Wikipedia's proof that $f : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$ is not uniformly continuous Prove that $f : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$ is not uniformly continuous


Now I actually ask this question, as Wikipedia's proof of this seems wrong to me.



Wikipedia's Proof:

$f$ is uniformly continuous on $\mathbb{R}$ if $\forall \epsilon &gt; 0$, there exists a $\delta &gt; 0$ for all $x_1, x_0 \in \mathbb{R}$ such that

$$|x_1 - x_0| &lt; \delta \implies |f(x_1) -f(x_0)| &lt; \epsilon$$

But $$\begin{aligned} 
f(x+ \delta) - f(x) &amp;= x^2 + 2x\delta +\delta^2 - x^2\\
&amp;= 2x\delta + \delta^2\\
&amp;= \delta(2x + \delta)\\
&amp;&gt; \epsilon \ \ \ \text{for suffiently large $x$} \  \ \ \square
\end{aligned}$$ 



Now what I don't get is that if we set $x_1 = x+ \delta$ and $x_0 = x$, then $|x_1 -x_0| = |x+ \delta - x| = |\delta| \not&lt; \delta$, so everything above should not apply as $|x+ \delta - x| \not &lt; \delta$ in the first place.

Is Wikipedia's proof correct? If so, then why would my argument above be incorrect?
You can replace the $|x - y| &lt; \delta$ with $|x - y | \leq \delta$ in the definition of uniform continuity  and not fundamentally change anything.  However, if it makes you feel comfier, you can fix it up by choosing a $\rho &lt; \delta$ and proceed exactly as shown, but with $f(x+ \rho) - f(x)$.



We can establish the equivalence of the two slightly different definitions as follows:

First, if there exists a $\delta$ such that $|x - y| &lt; \delta \implies |f(x) - f(y)| &lt; \varepsilon$, then there exists a $\rho$ such that $|x - y| \leq \rho \implies |f(x) - f(y)| &lt; \varepsilon$.  For example, $\rho = \delta / 2$.

On the other hand, if there exists a $\delta$ such that $|x - y| \leq \delta \implies |f(x) - f(y)| &lt; \varepsilon$, then of course it's also true that $|x-y| &lt; \delta \implies |f(x) - f(y)| &lt; \varepsilon$.



One can also change the $&lt; \varepsilon$ condition to $\leq \varepsilon$.  To establish this equivalence, first suppose that if we are given any $\varepsilon &gt; 0$, we can find a $\delta$ such that $|x - y| &lt; \delta \implies |f(x) - f(y)|&lt; \varepsilon$.  It follows immediately that $|x-y| &lt; \delta \implies |f(x) - f(y)| \leq \varepsilon$.

On the other hand, suppose that if we are given any $\varepsilon &gt; 0$, we can find a $\delta$ such that $|x-y| &lt; \delta \implies |f(x) - f(y)| \leq \varepsilon$. By choosing a $\delta$ that satisfies this condition for some $\varepsilon' &lt; \varepsilon$, we thus have a $\delta$ such that $|x-y| &lt; \delta \implies |f(x) - f(y)| &lt; \varepsilon$.



I suspect the reason why the strict less-than condition for both $\varepsilon$ and $\delta$ is prefered in analysis texts is because this emphasizes the more general topological definition of continuity.  Namely, the preimage of an open set is open.
If we put

$$f(x)=x^2,$$

$$x_n=\sqrt{n+1},$$

and $$y_n=\sqrt{n}.$$

then we have

$$\lim_{n\to+\infty}(x_n-y_n)=0$$

but

$$|f(x_n)-f(y_n)|=1$$

So, $\;\;f\;\;$is not uniformly continuous at $[0,+\infty)$.

If we take $\epsilon=0.5$ for example,

then $\forall \eta&gt;0$,

we will always find $x_n$ and $  y_n$ such that

$|x_n-y_n|&lt;\eta$ and

$|f(x_n)-f(y_n)|=1&gt;\epsilon$.