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Open, Closed, or neither?

  1. Jun 12, 2005 #1
    Question 1

    Let [itex]\mathcal{H} = \mathbb{C}^k[/itex], where [itex]\mathcal{H}[/itex] is a Hilbert space. Then let

    [tex]S = \left\{x : \sum_{i=1}^{k} |x_i| \leq 1 \right\}[/tex]

    be a subset of [itex]\mathcal{H}[/itex]. Is the subset [itex]S[/itex] open, closed or neither?



    Question 2

    Let [itex]\mathcal{H} = \mathbb{C}[/itex]. Then let

    [tex]S = \left\{\frac{1}{n} : n\in \mathbb{N}\right\}[/tex]

    be a subset of [itex]\mathcal{H}[/itex]. Is the subset [itex]S[/itex] open, closed or neither?



    Question 3

    Let [itex]\mathcal{H} = \mathbb{C}^2[/itex]. Then let

    [tex]S = \left\{(z,0) : z\in \mathbb{C}\right\}[/tex]

    be a subset of [itex]\mathcal{H}[/itex]. Is the subset [itex]S[/itex] open, closed or neither?



    Question 4

    Let [itex]\mathcal{H} = l^2[/itex]. Then let

    [tex]S = \left\{x : \sum_{i=1}^{\infty} |x_i|^2 < 1\right\}[/tex]

    be a subset of [itex]\mathcal{H}[/itex]. Is the subset [itex]S[/itex] open, closed or neither?



    Question 5

    Let [itex]\mathcal{H} = L^2([0,1])[/itex]. Then let

    [tex]S = \left\{f : f(t) \neq 0 \, \forall \, t \in [0,1]\right\}[/tex]

    be a subset of [itex]\mathcal{H}[/itex]. Is the subset [itex]S[/itex] open, closed or neither?
     
    Last edited: Jun 12, 2005
  2. jcsd
  3. Jun 12, 2005 #2
    Solution 1

    Intuitively, this set is closed because it contains its own boundary.
     
  4. Jun 12, 2005 #3
    Solution 2

    We can write out the subset as

    [tex]S=\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \}[/tex]

    In set notation...

    [tex]S = (0,1][/tex]

    That is, S is neither open nor closed.
     
  5. Jun 12, 2005 #4
    Solution 3

    Since [itex] (z,0) = 0 \, \forall \, z \in \mathbb{C}[/itex], then S is closed since the complement is open. ie since

    [tex]\mathbb{C}^2 \backslash \left(S = \{0\}\right)[/tex] is open.
     
  6. Jun 12, 2005 #5

    HallsofIvy

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    ??? I would interpret (0, 1] as the set of all real numbers between 0 and 1 (including 1 but not 0) not S.

    Of course, it is true that the set is neither open nor closed. It is not open because a neighborhood of 1/n, a disk in the complex plane centered on 1/n will contain numbers not in the set. It is not closed because the sequence has limit point 0 which is not in the set. (If 0 were included, the set would be closed.)
     
  7. Jun 12, 2005 #6

    HallsofIvy

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    I may be misunderstanding your notation. It is true that S is closed because the complement of S is open.
    But I don't understand your saying (z, 0)= 0 . (z, 0) (for all complex z) is topologically equivalent to the complex plane in the same way that the x-axis, y= 0 (all points (x, 0)), is topologically equivalent to the real line.

    And I really don't understand what you mean by [tex]\mathbb{C}^2 \backslash \left(S = \{0\}\right)[/tex]. What could S= {0} mean?
     
  8. Jun 13, 2005 #7
    Ha. I dont know what I was talking about ??? You are right of course, and what you wrote is exactly what I was thinking...

    I see. I wasnt sure what was meant by (z,0) - but from what you wrote I would guess that what the question means. I suppose I had a lucky guess then!?
     
    Last edited: Jun 13, 2005
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