Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
See also math.stackexchange.com/questions/67620/… –  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
1  
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

2 Answers 2

up vote 1 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 $$ |f(s) - f(t)| \le |f(s) - f(x)| + |f(x) - f(t)| \le 1 /n \ \text{whenever} \ s,t \in (x - \delta, x + \delta) $$

share|improve this answer

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 $x$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.