Let $(X,d_{X})$ be a metric space, and let $f:X\to\textbf{R}$ and $g:X\to\textbf{R}$ be uniformly continuous functions. Show that the direct sum $f\oplus g:X\to\textbf{R}^{2}$ defined by $f\oplus g(x) = (f(x),g(x))$ is uniformly continuous.
MY ATTEMPT
Let $\varepsilon/2 > 0$. Then there exist $\delta_{1} > 0$ and $\delta_{2} > 0$ such that for every $x,y\in X$ \begin{align*} \begin{cases} d_{X}(x,y) < \delta_{1} \Rightarrow |f(x) - f(y)| < \varepsilon/2\\\\ d_{X}(x,y) < \delta_{2} \Rightarrow |g(x) - g(y)| < \varepsilon/2 \end{cases} \end{align*}
Let us equip $\textbf{R}^{2}$ with the Euclidean metric. Since the following inequality holds \begin{align*} \sqrt{|f(x) - f(y)|^{2} + |g(x) - g(y)|^{2}} \leq |f(x) - f(y)| + |g(x) - g(y)| \end{align*} for every $\varepsilon > 0$ there corresponds a $\delta = \min\{\delta_{1},\delta_{2}\}$ such that for every $x,y\in X$ \begin{align*} \sqrt{|f(x) - f(y)|^{2} + |g(x) - g(y)|^{2}} \leq |f(x) - f(y)| + |g(x) - g(y)| < \varepsilon \end{align*} whenever $d_{X}(x,y) < \delta$, and the proposed result is valid.
Does anyone want to make any suggestion or critique? Any of them are welcome.
This is entirely correct, but I'd like to make 2 small nitpicks:
It's probably better to start the proof by saying "Let $\varepsilon > 0$" (and if you like you can proceed to say "so that $\varepsilon/2 > 0$"), because $\varepsilon$ is the thing you really need to pick arbitrarily.
It only makes sense to ask if a function is uniformly continuous when its domain and codomain are metric spaces: if you haven't specified a metric on $\mathbb{R}^2$, it doesn't even make sense to ask if $f \oplus g$ is uniformly continuous. So you don't get to make the choice of which metric to equip $\mathbb{R}^2$ with, but you're right that the Euclidean metric is (probably) what's intended here. You should rephrase your homework solution(?) to begin with "I will assume that $\mathbb{R}^2$ has the standard Euclidean metric", or something like that.