I have to prove that $f: \mathbb{Z} \to \mathbb{R}: x \mapsto x²$ is uniformly continuous (or not)

My attempt:

Suppose $f$ is not uniformly continuous.

Then $\exists \epsilon >0: \forall \delta >0: \exists x,y \in \mathbb{Z}: |x-y| < \delta \land |x² - y²| \geq \epsilon$

Choose such an $\epsilon > 0$ and $\delta = 1/2$

Then, there exists $x,y \in \mathbb{Z}$ such that both $|x-y| < 1/2 $ and $|x²-y²| \geq \epsilon$

But from $|x-y| < 1/2$, we find that $x=y$. Hence $|x²-y²| = 0$, contradicting the fact that $|x²-y²| \geq \epsilon$

Hence, $f$ must be uniformly continuous

Is this correct?


You are correct!

An alternative way would be to prove it directly instead of proving by contradiction.

To prove it directly, given $\varepsilon > 0$ we need to find $\delta > 0$ such that $|x^2 - y^2| < \varepsilon$ for all $x,y \in \mathbb{Z}$ satisfying $|x - y| < \delta$. Now, for any $\varepsilon > 0$, $\delta = 1/2$ works for the same reason as in your proof: $|x-y| < 1/2$ and $x,y \in \mathbb{Z}$ imply that $x=y$, and so $|x^2 - y^2| = 0 < \varepsilon$ for any $\varepsilon > 0$. In particular, $\delta$ is independent of $x$ and $y$, and so it is uniformly continuous.


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