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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 $\cap A_n$.

Thanks.

  • 0
    @BrianM.Scott: thanks. I've resolved it now.2012-04-28

2 Answers 2

5

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) $

3

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 $x$.

  • 0
    Actually, you can take $s=x$, since x-\delta < x < x+\delta.2012-04-29