I'm studying through Mendelson's Introduction to Topology and have a question on proving continuity.
First, let's define two distance functions. Let $x= (x_{1}, ..., x_{y})$ and $y=(y_{1}, ... , y_{n})$. Let $d(x,y)=\text{max} \big\{|x_{i} - y_{i}| \big\}$ and $d'(x,y) = \sqrt{\sum_{i=1}^{n} (x_{i} - y_{i})^{2}}$.
Now see the following theorem:
Theorem $3.5$ from Mendelson's Intro to Topology
What I don't understand is how can we compare and say that $$\sqrt{\sum_{i=1}^{n} (x_{i} - a_{i})^{2}} < \sqrt{n \delta^{2}}$$
Let's let $n=3$ to make this more clear. You are assuming that $d(x, a) < \delta$, so $\max \{|x_1 - a_1|, |x_2 - a_2|, |x_3 - a_3|\} < \delta$. This means that all three elements in the set, $|x_1 - a_1|, |x_2 - a_2|$, and $|x_3 - a_3|,$ are each less than $\delta$. Therefore
$$|x_1 - a_1|^2 < \delta ^2$$
$$|x_2 - a_2|^2 < \delta ^2$$
$$|x_3 - a_3|^2 < \delta ^2$$
Adding up all these inequalities gives
$$(x_1-a_1)^2 + (x_2-a_2)^2+(x_3-a_3)^2 < 3 \delta^2$$
Or
$$\sqrt{(x_1-a_1)^2 + (x_2-a_2)^2+(x_3-a_3)^2} < \sqrt{3 \delta^2}$$