I find a hard time understanding the above example of continuity of coordinate functions. What I understood from the example is that a coordinate function takes a point in $R^k$ and gives a real number; how is that gonna be continuous on $R^k$. May someone explain please?
On
The definition of continuity is the same as the one you are used to for single variate functions. If you want to keep distances in the range smaller than say $\epsilon>0$, then you can find a suitably small $\delta>0$ ball in the domain for which inputs in this ball map to outputs inside the $\epsilon>0$ ball.
Try it with a two variable example: If $$ \phi_1(x,y)=x $$ Then insuring $|x-x_0|+|y-y_0|\leq \delta$ (the 1 norm in the domain, norms on $\mathbb{R}^n$ are equivalent), then $$ |\phi_1(x,y)-\phi(x_0,y_0)|=|x-x_0|\leq \delta $$ the norm in the range being used above. So indeed, choosing $\delta=\epsilon$ meets any $\epsilon>0$ challenge.
$\phi_i$ just picks out the coordinate $x_i$.
It's just the statement that if distance between two vectors goes to $0$, then the distance between each of their coordinates also goes to $0$.
If you want, you could use the definition of the standard metric on $\mathbb{R}^n$ and then plug in what everything is:
$$ |\mathbf{x} - \mathbf{y}| = \sqrt{(x_1-y_1)^2 + ... + (x_n-y_n)^2 } $$ so
$$|\phi_i(\mathbf{x}) - \phi_i(\mathbf{y})| = |x_i - y_i|= \sqrt{(x_i-y_i)^2}$$
As you can see, just as Rudin says, $$|\phi_i(\mathbf{x}) - \phi_i(\mathbf{y})| \leq |\mathbf{x} - \mathbf{y}|$$
And so, given $\epsilon$, put $\delta = \epsilon$, if $|\mathbf{x} - \mathbf{y}|< \delta$, then $|\phi_i(\mathbf{x}) - \phi_i(\mathbf{y})| < \epsilon$