how to verify that f is continuous while not uniformly continuous itself.

Question

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

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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 $E$.

Answer 1

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.

Answer 2

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 $f(x_n) \to f(x)$.