A Candle in the Dark

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Banach Contraction Theorem

with 5 comments

Today, I came across a nice theorem which, If I understood correctly, proves the existence and uniqueness of points of a self map in a metric space, such that the map of that point is itself. The formal statement goes as follows.

Let be a non-empty closed subset of a complete metric space and be a map. Let be such that . Then for each , the iterative sequence converges to a fixed point x, which is unique.

To prove this, one needs to first show that
(i) for any . This shows that as
(ii) as . That is, is a Cauchy sequence in X. One can prove this with judicious use of the triangle inequality. Further, since is a complete metric space, also converges to the same in

Using the above, one can show that , and that this is unique.

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Written by parseval

March 28, 2007 at 6:10 pm

Posted in mathematics

5 Responses

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  1. Ah, that was interesting, although I should spend quite some more time on it finding some nice examples 🙂

    Talking of Mr.Banach, surely you have read about the Banach-Tarski paradox?
    (http://en.wikipedia.org/wiki/Banach-Tarski_Paradox)
    That quite shook up people’s ideas
    about the Axiom of Choice, and was
    quite a milestone in topology.

    Mohan K.V

    March 29, 2007 at 11:40 pm

  2. Yeah, I’ve read about it. The first time I came across the result was in Surely You’re Joking, Mr Feynman!, where he points out that you can’t cut an orange thinner than it’s atoms. It’s a really fascinating result.

    parseval

    April 1, 2007 at 9:03 pm

  3. I have two things to day,
    1. Well, that’s clearly a great use of Latex! Mr. Knuth will be pleased. Mimetex is just the killer! It’s brought the Latex environment on to the blogger board.

    2. Well, I am guessing this result can be extended to spaces without a metric. And looking at the entire operation as a group theoretic one, I’ve got a rather simple question, is this a monoid, a semi-group or a group itself?

    Vettius Carnaticae

    April 6, 2007 at 2:17 pm

  4. Just try to prove this for a Zariski space/topology.
    http://en.wikipedia.org/wiki/Zariski_topology

    May be it’s already proved. I wasn’t able to do it.

    Vettius Carnaticae

    April 6, 2007 at 2:25 pm

  5. I don’t think that the statement makes any sense without a norm, metric, or some other measure.

    Regarding, your group theory question, I don’t understand what you’re referring to as ‘this’.

    If you mean the vector space as the set together with the self map, then it’s neither, since the linear transform in not binary.

    parseval

    April 6, 2007 at 5:34 pm


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