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Why metric spaces?

You may have been introduced in your analysis class to the notion of convergence, but only about the convergence of real sequences. If we want to extend this idea for say vectors, complex numbers, functions, tensors, or even sequences of sequences, and to generalize it, we can define for every mathematical object the meaning of convergence. But as you may guess, this isn't very efficient and will quickly get tedious. The other choice is to get more abstract and define a space (which is like a set) that contain all of these objects, and to define convergence there. But since distance is important for convergence, we must be able in this space to determine distances between its points. And ladies and gentlemen, two classes of spaces fit our requirements, the first is the so-called metric spaces, and a more general class of spaces which are called topological spaces.

Another presentation made by T. W. Körner:

If I wish to travel from Cambridge to Edinburgh, then I may be interested in one or more of the following numbers.

  1. The distance, in kilometres, from Cambridge to Edinburgh ‘as the crow flies’.
  2. The distance, in kilometres, from Cambridge to Edinburgh by road.
  3. The time, in minutes, of the shortest journey from Cambridge to Edinburgh by rail.
  4. The cost, in pounds, of the cheapest journey from Cambridge to Edinburgh by rail.

Each of these numbers is of interest to someone and none of them is easily obtained from another. However, they do have certain properties in common which we try to isolate in the following definition:

Let \(X\) be a metric space with \(d:X^2\to\Bbb R\) a function with the following properties:

  1. \(d(x,y)\geqslant0\) for all \(x,y\in X\).
  2. \(d(x,y)=0\) only and only if \(x=y\).
  3. \(d(x,y)=d(y,x)\) for all \(x,y\in X\). (symmetry)
  4. \(d(x,y)+d(y,z)\geqslant d(x,z)\) for all \(x,y,z\in X\). (triangle inequality)

Note by حكيم الفيلسوف الضائع
3 years, 5 months ago

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Omar Nader - 3 years, 2 months ago

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