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Function Spaces

  1. Apr 5, 2005 #1
    Let T be an element of B[0,1] be the set V=[B[0,1];f(1)=2]. Prove that T is closed (in metric space B[0,1]).

    I am not sure if it is obvious since I am new to this stuff but B[0,1] is an open ball I believe.

    My question is how do I find the complement of V. If I could define B/T then I am hoping it will follow easily from the definitions of open set that this is an open set and my proof will be complete. Is B/T=[B[0,1]; f(1)>2 V f(1)<2]? Very confused.
  2. jcsd
  3. Apr 5, 2005 #2
    What is B[0,1], the set of functions on [0,1] into R? Are they continuous? What is the metric?
  4. Apr 5, 2005 #3


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    standard stuff would be something like this:

    1) B[0,1] is the set of continuous functions from [0,1] to the reals

    or maybe because there is a B there, it is just bounded functions.

    2) the topology on B[0,1] is given by say the sup norm, i.e. the distance from f to g is the furthest apart any two values f(t), g(t) ever get for t in [0,1].
    3) then prove the evaluation map taking f to f(1) is continuous.

    4) then the inverse image of 2 under the evaluation map is clsoed since 2 is closed in R.
  5. Apr 5, 2005 #4
    Thanks a lot mathwonk
  6. Apr 9, 2005 #5
    What if I were to try to solve this by taking the complement of V and using open or closed balls? Is there any way to prove it this way? B[0,1] is the set of bounded functions by the way.
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