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Define $g:\mathbb{R}^2 \to \mathbb{R}^2 \times \mathbb{R}^2$ by $g(x,y) = \left((x,y), (x,y)\right)$. Prove that $g$ is continuous. This exercise came out of the Continuity section of Mendelson's Introduction to Topology. I was first confused because it doesn't specify which metric is being used, but my professor suggested the max metric. So I went ahead and tried the following:

Let $\vec{x} = (x_1,x_2) \in \mathbb{R}^2$ and $\varepsilon > 0$ be given. Assume $d(\vec{x},\vec{y}) = \max_{1\leq i \leq n}\{|x_i-y_i|\} < \delta$. Then $d(g(\vec{x}),g(\vec{y})) = \max_{1\leq i \leq n}\{|g(\vec{x})-g(\vec{y})|\} = \max_{1\leq i \leq n}\{|(\vec{x},\vec{x})-(\vec{y},\vec{y})|\}$.

Here, I get confused because of all the parentheses going on. Can ordered pairs of vectors be subtracted component-wise? i.e. $(\vec{x},\vec{x})-(\vec{y},\vec{y})=(\vec{x}-\vec{y},\vec{x}-\vec{y}) = ((x_1-y_1, x_2-y_2),(x_1-y_1, x_2-y_2))$? If this is true, then what is $|((x_1-y_1, x_2-y_2),(x_1-y_1, x_2-y_2))|$?

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    Side note: this map is a special case of a general construction that goes under the name _diagonal map_. I'm sure you can find general topological proofs if you search for "diagonal map is continuous" on your favourite search engine. – 2017-02-09
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    Interesting! I couldn't visualize the result in $\mathbb{R}^4$, but thinking about $\mathbb{R} \to \mathbb{R} \times \mathbb{R}$ illuminates the diagonal-ness of it. However, when searching for proofs involving the continuity of diagonal maps in topology, I can't seem to find any involving $\varepsilon$-$\delta$ proofs--just ones involving open sets. – 2017-02-09
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    It is easier to identify $\mathbb{R}^2\times\mathbb{R}^2$ with $\mathbb{R}^4$. Then, you can see that you don't need to worry about parentheses as you do. – 2017-02-09

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This can be heavily generalized. Let $X$ be a topological space and consider the function

$$\Delta:X\to X\times X$$ $$\Delta(x)=(x,x)$$

In order to show that $\Delta$ is continous it is enough to show that $\Delta^{-1}(U\times W)$ is open in $X$ for any open $U,W\subseteq X$. So let $x\in \Delta^{-1}(U\times W)$. Then $(x,x)=\Delta(x)\in U\times W$ so $x\in U\cap W$. Note that $U\cap W$ is open and also

$$\Delta(U\cap W)\subseteq (U\cap W)\times (U\cap W)\subseteq U\times W$$

In particular $U\cap W$ is an open neighbourhood of $x$ contained in $\Delta^{-1}(U\times W)$. This completes the proof.