul 19 at 17:32












@Robert. How can BCT will help here?
– abcdmath
Jul 19 at 17:56




@Robert. How can BCT will help here?
– abcdmath
Jul 19 at 17:56












Read the comments in here math.stackexchange.com/questions/67620/…
– Robert Wolfe
Jul 19 at 19:26




Read the comments in here math.stackexchange.com/questions/67620/…
– Robert Wolfe
Jul 19 at 19:26















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