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

Thanks.

hbghlyj
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Linda
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  • See also http://math.stackexchange.com/questions/67620/set-of-continuity-points-of-a-real-function – 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
  • @BrianM.Scott. So So, for $s,t \in (x-\delta, x+\delta)$, $|f(s)-f(t)|\lt 1/n \lt \epsilon$. But I want to show that $|f(x')-f(x)|\lt \epsilon$ if $x\in (x'-\delta, x'+\delta)$ and I'm confused setting it up. – Linda Apr 28 '12 at 16:31
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    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
  • @BrianM.Scott: thanks. I've resolved it now. – Linda Apr 28 '12 at 16:50
  • https://math.stackexchange.com/a/3757099/770687 – user770687 Aug 18 '20 at 21:09

2 Answers2

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

Saaqib Mahmood
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user29999
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5

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 $x$.

Siminore
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