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Question: Is there any function $f:\mathbb{R}\to\mathbb{R}$ that is continuous precisely on the rational points of $\mathbb{R}$?

Solution: Let $f:\mathbb{R}\to \mathbb{R}$ be any arbitrary function. Give a positive integer $n$, let $U_n$ be the union of the elements of the following collection:
$$\big\{U:U \; \text {is open in} \;\mathbb{R}\;\&\; \text {diam}(f(U))<\frac{1}{n}\big \}$$ Then $U_n$ is open in $\mathbb{R}$ 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.

lulu
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abcdmath
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  • If $f$ is continuous at $y$, then for all $\epsilon>0$ there is $\delta_\epsilon$ such that ${x:\ |x-y|<\delta_\epsilon}\subset{x:\ |f(x)-f(y)|<\epsilon}$. Since ${x:\ |f(x)-f(y)|<\epsilon}$ are nested in each other when you decrease the $\epsilon$, it is enough to check the definition for $\epsilon=1/n$. Observe that ${x:\ |x-y|<\delta_{1/n}}$ will be an element of ${U:\ U\text{ is open and }diam(f(U))<1/n}$. Therefore, $y\in U_n$ for all $n$. If $f$ is discontinuous at $z$, then for all neighborhood $U$ of $z$, $f(U)$ remains large. Therefore $z\notin U_n$ for some $n$. –  Jul 19 '18 at 17:31
  • Noticing density might help. And knowing an alternative formulation of the Baire category theorem will help as well. –  Jul 19 '18 at 17:32
  • @Robert. How can BCT will help here? – abcdmath Jul 19 '18 at 17:56
  • Read the comments in here https://math.stackexchange.com/questions/67620/set-of-continuity-points-of-a-real-function –  Jul 19 '18 at 19:26

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