I have the following problem:

Let $(S,d)$ and $(T,e)$ be two metric spaces. Their product space has underlying set $$S\times T=\{(s,t)|s\in S,t\in T\}$$ and metric $$m\big((s_1,t_1),(s_2,t_2)\big)=\sqrt{(s_1-s_2)^2+(t_1-t_2)^2}.$$ Verify that $(S\times T,m)$ is a metric space.

As for proving the triangle inequality, I made the following attempt:

\begin{align*} m\big((s_1,t_1),(s_3,t_3)\big)&=\sqrt{(s_1-s_3)^2+(t_1-t_3)^2}\\ &=\sqrt{(s_1-s_2+s_2-s_3)^2+(t_1-t_2+t_2-t_3)^2}\\ &\leq\sqrt{(s_1-s_2+s_2-s_3)^2}+\sqrt{(t_1-t_2+t_2-t_3)^2}\\ &\leq\sqrt{(s_1-s_2)^2+(t_1-t_2)^2}+\sqrt{(s_2-s_3)^2+(t_2-t_3)^2}\\ &=m\big((s_1,t_1),(s_2,t_2)\big)+m\big((s_2,t_2),(s_3,t_3)\big) \end{align*}

My friend has told me that I am making too large of logical jumps in the "less than or equal to" steps. Can anyone point out my error?

I know I need to use the established metrics, but I don't know how.

  • $\begingroup$ First of all $+/-$ may not make any sense in $S$ and $T$. You need to only use the metrics $d$ and $e$. $\endgroup$ – Forever Mozart Dec 5 '14 at 4:30
  • $\begingroup$ @Tom the $+/-$ issue was something I had ran into and wondered about. $S$ and $T$ have the metrics $d$ and $e$ respectively. Are you saying there is a way include them in the definition of $m$? Does $d$ have something to do with $s_1-s_2$? $\endgroup$ – user155385 Dec 5 '14 at 4:33
  • $\begingroup$ Yes. We know how to add and subtract real numbers. Try replacing $s_1-s_2$ with $d(s_1,s_2)$ and $t_1-t_2$ with $e(t_1,t_2)$, for instance. $\endgroup$ – Forever Mozart Dec 5 '14 at 4:37
  • $\begingroup$ You must have $m((s_1,t_1),(s_2,t_2)) = \sqrt{d(s_1,s_2)^2+e(t_1,t_2)^2}$, or something similar. $S,T$ are metric spaces, not vector spaces. $\endgroup$ – copper.hat Dec 5 '14 at 4:39
  • $\begingroup$ Am I allowed to say that $d(s_1,s_2)=s_1-s_2$ and just replace $s_1-s_2$ with $d$ in the definition of $m$ $\endgroup$ – user155385 Dec 5 '14 at 4:45

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.

  • $\begingroup$ thanks. I understand this. However, the problem in the book gives the metric $m$ as I defined it with the sum, difference, and square. This leaves me confused as why the book did this. It asks me to prove the space is a metric space with that metric. $\endgroup$ – user155385 Dec 10 '14 at 1:30
  • $\begingroup$ Your definition makes sense if $S$ and $T$ are spaces of real numbers. If you assume that $S$ and $T$ are arbitrary vector spaces, then the subtractions $s_1 - s_2$ and $t_1 - t_2$ make sense, but the squares $(s_1 - s_2)^2$ and $(t_1 - t_2)$ do not, even if you assume that you are working with vectors of real numbers, e.g. $\mathbb{R}^2$. I'm guessing that the author of the question made an error and intended to give the definition I used. $\endgroup$ – Bungo Dec 10 '14 at 1:35
  • $\begingroup$ Thank you, excellent answer. $\endgroup$ – user155385 Dec 10 '14 at 2:03

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