How Do You Prove a Function is Not Uniformly Continuous?

[Incand]

Homework Statement

Let $f:X \to Y$. Show that
$f$ not uniform continuous on $X$ $\Longleftrightarrow$ $\exists \epsilon > 0$ and sequences $(p_n), (q_n)$ in $X$ so that $d_X(p_n,q_n)\to 0$ while $d_Y(f(p_n),f(q_n))\ge \epsilon$.

Homework Equations

Let $f:X\to Y$. We say $f$ is uniform continuous on $X$ if $\forall \epsilon >0 \exists \delta > 0$ so that
$d_Y(f(x),f(y))< \epsilon$ $\forall x,y\in X$ for which $d_X(x,y)< \delta$.

The Attempt at a Solution

I was hoping someone could take a look at my proof and check if it's correct or not. I found the second part especially hard to formulate so I'm mostly unsure about that part.

Starting with $\Longleftarrow$
$d(p_n,q_n) \to 0$ means by definition that
$\forall \delta > 0$ $\exists N$ so that $d(p_n,q_n) < \delta$ for $n \ge N$. But then $f$ can't be uniform continuous since the points $p_n, q_n \in X$ and $d_Y(f(p_n),f(q_n))\ge \epsilon_0$ we have a counter example to
$d(x,y)<\delta \Longrightarrow d(f(x),f(y)< \epsilon$ if we take a $\epsilon < \epsilon_0$.

$\Longrightarrow$
That $f$ is not uniform continuous implies that $\exists \epsilon >0$ so that $\forall \delta > 0$ there exists $x,y\in X$ so that
$d_X(x,y)< \delta \Longrightarrow d_Y(f(x),f(y))\ge \epsilon$. Let's select some $x,y$ satisfying this and set $p_n = x$ and $q_n =y$ and we're done.

 

[stevendaryl]

$\Longrightarrow$
That $f$ is not uniform continuous implies that $\exists \epsilon >0$ so that $\forall \delta > 0$ there exists $x,y\in X$ so that
$d_X(x,y)< \delta \Longrightarrow d_Y(f(x),f(y))\ge \epsilon$. Let's select some $x,y$ satisfying this and set $p_n = x$ and $q_n =y$ and we're done.

You're not finished. You haven't defined the sequences $p_n$ and $q_n$.

Maybe it helps to look at a concrete case: $f(x) = x^2$. This is not uniformly continuous. Pick $\epsilon = 1$. Then for any $\delta > 0$, we can find an $x$ and $y$ such that $|x-y| < \delta$, but $|f(x) - f(y)| > \epsilon$. You just let $x = \frac{1}{\delta}$, and let $y = x+\frac{\delta}{2}$. Then even though $|x-y| < \delta$, $|x^2 - y^2| = 1 + \frac{\delta^2}{4} > \epsilon$.

So it's not uniformly continuous. But how does your argument show that there is a sequence $p_n$ and a sequence $q_n$ such that $|p_n - q_n| \rightarrow 0$ but $|(p_n)^2 - (q_n)^2| > \epsilon$?

(It's not hard to come up with such a sequence, but it doesn't seem to follow immediately from your proof.)

 

[Incand]

Good catch there! I tried to fix the problem:
So we have $d_X(x,y)<\delta \Longrightarrow d_Y(f(x),f(y))\ge \epsilon$.
Set $N= floor(1+1/\delta)$ Then $d_X(x_n,y_n)\le 1/n < \delta$ for $n\ge N$ implies $d_Y(f(x),f(y))\ge \epsilon$.
So for each $n$ select $x_n,y_n$ satisfying the above which will then form two sequences $x_n$ and $p_n$ for which $d_X(x_n,y_n)< 1/n \to 0$.

For your example we could set $p_n = n+1/n$ and $q_n = n$ then $|p_n-q_n| = 1/n \to 0$ while $|(n+1/n)^2-n^2| = 2+1/n^2 \to 2$.

 

[Incand]

In the proof above we can shift the indexes to start counting from $1,2,\dots$ by setting $p_1 = x_N, p_2 = x_{N+1}, \dots$