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Property of conitnuity?

  1. Aug 31, 2005 #1
    How do I prove that when a function f is continuous on a closed bounded interval [a,b], it is uniformly continuous on that interval?

    Actually, I have found some proofs to this but I have not tackled about compact, Heine-Borel theorem, metric spaces, sequences and series, etc. in my class..
  2. jcsd
  3. Aug 31, 2005 #2


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    The proof I have uses Heine-Borel, so I assume you won't be interested in that one.

    I do doubt thought that you would be able to rigourously proof it without any of the concepts or theorems you mentioned, unless you're just interested in explaining it, the difference between 'normal' and 'uniform' continuity.
    Last edited: Aug 31, 2005
  4. Aug 31, 2005 #3

    matt grime

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    Suppose it is not unifromly continuous, construct a sequence with no limit point in [a,b] surely you know enough abuot sequences to be able to konw that that is impossible? (no one does continuity before sequences, surely?)
  5. Sep 1, 2005 #4
    Thank you for the help...
  6. Sep 1, 2005 #5


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    you can make matt's proof concrete by the trick of subdivision. I.e. on the interval [0,1] if you have an infinite collection of points then there is an infinite number of them in some subinterv of length 1/10, say between .1 and .2.

    Then ther is also an infinite num=ber of them between say .11 and .12.

    and so on. Eventually (after an infinite number of steps) you construct an infinite decimkal, i.e. a real number, with an infinite number of these points in every neighborhood.

    Since the infinite sequence was chosen in the beginning to be points hwere the function was "less and less continuous" i.e. needing bigger and bigger delta for a given epsilon, you deduce that your function is not continuous at all at the coinstructed point. a contradiction.
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