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Definition of a continuous function

  1. Jun 12, 2010 #1

    nicksauce

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    I am reading Schutz's "Geometrical methods of mathematical physics". He writes: "A map f:M->N is continuous at x in M if any open set of N containing f(x) contains the image of an open set of M." However, it seems to me that a more appropriate definition would be "... contains the image of a neighbourhood of x". Am I right, or am I missing something obvious?
     
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  3. Jun 12, 2010 #2

    arildno

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    I would think that with Schutz's definition, you'll be able to PROVE that neighbourhoods around "x" must be included in those "open sets of M".

    Note, for example, that a discontinuity of value at f(x) will prevent the existence of there being any open set of N about it. I think..
     
  4. Jun 12, 2010 #3

    nicksauce

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    Well the counter-example of the Schutz definition I'm thinking of would look something like this: http://imgur.com/ppE5t [Broken]

    But maybe I need to think about it some more...
     
    Last edited by a moderator: May 4, 2017
  5. Jun 12, 2010 #4
    Counterexample perhaps from R to R:
    [tex]
    f(x) = \begin{cases}
    1 & \text{if }x \in \mathbb{Q}\text{ and }x \geq 0, \\
    0 & \text{if }x \notin \mathbb{Q}\text{ and }x \geq 0, \\
    1 & \text{if }-1 < x < 0, \\
    0 & \text{if }x \leq -1.
    \end{cases}
    [/tex]​
    According to the given definition, f(x) is continuous everywhere.
     
  6. Jun 12, 2010 #5
    What if x is an isolated point of M?

    Was the definition not so written to cater for this eventuality?
     
  7. Jun 13, 2010 #6
    No, that's not the issue. Clearly if you consider continuity at x and then forget to require an open set about x to be contained in an arbitrarily small open set about f(x), you have a worthless definition. I'm pretty sure the author meant to say "A map f:M->N is continuous at x in M if any open set of N containing f(x) contains the image of an open set of x in M."
     
  8. Jun 13, 2010 #7
    That would satisfy nicksauce' qualms, without introducing a neighbourhood.

    However the original question was should a neighbourhood be introduced and I think I answered that point since I don't think an isolated point has a neighbourhood, although it is defined to be an open set, thereby satisfying your form of words.

    I found, like others, it quite hard to get my head round this particular statement of continuity, which differs subtly from the topological one I am more used to.

    So I think you may have cracked it.
     
  9. Jun 13, 2010 #8
    This is a simple counter example.

    [tex]
    f:\mathbb{R}\to\mathbb{R}
    [/tex]

    [tex]
    f(x) = x,\quad\quad x\neq 0
    [/tex]

    [tex]
    f(0) = 1
    [/tex]

    With Euclidian topology, [itex]f[/itex] is not continuous at [itex]x=0[/itex]. Now choose an arbitrary open set [itex]V\subset\mathbb{R}[/itex] such that [itex]1\in V[/itex] (that means [itex] f(0)\in V[/itex]). You can find an open set [itex]]1-\epsilon,1+\epsilon[[/itex], such that [itex]f(]1-\epsilon,1+\epsilon[)\subset V[/itex], so [itex]f[/itex] should be continuous at [itex]x=0[/itex] according to the strange definition now.
     
  10. Jun 13, 2010 #9

    Landau

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    It is a mistake. But Schutz says it right at the following page (8):

     
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