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I have recently started reading Kosniowski's A First Course in Algebraic Topology; my background on Calculus, Algebra and Vectors is vast enough to understand many of the concepts shown in the first few pages. The very first exercises, though, have proven way more interesting (and difficult) than expected.

Kosniowski defines a metric as a function $d(x, y): A \times A \rightarrow \mathbb{R}$ such that:

  1. $d(x, y) = 0 \iff x=y$
  2. $d(a, b) + d(a, c) \ge d(b, c)\qquad\forall \space a, b, c \in A $

Then leaves the other two usual properties (positivity and symmetry) to be derived as an exercise, which is pretty easy to do.

In the exercises section, there is a really interesting exercise which is similar (if not identical) to the usual triangular inequality for vectors in $\mathbb{R}^n$: $$d(x, y) = \biggl(\sum_{i=1}^{n}{(x_i-y_i)^2}\bigg)^{\frac12} $$ Showing property 1 is fairly easy. Showing that triangular inequality holds, however, is a bit trickier to me - it is still intuitive. My attempts so far:

  • By induction, I can easily prove that it holds for $n=1$ (usual euclidean triangular inequality). By taking advantage of the inequality: $$ \sqrt{a^2} + \sqrt{b^2} \ge \sqrt{a^2 + b^2} $$ I could write: $$ \biggl(\sum_{i=1}^{n-1}{(x_i-y_i)^2}\bigg)^{\frac12} + |x_n+y_n| \ge \biggl(\sum_{i=1}^{n}{(x_i-y_i)^2}\bigg)^{\frac12} $$ But this method fails when one tries to use the above to prove the statement for every $n>1$, assuming the inequality holds for $n-1$. Or, at least, it failed for me.
  • By squaring both sides of the inequality and solving, which led me to proving the inequality only for some cases (when $a < 0, bc \le 0$, for example).

How does one go about solving this problem?

Andres Mejia
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Niki Di Giano
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    This is Euclidean metric http://mathworld.wolfram.com/EuclideanMetric.html ... there should be tons of materials on internet. – rtybase Sep 13 '17 at 17:28
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    Use this – Andres Mejia Sep 13 '17 at 17:36
  • Symmetry cannot be derived from 1. and 2. Let $A={x,y}$ with $x\ne y.$ Let $d(x,x)=d(y,y)=0.$ Let $d(x,y)=1$ and $d(y,x)=2.$ Then 1. and 2. are satisfied – DanielWainfleet Sep 13 '17 at 20:27
  • Either your statement is true, or Kosniowski has written an unsolvable exercise.

    My attempt (using only 1 and 2): $$d(x, y) + d(x, x) \ge d(y, x)$$ $$d(y, x) + d(y, y) \ge d(x, y)$$ $$\implies d(x, y) = d(y, x)$$

     – Niki Di Giano Sep 13 '17 at 20:35
  • In fact, the first of those three statements (that must be true in order for the function to be a metric) is not true in your case, since one plus zero is surely not greater than two... – Niki Di Giano Sep 14 '17 at 00:01

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You have to show the inequality $$\sqrt{ \sum a_i^2} + \sqrt{ \sum b_i^2} \ge \sqrt { \sum (a_i + b_i)^2}$$

It does not work by induction, you have to square and examine some difference, which can be shown to be a sum of squares. Try it for $n=2$ first.

orangeskid
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