Proving $x^3$ is uniformly continuous over $\mathbb R$ How do I choose the $X_n$ and $Y_n$ sequences to prove if this is uniformly continuous? Any help please no clue what to choose as the sequences.
 A: Choosing sequences you can't prove uniform continuity. You can just disprove it using sequences. Because there are uncountable sequences converging to the same limit.and if you get two sequences that satify
Uniform continuity criteria that doesnt mean that that the function is Uniformly continuous,this just shows that the sequences you took specifically worked for you but you cant say that for the uncountable sequences that you havn't checked..and which don't satisfy the criteria....and $f(x)=x^3$ is not uniformly continuous on R
A: Note that $x^3-y^3=(x-y)(x^2+xy+y^2)$
Now, take $\epsilon =1/2,\ x_n=n+1/n$ and $y_n=n-1/n$.
Then for any $\delta >0$ we may choose $N\geq 1$ so large that $2/N<\delta $. Then,
$\vert x_N-y_N\vert =2/N<\delta $ and yet 
$\vert x^3-y^3\vert =\vert (x_N-y_N)(x_N^2+x_Ny_N+y_N^2)\vert =\frac{2}{N}\cdot (3N^2+\frac{1}{N^{2}})>N>1>\epsilon.$
A: It is not uniformly continuos. The intuition is that $x^3$ keeps increasing at a faster rate; so if you need to "restrict " the values of the function in a neighborhood of a point, you need $\delta$ progressively smaller to make up the fact that $x^3$ tends to "escape faster". Since $\delta$ must depend on the point where you evaluate continuity, then it's not uniformly continuos.
Question

Then why is $x^3$ uniformly continuos on any compact interval like
  $[0,1]$? (Heine Borel theorem). Wouldn't you still need to take
  $\delta$ progressively smaller?

Yes, indeed. But since in this case you're dealing with a compact interval, you can find a $\delta$ that works for every point. (Basically you take the "smallest" of the delta; the infimum is a minimum and you can take a $\delta$ that works)
But on $\mathbb R$, this infimum exists, but it's $0$! and of course it's not a minimum.
This is an intuitive explanation :) To prove it rigorously refer to the other answers 
A: HINT:
Let $x_n=n$ and $y_n=n+1/n$ for integer valued $n\ge 1$.  Then, show that 
$$|x_n^3-y_n^3| \ge \epsilon =1$$
