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I am trying to solve this exercise from Royden's 3rd edition.

The question is as follows: Let $f$ be a real-valued function defined for all real numbers. Show that the set of points at which $f$ is continuous is a $G_{\delta}$.

Let $$A_n = \{y : \text{there is a }~\delta_y \gt 0 : |f(s)-f(t)|\lt 1/n ~ \text{whenever}~ s,t \in (y-\delta, y+\delta)\}\;.$$

Then by the definition of open sets, $A_n$ is open.

To complete the proof, I need help in showing that $f$ is continuous at say $x$ if and only if $x\in \cap A_n$.

If $f$ is continuous at $x$, the there is a $\delta \gt 0$ such that $|f(x) - f(a)| \lt 1/n$ whenever, $x\in (a-\delta, a+\delta)$. so $x \in A_n$ son it must be in $\cap A_n$.


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See also… – Zev Chonoles Apr 28 '12 at 15:43
Have you made any progress at all on either of the two implications? Showing that if $f$ is continuous at $x$, then $x\in\bigcap_nA_n$ is pretty straightforward. – Brian M. Scott Apr 28 '12 at 15:44
@BrianM.Scott I just figured out that part. see edit. How about the other direction? – Linda Apr 28 '12 at 15:48
Your edit is almost right. What you should say is that for each positive integer $n$ there is a $\delta>0$ such that etc. For the other direction, assume that $x\in\bigcap_nA_n$, and let $\epsilon>0$. Then there is a positive integer $n$ such that $1/n<\epsilon$, so ... ? – Brian M. Scott Apr 28 '12 at 15:48
You’re assuming that $x\in\bigcap_nA_n$. You need to show that for any $\epsilon>0$ there is a $\delta>0$ such that $|f(x')-f(x)|<\epsilon$ whenever $x'\in(x-\delta,x+\delta)$, so let $\epsilon>0$; there is a positive integer $n$ such that $1/n<\epsilon$. Now $x\in A_n$, so by the definition of $A_n$ there is a $\delta_x>0$ such that $|f(x')-f(x)|<1/n<\epsilon$ whenever $x'\in(x-\delta_x,x+\delta_x)$, and that’s exactly what you need. – Brian M. Scott Apr 28 '12 at 16:44
up vote 2 down vote accepted

If $f$ is continuous in $x$ there exist $\delta_x$ such that $$ |f(s) - f(x)| \ < \dfrac{1}{2n} \ \text{whenever} \ s \in (x - \delta, x + \delta) $$ Hence if $s,t \in (x - \delta, x + \delta)$ we have $$ |f(s) - f(t)| \le |f(s) - f(x)| + |f(x) - f(t)| \le 1 /n \ \text{whenever} \ s,t \in (x - \delta, x + \delta) $$

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A point $x \in \bigcap_n A_n$ iff, for every $\varepsilon>0$ there exists $\delta>0$ such that $|f(t)-f(s)|<\varepsilon$ whenever $x-\delta < s \leq t < x+\delta$. But this condition, via some triangular inequality, is simply the definition of continuity at the point $x$.

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Can you help with that part? – Linda Apr 28 '12 at 16:28
Actually, you can take $s=x$, since $x-\delta < x < x+\delta$. – Siminore Apr 29 '12 at 8:30

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