Exercise 2.3.5

Question

Let $\left(X, d_{X}\right)$ be a metric space, and let $f: X \rightarrow \mathbb{R}$ and $g: X \rightarrow \mathbb{R}$ be uniformly continuous functions. Show that the direct sum $f \oplus g: X \rightarrow \mathbb{R}^{2}$ defined by $f \oplus g(x):=(f(x), g(x))$ is uniformly continuous.

Prakash Anand · Accepted Answer

\begin{proof} Since $f, g$ are uniformly continuous we know that for all $x_{0} \in X$ given $\epsilon^{\prime}>0$ there exists $\delta^{\prime}>0$ such that if $d_{X}\left(x, x_{0}\right)<\delta^{\prime}$, then $d_{\mathbb{R}}\left(f(x), f\left(x_{0}\right)\right)<\epsilon^{\prime}$, and $d_{\mathbb{R}}\left(g(x), g\left(x_{0}\right)\right)<\epsilon^{\prime}$. So given $\epsilon>0$ take $\epsilon^{\prime}=\epsilon / \sqrt{2}$, and $\delta=\delta^{\prime}$. If $d_{X}\left(x, x_{0}\right)<\delta^{\prime}=\delta$, then from the uniform continuity of $f, g$ we get that: $$d_{\mathbb{R}}\left(f(x), f\left(x_{0}\right)\right)=\left|f(x)-f\left(x_{0}\right)\right|<\epsilon^{\prime}= \frac{\epsilon}{\sqrt{2}}$$ and $$d_{\mathbb{R}}\left(g(x), g\left(x_{0}\right)\right)=\left|g(x)-g\left(x_{0}\right)\right|<\epsilon^{\prime}=\frac{\epsilon}{\sqrt{2}}.$$ Thus, $$\left(f(x)-f\left(x_{0}\right)\right)^{2}+\left(g(x)-g\left(x_{0}\right)\right)^{2}<\epsilon^{2}.$$ Hence, $$\sqrt{\left(f(x)-f\left(x_{0}\right)\right)^{2}+\left(g(x)-g\left(x_{0}\right)\right)^{2}}<\epsilon.$$ But, $$\sqrt{\left(f(x)-f\left(x_{0}\right)\right)^{2}+\left(g(x)-g\left(x_{0}\right)\right)^{2}}=d_{\mathbb{R}^{2}}\left((f(x), g(x)),\left(f\left(x_{0}\right), g\left(x_{0}\right)\right)\right)<\epsilon .$$ Therefore $f \oplus g$ is uniformly continuous. \end{proof}

Exercise 2.3.5

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Exercise 2.3.5

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