Examine if the function $f(x)=x^2, \;\; x \in \bigcup \limits_{n=1}^{\infty}\left [ 2n, 2n+1 \right ]$ is uniformly continuous.
Solution
No, it is not. Let $A=\bigcup \limits_{n=1}^{\infty}\left [ 2n, 2n+1 \right ]$, then for each $n \in \mathbb{N}$ we have that $2n+\frac{1}{n} \in A$. However,
$$\lim_{n \rightarrow +\infty}\left ( 2n+ \frac{1}{n} - 2n \right )=0 \neq 4 = \lim_{n \rightarrow +\infty} \left [ \left ( 2n+\frac{1}{n} \right )^2 -(2n)^2 \right ]$$
meaning that $f$ is not uniformly continuous in $A$.
Solution
No, it is not. Let $A=\bigcup \limits_{n=1}^{\infty}\left [ 2n, 2n+1 \right ]$, then for each $n \in \mathbb{N}$ we have that $2n+\frac{1}{n} \in A$. However,
$$\lim_{n \rightarrow +\infty}\left ( 2n+ \frac{1}{n} - 2n \right )=0 \neq 4 = \lim_{n \rightarrow +\infty} \left [ \left ( 2n+\frac{1}{n} \right )^2 -(2n)^2 \right ]$$
meaning that $f$ is not uniformly continuous in $A$.
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