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Homework Help: Determine the following sets

  1. Sep 29, 2010 #1
    I hate to do this, but I've actually answered the questions. It's just that it seems strange to ask the student questions with such answers, as well as giving so much space to answer something so simple, I feel like I've done something wrong.

    1. The problem statement, all variables and given/known data
    Let set Q represent all rational numbers

    [tex]\partial[/tex](Q[tex]^{int}[/tex]) = ?
    [tex]\partial[/tex]((Q[tex]^{c}[/tex])[tex]^{int}[/tex]) = ?
    [tex]\overline{(Q^{int})}[/tex] = ?

    2. Relevant equations
    [tex]\partial[/tex]Q is the boundary of Q
    Q^int is the set of interior points of Q
    [tex]\overline{Q}[/tex] is the union of Q's boundary and Q
    Q^c is Q's complement (in this case the set of irrational numbers)

    3. The attempt at a solution

    I answered the empty set for all three. The interior of Q must be the empty set because any "ball" of values around Q will contain an unlimited number of both rational and irrational numbers. Thus there are no interior points. Now the boundary of the empty set would also be the empty set no?

    Then the question is asked again but for the complement of Q. With the same reasoning, any ball around any irrational number in Q^c will contain rational numbers, so the set of internal points for Q^c is also the empty set. Boundary of which is again the empty set.

    Finally the union of the boundary of Q^int and Q^int, must be the empty set because they both are empty.

    So are my reasonings correct? I think they are, but it seems weird for my professor to ask these questions.
  2. jcsd
  3. Sep 29, 2010 #2


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    Right on all counts.
  4. Sep 29, 2010 #3
    Okay thanks. I thought as much.

    Would you mind me asking one more?

    Graph of sin(1/x). This graph is subset of R^2 called G.

    Find G^int, Boundary of G, and closure of G

    G^int is empty set because the set G is a 1 dimensional line on a 2 dimensional plane, therefore every ball around a point on G contains points not on G.

    Boundary of G is G because every point on G is a boundary point (since any ball around any point on G contains both points in G and not in G.

    Closure of G is also G, easily shown by it's definition.

    I'm pretty sure I'm right, but once again, these seem like strange questions for the professor to ask. I mean, any one dimensional function could have produced the same answers, so why chose sin(1/x)?
  5. Sep 29, 2010 #4


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    The question isn't that trivial this time. Think harder about the graph near x=0. Could, for example, (0,1/2) be a boundary point?
  6. Sep 29, 2010 #5
    Let me see if I've got it straight.

    The line on the y axis from y=-1 to y=1 are also boundary points because for any ball from those points of any radius r there is a value for which x has a value in sin(1/x) and thus exists in G?

    Is that what you meant?
  7. Sep 29, 2010 #6


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    Yes, any ball around those points contains points in the graph. Away from x=0, your previous description is right. But points with x=0 need special attention.
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