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According to the classroom notes "Uniformly Continuous Linear Set" in American Mathematical Monthly, Vol. 62. No. 8(Oct., 1955) pp. 579-580, Author: Norman Levine, DOI: 10.2307/2307254.

In the proof of theorem 1,how to verify that f is continuous while not uniformly continuous itself?

thanks

Relevant part of the proof (in this proof $E$ is a linear set of points):

If $E$ is not closed, there is a point $c$ which is a limit point for $E$, but which does not belong to $E$. Then, without loss of generality, there exists a sequence of points $p_i$ in $E$ such that $p_i<p_{i+1}$ and $\lim p_i=c$.

Define $$f\equiv \begin{cases} 1 & \text{for }x=p_{2i-1} \\ 0 & \text{for }x=p_{2i} \end{cases} $$ and linear between $p_i$ and $p_{i+1}$, and let $f(x)=1$ for $x>c$ and for $x\le p_1$. It is easy to verify that $f(x)$ is continuous on $E$, but not uniformly continuous on $E$.

Martin Sleziak
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user90402
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  • try to show that for given $\epsilon>0$ the $\delta$ you get depends upon the point $x$(at which you want to see continuity) and also on $\epsilon$ – Mathronaut Aug 15 '13 at 07:41
  • what do you mean by uniformly continous – Suraj M S Aug 15 '13 at 07:42
  • @NeerajBhauryal I've been trying to prove it for a couple of hours but I'm still confused. Can you just hint me the way to show it please. – user90402 Aug 15 '13 at 08:38
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    @user90402 Please, try to give sufficient context to your posts. It is not good, if the large part of your post depends on an external source. (Which is, to my knowledge, not freely available.) This time, I've tried to include the necessary details. – Martin Sleziak Aug 15 '13 at 09:45

2 Answers2

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This is basically expanding on what @Neeraj said : Since $p_n \to c$, it is Cauchy, and hence for any $\delta > 0$, there exists $N_{\delta} \in \mathbb{N}$ such that $$ |p_{2i+1} - p_{2i}| < \delta \quad \forall i \geq N_{\delta} $$ Now suppose $f$ were uniformly continuous, then for $\epsilon = 1/2 > 0$, there is a $\delta > 0$ such that $$ |x-y| < \delta \Rightarrow |f(x) - f(y)| < 1/2 $$ But for this delta, choose $x = p_{2N_{\delta}+1}$ and $y = p_{2N_{\delta}}$, then $$ |x-y| < \delta, \text{but} |f(x) - f(y)| = 1 > 1/2 $$ Hence, $f$ cannot be uniformly continuous.

As for continuity, you can use the fact that $f$ is continuous iff whenever $x_n \to x$, then $f(x_n) \to f(x)$.

Prahlad Vaidyanathan
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A piecewise linear function is continuous - this shows that the function $f$ is continuous at each point $x<c$, $x\in E$. For $x>c$ the function $f$ is constant. Thus we have continuity on $E$ (=continuity at each point $x\in E$).

If the function were uniformly continuous on $E$, then a uniformly continuous extension of $\overline E$ would exists; see Continuous extension of a uniformly continuous function from a dense subset.

But no extension of the function $f$ can be continuous at the point $c$, since it attains the values 0 and 1 arbitrarily close to the point $c$.

It can be relatively easily shown that it is not uniformly continuous directly from the definition - if you look at the consecutive points from the sequence $p_i$, you get arbitrarily close points with "jump" of height 1 between their values.

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Martin Sleziak
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