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Open/closed set

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data

    If [itex]f:\mathbb{R}\to\mathbb{R}[/itex] and [itex]g:\mathbb{R}\to\mathbb{R}[/itex] are continuous functions, give examples to show that the set [itex]\{ (f(x),g(x)) : x\in\mathbb{R} \}[/itex] might or might not be closed in [itex]\mathbb{R}^2[/itex].

    3. The attempt at a solution

    Letting [itex]f(x)=g(x)=0[/itex] gives the set equal to [itex]\{ (0,0) \}[/itex], a singleton and singleton sets are closed.

    What functions would make the set open?
     
  2. jcsd
  3. Nov 8, 2011 #2

    Dick

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    That's a pretty trivial example. You should probably think about it a little more before someone just gives you the answer. Suppose f(x) and g(x) have a horizontal asymptote as x->inf. There's other ways it can fail to be closed as well, but that should get you started.
     
  4. Nov 9, 2011 #3
    Well if the functions have horizontal asymptotes they won't be continuous on the real line for a start.

    If we take f(x) a constant function, what condition must g(x) satisfy in order that the set isn't closed?
     
  5. Nov 9, 2011 #4

    micromass

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    One VERY IMPORTANT THING:

    Open is NOT the same as "not closed".

    You don't need to find a set that is open, you need to find a set that is not closed. These are completely different questions!!!!
     
  6. Nov 9, 2011 #5

    Dick

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    I said HORIZONTAL asymptote. Why do you think that means it wouldn't be continuous?
     
  7. Nov 9, 2011 #6
    Sorry, of course it can be continuous.

    Is the set [itex]\{ (0,x):x\in\mathbb{R} \}[/itex] not closed?
     
  8. Nov 9, 2011 #7

    Dick

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    It doesn't look closed to me. What do you think? Keep micromass's comment in mind too. A set that is 'not closed' doesn't have to be open.
     
  9. Nov 9, 2011 #8

    Dick

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    I changed my mind. I thought you were trying to write an open subsegment of a straight line. You're aren't, are you? Your set is the x-axis. It's closed in R^2. How would you show that?
     
  10. Nov 9, 2011 #9
    I've just realised that I'm confusing 'open' and 'not closed'.

    With [itex]f[/itex] a constant function, I need to find [itex]g[/itex] such that [itex]g(\mathbb{R})[/itex] is not closed. What is an example a function from R to R that isn't closed?
     
  11. Nov 9, 2011 #10

    Dick

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    So you aren't going to try the horizontal asymptote suggestion???
     
  12. Nov 9, 2011 #11
    [itex]g(x)=e^x[/itex].

    Then [itex]\{(0,e^x) :x\in\mathbb{R} \}[/itex] is not closed is it?
     
  13. Nov 9, 2011 #12

    Dick

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    No, it isn't. Why isn't it closed?
     
  14. Nov 9, 2011 #13
    The boundary of [itex]A = \{(0,e^x) : x\in \mathbb{R} \}[/itex] is [itex]\partial A =\{(0,0)\}[/itex] and [itex](0,0) \notin A[/itex] so [itex]A[/itex] doesn't contain all its boundary and is therefore closed.
     
  15. Nov 9, 2011 #14

    Dick

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    Good answer! Except that the boundary of A is actually all of A AND (0,0). Every point in A is also a boundary point of A. But the critical thing is that (0,0) is on the boundary of A but not in A.
     
    Last edited: Nov 9, 2011
  16. Nov 9, 2011 #15
    Would you express the boundary as [itex]\partial A = A\cap \{(0,0)\}[/itex]?
     
  17. Nov 9, 2011 #16

    Dick

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    Certainly not, it's a union not an intersection.
     
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