Let R be the set of all real numbers and f: R | vedclass

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$f$ is both one-one and onto

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(C) Given $|f(x) - f(y)| \geq |x - y|$ for all $x, y \in R$.First,we show $f$ is one-one. Suppose $f(x_1) = f(x_2)$ for some $x_1, x_2 \in R$. Then $|f(x_1) - f(x_2)| = 0$. From the given inequality,$0 \geq |x_1 - x_2|$,which implies $|x_1 - x_2| = 0$,so $x_1 = x_2$. Thus,$f$ is one-one.Next,we show $f$ is onto. Since $f$ is continuous and one-one,$f$ must be strictly monotonic. If $f$ is strictly increasing,then $f(x) - f(y) \geq x - y$ for $x > y$. As $x 	o \infty$,$f(x) 	o \infty$,and as $x 	o -\infty$,$f(x) 	o -\infty$. If $f$ is strictly decreasing,then $f(y) - f(x) \geq x - y$ for $x > y$,which implies $f(x) - f(y) \leq -(x - y)$. As $x 	o \infty$,$f(x) 	o -\infty$,and as $x 	o -\infty$,$f(x) 	o \infty$. In both cases,the range of $f$ is $(-\infty, \infty) = R$. Thus,$f$ is onto.Therefore,$f$ is both one-one and onto.

Let $R$ be the set of all real numbers and $f: R \rightarrow R$ be a continuous function. Suppose $|f(x) - f(y)| \geq |x - y|$ for all real numbers $x$ and $y$. Then,

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