Proving that a function not uniformly continuous on $\mathbb{N}$ is not uniformly continuous on $\mathbb{R}$

Question

I am trying to prove that $f(x) = x^3$ is not uniformly continuous on $\mathbb{R}$ using the Sequential Criterion for Absence of Uniform Continuity. So, I found the sequences $(x_n) = n$ and $(y_n) = n+1/n$ where $n \in \mathbb{N}$ which satisfy the Sequential Criterion for Absence of Uniform Continuity, so, this means that $f(x) = x^3$ is not uniformly continuous on $\mathbb{N}$.

However, I have to show that $f(x) = x^3$ is not uniformly continuous on $\mathbb{R}$. Would it be valid if I just take $(x_n) = n$ and $(y_n) = n+1/n$ where $n \in \mathbb{R}$ instead of the sequences I currently have?

Sequential Criterion for Absence of Uniform Continuity:

a function $f:A \rightarrow $ R fails to be uniformly continuous on A iff there exists a particular $\epsilon_0$>0 and two sequences ($x_n$) and ($y_n$) in A, satisfying $|x_n -y_n| \rightarrow 0$, but $|f(x_n) - f(y_n)|\ge \epsilon_0$

Answer 1

Your sequential criterion does not restrict the conclusion to only the union of the elements in the sequence. You can pick $A=\mathbb R$ and the condition is still satisfied, so it fails to be uniformly continuous on $\mathbb R$ as well.

Answer 2

"so, this means that $f(x) = x^3$ is not uniformly continuous on $\mathbb{N}$." No! Every function on $\mathbb N$ is uniformly continuous on $\mathbb{N}.$ Proof: Given any $\epsilon>0,$ take $\delta = 1/2.$