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Open and Closed Set with Compactness

  1. Jun 8, 2012 #1
    Give an example of:

    a) a closed set S[itex]\subset[/itex]ℝ and a continous function f: ℝ-->ℝ such that f(S) is not closed;

    b) an open set U[itex]\subset[/itex]ℝ and a continous function f: ℝ-->ℝ such that f(U) is not open


    Solution:

    a) e^x b) x^2

    here's my problem, this is what was given in the solutions, but I don't see how the two respective answers are an open and closed set. The continuity isn't a problem, but the open and close concept is bugging me. How is e^x closed if I can pick any value of "x"? Like my interpretation of closed was any stated interval where you may have to say [a,b] is the interval also topologically I viewed it as the closure = union of the interior points an boundary points.
     
  2. jcsd
  3. Jun 8, 2012 #2
    What is the closure of ℝ? Thus, ℝ is closed. If f(x)=e^x, what is f(ℝ)?

    ℝ is also open. If f(x)=x^2, what is f(ℝ)?



    What do you mean when you say e^x is closed?
     
  4. Jun 8, 2012 #3


    I'm going to hate myself because it's probably really simple, but I think it may be that I don't even understand the question.

    So the closure of ℝ has to be ℝ itself. as for f(ℝ) i take it you mean I'm just taking ℝ and mapping it its image in this case f(ℝ) = ℝ. But then if I'm right in what your saying, all it is, is that I'm taking a number (any number) in ℝ and applying e^x to it. But how does it show "not open"
     
  5. Jun 8, 2012 #4
    The "answers" that they gave you do not, in fact, answer the questions that were asked, so you are justified in your confusion. The original questions asks for a set and a function, and the answer key only gives you a function.
    So that we are clear on notation, the image of a set X under a function f is f(X): the set of all elements y such that y = f(x) for some x in X.
    For the first question, you need a set S whose image f(S) is not closed for a specific continuous function f. The answer key claims that the function f(x) = e^x gives us this property for some set S, so it is still up to us to provide that set. If we use S = R, then we see that f(R) is the set of all positive real numbers, not R. The exponential function is not 0 for any real number and is never negative. Is the set of all positive real numbers closed relative to R?
    An easy way to check this is to see if the set contains its closure, as a set is closed if and only if it contains its closure.
    If f(R) were R, we could not use it, because R is closed, and we are looking for an image that is not closed.
    The second function maps U = R to the set of all non-negative real numbers. Is this set open relative to R? Why or why not?
     
    Last edited: Jun 8, 2012
  6. Jun 8, 2012 #5


    Ok so from your explanation then since the exponential function can never be 0 or negative this means that the image of ℝ i.e f(ℝ) will not contain the "whole" group of Real Numbers since 0 and the negatives are included in this group as such it cannot be closed.

    But from what I just said, that would mean that in the second question f(U) would also have to be open because it does not include the negative Real numbers either after the function, x^2, is applied to it.
     
  7. Jun 8, 2012 #6

    vela

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    No... The interval [0,1] is closed, but it obviously doesn't contain all of R. What does it mean when you say a set is open or closed? Once you review that, you can answer the following: The function ex maps R to (0, ∞). Is that interval closed?
     
  8. Jun 8, 2012 #7


    Set open = the set contains all of its interior points only

    Set Closed = the set contains its interior points and its boundary points.


    Oh.....since ex can never be 0 that means the boundary point "0" is not a part of the set. Therefore f(S) cannot be closed. Since S = ℝ is closed applying the function to S gives f(S), but since we picked our function to be ex, the value 0 is not present in f(s) which we would need for it to be closed.


    in the second case x2 maps to the [0,∞] because every value of the reals can be obtained i.e, the boundary points. So then this means it's closed.
     
  9. Jun 8, 2012 #8
    Your proof would be more complete if you completely describe the image, not just mention some point that is not in the image. But It looks like you have the idea.


    The notation is [0,∞).
     
  10. Jun 8, 2012 #9



    In the first one what am I missing to describe the image? and thanks everybody for the help.
     
  11. Jun 9, 2012 #10
    Simply saying 0 is not in f(S) does not suffice. The empty set does not contain 0 but it is closed. You need to explicitly say what f(S) is.

    And you're welcome, and I would like to thank everyone also, for being there to expand when I was not sure where to start.
     
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