I am trying to prove that $f(x)=\sqrt{x}$ is uniformly continuous on $[0,\infty)$. I would like to ask, if my proof is (1) watertight and correct and (2) suffices as a general result for any such function $f$ that is continuous on a closed subset of $\mathbf{R}$.
Also, I wonder if there's an alternative way to prove it.
[Abbott, 4.4.7] Prove that $\displaystyle f( x) =\sqrt{x}$ is uniformly continuous on $\displaystyle [ 0,\infty )$.
Proof.
Let $\displaystyle c\in [ 0,\infty )$. $\displaystyle f( c)$ exists and is well-defined. Since, $\displaystyle f( x) =\sqrt{x}$ is continuous on $\displaystyle \mathbf{R}$, $\displaystyle \lim _{x\rightarrow c} f( x) =f( c)$.
By definition of continuity, for all $\displaystyle \epsilon >0$, there exists $\displaystyle \delta >0$, such that for all points $\displaystyle x\in [ 0,\infty )$ satisfying $\displaystyle | x-c| < \delta $, we have $\displaystyle | f( x) -f( c)| < \epsilon $.
We proceed by contradiction. Assume that $\displaystyle f( x)$ is not uniformly continuous on $\displaystyle [ 0,\infty )$ and there exists an $\displaystyle \epsilon _{0} >0$ and two sequences $\displaystyle ( x_{n})$ and $\displaystyle ( y_{n})$ such that $\displaystyle | x_{n} -y_{n}| \rightarrow 0$ whilst $\displaystyle | f( x_{n}) -f( y_{n})| >\epsilon _{0}$.
From continuity of $\displaystyle f$, there exists $\displaystyle \delta _{1} >0$, such that for all $\displaystyle | x_{n} -c| < \delta _{1}$, $\displaystyle f( x_{n}) \in V_{\epsilon _{0} /2}( f( c))$ , that is $\displaystyle | f( x_{n}) -f( c)| < \epsilon _{0} /2$.
There exists $\displaystyle \delta _{2} >0$, such that for all $\displaystyle | y-c| < \delta _{2}$, $\displaystyle f( y_{n}) \in V_{\epsilon _{0} /2}( f( c))$, that is $\displaystyle | f( y_{n}) -f( c)| < \epsilon _{0} /2$.
Let us compute the distance $\displaystyle | f( x_{n}) -f( y_{n})| $. We have:
\begin{equation*} \begin{array}{ r l c } | f( x_{n}) -f( y_{n})| & =| f( x_{n}) -f( c) +f( c) -f( y_{n})| & \\ & \leq | f( x_{n}) -f( c) |+|f( y_{n}) -f( c) | & \left\{\text{Triangle Inequality}\right\}\\ & < \frac{\epsilon _{0}}{2} +\frac{\epsilon _{0}}{2} =\epsilon _{0} & \end{array} \end{equation*} This contradicts our initial assumption, and therefore it must be false. $\displaystyle f$ is uniformly continous.
