Prove that the product space is a metric space.

Solution 1:

First, as mentioned in the comments, your definition of $m$ assumes that you can add, subtract, and square $s_i$ and $t_i$. These operations are not defined for a general metric space. A suitable definition would be $$m((s_1,t_1), (s_2,t_2)) = \sqrt{d(s_1,s_2)^2 + e(t_1,t_2)^2}$$ The rest of this answer assumes this definition.

Note that for any positive numbers $a$ and $b$, the inequalities $a < b$ and $a^2 < b^2$ are equivalent. Therefore, proving the triangle inequality $$m((s_1,t_1), (s_3,t_3)) \leq m((s_1,t_1), (s_2,t_2)) + m((s_2,t_2), (s_3,t_3))$$ is equivalent to proving $$m((s_1,t_1), (s_3,t_3))^2 \leq [m((s_1,t_1), (s_2,t_2)) + m((s_2,t_2), (s_3,t_3))]^2$$ or equivalently, $$d(s_1,s_3)^2 + e(t_1,t_3)^2 \leq d(s_1,s_2)^2 + e(t_1,t_2)^2 + 2\sqrt{d(s_1,s_2)^2 + e(t_1,t_2)^2}\sqrt{d(s_2,s_3)^2 + e(t_2,t_3)^2} + d(s_2,s_3)^2 + e(t_2,t_3)^2$$ Let us call this inequality 1.

Now, since $d$ and $e$ are metrics, we know that $$d(s_1,s_3)^2 \leq [d(s_1,s_2) + d(s_2,s_3)]^2 = d(s_1,s_2)^2 + 2 d(s_1,s_2)d(s_2,s_3) + d(s_2,s_3)^2$$ and $$e(t_1,t_3)^2 \leq [e(t_1,t_2) + e(t_2,t_3)]^2 = e(t_1,t_2)^2 + 2 e(t_1,t_2)e(t_2,t_3) + e(t_2,t_3)^2$$ Adding these two inequalities gives us $$d(s_1,s_3)^2 + e(t_1,t_3)^2 \leq d(s_1,s_2)^2 + 2 d(s_1,s_2)d(s_2,s_3) + d(s_2,s_3)^2 + e(t_1,t_2)^2 + 2 e(t_1,t_2)e(t_2,t_3) + e(t_2,t_3)^2$$ Comparing this with inequality 1, we see that it will suffice to prove that $$d(s_1,s_2)d(s_2,s_3) + e(t_1,t_2)e(t_2,t_3) \leq \sqrt{d(s_1,s_2)^2 + e(t_1,t_2)^2}\sqrt{d(s_2,s_3)^2 + e(t_2,t_3)^2}$$ Let us call this inequality 2.

The left hand side is the dot product between the 2-dimensional real vectors $v = (d(s_1,s_2), e(t_1,t_2))$ and $w = (d(s_2,s_3), e(t_2,t_3))$. Let's call this dot product $v \cdot w$ for short.

The right-hand side is the product of the lengths of these vectors: $$\|v\|\|w\| = \sqrt{d(s_1,s_2)^2 + e(t_1,t_2)^2}\sqrt{d(s_2,s_3)^2 + e(t_2,t_3)^2}$$

The Cauchy Schwarz inequality is precisely the tool we need: it tells us that $$|v \cdot w| \leq \|v\|\|w\|$$ This along with the fact that $v \cdot w \leq |v \cdot w|$ (a real number is no larger than its absolute value) tells us that $$v \cdot w \leq \|v\|\|w\|$$ Substituting the definitions of $v$ and $w$ gives us exactly inequality 2, which as argued above, suffices to prove inequality 1, which in turn is equivalent to the triangle inequality.