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In elementary plane geometry, i.e. Euclidean geometry，there exists a statement as follows:

The straight line segment is the shortest among all of the lines connecting two fixed points.

This is often taught as an axiom, but it's not included in the Elements of Geometry by Euclid. Of course, this can be proven by variational method, but is there something of a vicious circle?

mengdie1982
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  • Please, add a reference to this "often taught as an axiom": In the context of Euclid's axioms, this is not even a meaningful statement. – Moishe Kohan Nov 14 '19 at 16:05
  • @MoisheKohan I can understand what you say. In reality, the length of a general curve has no good definition in Euclid' geometry. But in China's junior school book of mathematics, this is taught like this,indeed. – mengdie1982 Nov 14 '19 at 16:11
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    What do you mean by 'lines'? If it means straight lines than there is only one straight line connecting any two fixed points. If it means continuous (smooth?) curves, you first need to define a length for a curve before this statement can make sense. – quarague Nov 14 '19 at 16:11
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    The triangle inequality (which does appear in Euclid) tells us that the straight line segment is the shortest path made up of line segments connecting two fixed points. – Misha Lavrov Nov 14 '19 at 16:19
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    The shortest-distance property is a theorem in Euclidean plane geometry. The Triangle Inequality ("A side of a triangle is shorter than the sum of the other two sides") is essentially a first pass at the result, which needs to be refined for arbitrary paths (ultimately using Calculus). When we start playing with geometries on other surfaces (spheres, pseudospheres, etc), our fondness for that property can help identify what paths should interpreted as "lines"; in that sense, the property becomes somewhat axiomatic. – Blue Nov 14 '19 at 16:21
  • Among the masses, as opposed to among mathematicians, this is often treated as if it were axiomatic. – Michael Hardy Nov 14 '19 at 16:23
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    The triangle inequality tells us that the straight line in one (amongst maybe other) shortest paths. And indeed, if the norm of the vector space isn't strict (i.e., from $|x+y|=|x|+|y|$ follows that $x$ and $y$ are positive linearly dependent) there may be more shortest paths as in the taxi cab norm. – Michael Hoppe Nov 14 '19 at 17:43
  • In this case, I would discard your Chinese sources as insufficiently rigorous. Take a look here and see if this discussion answers your question. For myself, I would like to add that different areas of modern mathematics have different approach to the notion of a line segment: None of these have "shortest path" as an axiom, but some will have it as a definition of a line segment. – Moishe Kohan Nov 14 '19 at 18:10

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The triangle inequality tells us that the straight line in one (amongst maybe other) shortest paths. And indeed, if the norm of the vector space isn't strict (i.e., from $\|+\|=\|\|+\|\|$ follows that $x$ and $y$ are positive linearly dependent. There may be more shortest paths between two points as in the taxi cab norm.

Moreover, length in real normed vector space $V$ the length of a path is usually defined as the infimum of all inscribed polygonal paths. Let $c_{pq}\colon[0,1]\to V$ with $c(0)=p$ and $c(1)=q$ the path that connects $p$ and $q$, defined by $t\mapsto p+t(q-p)$ and let the norm be strict. Now it can be shown that any shortest $c$ defined on an interval $[a,b]$ between $p$ and $q$ (which isn't constant on any non-degenerated interval where it's defined) is derived by an orientation preserving, increasing homeomorphism $\phi\colon [a,b]\to[0,1]$ such that $c=c_{pq}\circ \phi$.

Michael Hoppe
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