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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. Show that the points at which $f$ is continuous is a $G_{\delta}$ set.

$$A_n = \{ x \in \mathbb{R} | x \in B(x,r) \text{ open }, f(x'')-f(x')<\frac{1}{n}, \forall x',x'' \in B(x)\}$$

I saw that this proof was already on here, but I wanted to confirm and flesh out more details.

"$\Rightarrow$" If f is continuous at $x$, then $f(x'')-f(x')<\frac{1}{n}$ for $x'',x' \in B(x, r_{n})$. That is, there is a ball of radius $r$ where $r$ depends on $n$. Then $x \in A_n$ and thus $x \in \cap A_n$.

"$\Leftarrow$" If $x \in \cap A_n$, then there is an $\epsilon > 0$ and a $\delta > 0$ such that $x' , x'' \in B(x, \delta_n)$ for all $n$ and $$|f(x'')-f(x')|<\epsilon.$$ Take $\epsilon = \frac{1}{n}$.

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    If $f$ is continuous, there’s nothing to prove, so you must have misstated something here. Also, the definition of $A_n$ doesn’t make sense, because you’ve not defined $B(x)$.2012-10-12
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    Well I want to prove that the set of all points which f is continuous is a G_\delta set.2012-10-12
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    But if $f$ is continuous, that’s automatic: the set of points of continuity of $f$ is $\Bbb R$, which is not just a $G_\delta$, but even an open set.2012-10-12
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    Ahh...I see what's confusing. $B(x)$ is an open set containing $x$ of some radius $r$. I'll edit the post for that clarity.2012-10-12
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    That fixes the problem with $A_n$, but not with what you’re trying to prove. Once again: if you assume that $f$ is continuous, then there is nothing to prove, because the set of points of continuity is all of $\Bbb R$, which is trivially a $G_\delta$.2012-10-12
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    Well $A_n$ is an open set by how we defined it. So $\mathscr{A}$ - the collection of $A_n$'s, is a collection of open sets. To show that it is $G_\delta$ I just need to show that it is a countable union. Is all that is left to show that $\cap \mathscr{A}$ is a countable intersection.2012-10-12
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    Stop. You’re missing the point. Forget $A_n$. **If** $f$ **is continuous, it is continuous at every point of** $\Bbb R$. $\Bbb R$ is a $G_\delta$ in $\Bbb R$. Done. If this is a problem from somewhere, re-read it to check the hypotheses: I’m willing to bet that they **don’t** include continuity of $f$.2012-10-12
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    I'll add links to related questions (You wrote yourself that you saw the proof at MSE, the links might be useful for other users.) For example: [Set of continuity points of a real function](http://math.stackexchange.com/questions/67620/set-of-continuity-points-of-a-real-function) and [How to show that the set of points of continuity is a $G_\delta$?](http://math.stackexchange.com/questions/138072/how-to-show-that-the-set-of-points-of-continuity-is-a-g-delta).2012-10-12
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    @BrianM.Scott You're right. I'm copying it wrong. I'll make the edit right away. $f$ is just suppose to be a function - not necessarily continuous.2012-10-12

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The definition of $A_n$ is a bit confusing. I think that what you want here is to let

$$A_n=\left\{x\in\Bbb R:\exists r_n(x)>0\,\forall x',x''\in B\big(x,r_n(x)\big)\left(\left|f(x'')-f(x')\right|<\frac1n\right)\right\}\;.$$

You definitely want the absolute values, and you need to say that it’s the points $x$ for which such a neighborhood $B\big(x,r_n(x)\big)$ exists. You don’t have to indicate explicitly the dependence of $r$ on $n$ and $x$ as I did here, but it doesn’t hurt, especially when you’re learning.

Now let $G=\bigcap_{n\in\Bbb Z^+}A_n$, and let $C=\{x\in\Bbb R:f\text{ is continuous at }x\}$. You need to show three things:

  1. Each $A_n$ is open.
  2. $C\subseteq G$. This is your ‘$\Rightarrow$’.
  3. $G\subseteq C$. This is your ‘$\Leftarrow$’.

You omitted (1) altogether, but it’s not hard: just show that if $x\in A_n$, then $B\big(x,r_n(x)\big)\subseteq A_n$, and conclude that $A_n=\bigcup_{x\in A_n}B\big(x,r_n(x)\big)$ and hence is open.

You’ve essentially got (2), but it could be stated much more clearly. Suppose that $x\in C$ and $n\in\Bbb Z^+$. Then there is an $r_n(x)>0$ such that $|f(x')-f(x)|<\frac1{2n}$ for all $x'\in B\big(x,r_n(x)\big)$. But then by the triangle inequality $$|f(x'')-f(x')|\le|f(x'')-f(x)|+|f(x)-f(x')|<\frac1n$$ for all $x',x''\in B\big(x,r_n(x)\big)$, so $x\in A_n$. And since $n\in\Bbb Z^+$ was arbitrary, $x\in G$.

Much the same applies to (3). Suppose that $x\in G$, and let $\epsilon>0$ be arbitrary. There is an $n\in\Bbb Z^+$ such that $\frac1n\le\epsilon$, and $x\in A_n$, so $|f(x')-f(x)|<\frac1n\le\epsilon$ for all $x'\in B\big(x,r_n(x)\big)$, i.e., for all $x'$ such that $|x'-x|, and it follows immediately that $x\in C$.

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    When you show that each $A_n$ is open, why do you need to conclude that $A_n=\bigcup_{x\in A_n}B\big(x,r_n(x)\big)$? Isn't it enough to show that if $x\in A_n$, then $B\big(x,r_n(x)\big)\subseteq A_n$?2015-01-11
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    @user144840: It depends on exactly how you’ve defined *open set* in a metric space and what equivalences you’ve already proved. Here I chose to use the definition that a set is open iff it’s a union of open balls (i.e., taking the open balls as a base for the topology) and not to assume that any equivalences had been proved. In essense I gave the proof that a set open by your definition is open by the one that I was using.2015-01-11