Sequential Criterion for Non-Uniform Continuity

Detailed Solution

The key idea is the sequential characterization of uniform continuity. A function $f:\mathbb R\to\mathbb R$ is uniformly continuous on $\mathbb R$ if and only if for every pair of sequences $(x_n)$ and $(y_n)$ in $\mathbb R$ satisfying $$ |x_n-y_n|\to0, $$ we have $$ |f(x_n)-f(y_n)|\to0. $$ Since $f$ is assumed to be **not uniformly continuous**, the contrapositive of this characterization gives the existence of sequences $(x_n)$ and $(y_n)$ such that $$ |x_n-y_n|\to0 $$ but $$ |f(x_n)-f(y_n)|\not\to0. $$ More explicitly, because uniform continuity fails, there exists some $\varepsilon_0>0$ such that for every $\delta>0$, there are $x,y\in\mathbb R$ satisfying $$ |x-y|<\delta $$ but $$ |f(x)-f(y)|\ge\varepsilon_0. $$ Taking $\delta=1/n$, we can choose $x_n,y_n\in\mathbb R$ such that $$ |x_n-y_n|<\frac1n $$ and $$ |f(x_n)-f(y_n)|\ge\varepsilon_0. $$ Therefore, $$ |x_n-y_n|\to0, $$ while $$ |f(x_n)-f(y_n)|\not\to0, $$ because it is always at least $\varepsilon_0>0$. Hence option B must be true. Option A is false. A continuous function can be bounded and still fail to be uniformly continuous on $\mathbb R$. For example, $$ f(x)=\sin(x^2) $$ is bounded on $\mathbb R$ but is not uniformly continuous. Option C is not the required characterization of failure of uniform continuity. In fact, the conclusion $$ |f(x_n)-f(y_n)|\to0 $$ is precisely the sequential behavior associated with uniform continuity, not its failure. Moreover, as written, C is too weak because one can trivially take $x_n=y_n$. Option D is false because the hypothesis already assumes that $f$ is continuous on all of $\mathbb R$. Failure of uniform continuity does not imply failure of ordinary continuity. Thus, the sequential obstruction to uniform continuity is exactly the existence of two sequences whose arguments become arbitrarily close while their function values remain separated by a positive amount. Therefore, $$ \boxed{\mathrm{B}}. $$