Is x^2 Not Uniformly Continuous on the Real Numbers?

[Lee33]

Homework Statement

Show that the function $x^2$ is not uniformly continuous on $\mathbb{R}$

Homework Equations

Delta - Epsilon Definition:

$\exists \epsilon > 0, \ \forall \delta >0, \exists x \in S [|x-x_0|< \delta \text{and} |x^2 - x_0^2| \ge \epsilon ].$

The Attempt at a Solution

I am confused on how they got $\epsilon$ and $\delta$. It seems they found it through thin air.

What I did was:

Let $\delta >0$ and $|x-x_0| < \delta$ then for a given $\epsilon >0$ we have that $|f(x) - f(x_0)| = |x^2 - x_0^2| = |x - x_0||x+x_0| = \delta|x+x_0| = \delta|x-x_0 +2x_0| = \delta[|x-x_0| + |2x_0|] = \delta^2 + \delta|2x_0| ...$

I googled this and find out they chose $\epsilon = 1$ and $x= x_0 + \delta/2$ and $x_ 0 = 1/\delta$. I am not sure how they got that? And this happens a lot in real analysis, they choose an $\epsilon$ and $\delta$ that I have no idea how they got. It's like they got it from thin air. I know those $\epsilon$ and $\delta$ work but they never provide information on how they obtained it.

Can anyone help me rigorously show how to solve problems like this?

 

[mighty2000]

Thank you for your question. I understand your confusion about how to choose appropriate values for $\epsilon$ and $\delta$ in the proof for non-uniform continuity of the function $x^2$ on $\mathbb{R}$. Let me walk you through the steps to help you better understand the reasoning behind the choices of $\epsilon$ and $\delta$.

First, let's recall the definition of uniform continuity. A function $f$ is uniformly continuous on a set $S$ if for any given $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x, y \in S$, if $|x-y| < \delta$, then $|f(x)-f(y)| < \epsilon$. In other words, for any given $\epsilon$, we need to find a $\delta$ that works for all points in the set $S$.

Now, let's apply this definition to the function $x^2$ on $\mathbb{R}$. We want to show that this function is not uniformly continuous, which means that there exists an $\epsilon > 0$ such that for all $\delta > 0$, we can find points $x, y \in \mathbb{R}$ such that $|x-y| < \delta$ but $|f(x)-f(y)| \geq \epsilon$.

To prove this, we need to choose a specific value for $\epsilon$ and then find a corresponding $\delta$ that works for all points in $\mathbb{R}$. In this case, we can choose $\epsilon = 1$. Why did we choose this value? We want to show that as we get closer and closer to $x_0$, the difference between $f(x)$ and $f(x_0)$ can get arbitrarily large. By choosing $\epsilon = 1$, we are essentially saying that no matter how small we make $\delta$, we can always find points $x, y$ such that $|x-y| < \delta$ but $|f(x)-f(y)| \geq 1$. This will demonstrate that the function is not uniformly continuous.

Now, let's find a corresponding $\delta$ that works for all points