Proof that $x^2$ is not uniformly continuous on $\mathbb R$ This is my attempt. My instructor said it was incorrect. I would like to know exactly where my mistake is.
Suppose $x^2$ is uniformly continuous on $\mathbb R$.
So, $\forall \epsilon > 0 \ \exists \delta > 0 \ \forall x,y \in \mathbb R \ s.t. \ if \ |x - y| < \delta \ \text{then} \ |f(x) - f(y)| < \epsilon$
In this case, $|x - y| < \delta \ \text{and} \ |x^2 - y^2| < \epsilon$
$|(x - y)(x + y)| < \epsilon$
$|(x - y)| |(x + y)| < \epsilon$
$\delta|x + y| < \epsilon$
Notice, $|x + y|$ is unbounded on the real line. So $x$ and $y$ can be made arbitrarily large. So there will not always be a $\delta$ that will make all values of $x$ and $y$ less than an arbitrarily small $\epsilon$.
Therefore, $x^2$ is not uniformly continuous on $\mathbb R$.
 A: The intuition is not bad, but it is incorrect. From $|(x-y)(x+y)|<\varepsilon$ you cannot go to $\delta|x+y|\le\varepsilon$. Indeed, by applying $|x-y|<\delta$, you have $|(x-y)(x+y)|<\delta|x+y|$. So you have that one expression (say $|(x-y)(x+y)|$) is bounded by two different estimates ($\delta|x+y|$ and $\varepsilon$) but you cannot conclude anything about wich one is bigger.
A correct approach is trying to find a concrete value of $\varepsilon$ for wich for any $\delta, you have the oppose inequality.
For example, take $\varepsilon=1$ and try to prove that for any $\delta>0$ you always can find $x,y\in\Bbb R$ such that $|x-y|<\delta$ and $|x^2-y^2|\ge 1$. And to do that, your intuition about the factor $(x+y)$ will helps you.
A: There is no single mistake. Even if every step were correct, which is not the case, this doesn’t have the right logical structure to be a proof that the function $f(x)=x^2$ is not uniformly continuous. For that you need to show that there is some $\epsilon>0$ such that no matter what $\delta>0$ you choose, there are points $x_\delta,y_\delta\in\Bbb R$ such that $|x_\delta-y_\delta|<\delta$, but $|x_\delta^2-y_\delta^2|\ge\epsilon$. For example, you could let $\epsilon=1$ and for each $\delta>0$ show how to find points $x_\delta$ and $y_\delta$ such that $|x_\delta-y_\delta|<\delta$, but $|x_\delta^2-y_\delta^2|\ge 1$.
I mentioned that not every step is correct: in particular, $|x-y|\cdot|x+y|<\epsilon$ and $|x-y|<\delta$ do not imply that $\delta|x+y|<\epsilon$.
That said, I suspect that you do have a reasonable intuitive understanding of why $f(x)=x^2$ is not uniformly continuous on $\Bbb R$.
A: Your intuition as to why it's not uniformly continuous is correct, but your proof doesn't make any sense. Start with the definition of not uniformly continuous, which is what you're trying to prove.
There exists $\epsilon_0>0$ such that for all $\delta>0$ there exists $x,y$ with $|x-y|<\delta$, but $|f(x)-f(y)|\geq \epsilon_0$.
In this case, just choose $\epsilon_0=1$. Then for every $\delta$, try to come up with $x,y$ such that $|x-y|<\delta$, but $|x-y||x+y|\geq 1$. Let $x$ be arbitrary and pick $y=x+\delta/2$. Then $|x-y|=\delta/2<\delta$ is satisfied and
$$
|x-y||x+y|=\frac{\delta}{2}\left(2x+\frac{\delta}{2}\right).
$$
If $x$ is sufficiently large....
A: Do it with sequences
Let
$$x_n=n \;\; and\;y_n=n+\frac 2n$$
we have
$$\lim_{n\to+\infty}|x_n-y_n|=0$$
But
$$\lim_{n\to+\infty}|x_n^2-y_n^2|=4\ne0$$
thus
$x\mapsto x^2 $ is not uniformly continuous at $\Bbb R$.
We proved that with $\epsilon=3 $ and any $ \eta>0 $, we take $ n $ such that $ \frac 1n<\eta $ and check that
$$|x_n-y_n|=\frac 1n <\eta$$
but
$$|x_n^2-y_n^2|=4+\frac{4}{n^2}>3$$
