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Math Help - Set Inclusion

  1. #1
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    Set Inclusion

    Hi

    The two sets V_{1}, V_{2} are subsets of the Euclidian space \mathbb{R}^n.

    I need to show that

    (1) V_{1} \cup V_{2} is open.

    (2) V_{1} \cap V_{2} is open.

    Definion:

    I know according to the definition that a subset V of \mathbb{R}^n is said to be open, if there for every point x \in V exist a real number \epsilon > 0 given every point y \in \mathbb{R}^n. Then |x - y| < \epsilon and y \in V

    Solution:

    (1) and (2)

    Then if V_{1}, V_{2} both are open subsets of \mathbb{R}^n by definition above. Then if there exist a point x and y in both subsets, then if there still exist a epsilon > 0, then their respective Union and Intersection is also open according to the definition above.

    How does that sound? Or do I need to above something more concrete?

    Or do I need to add something about the union of two open sets are closed? Bu if the union of two subsets are closed, then how possible can it be open?

    Best Regards.
    Billy
    Last edited by Billy2007; December 9th 2006 at 12:03 PM.
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  2. #2
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    Quote Originally Posted by Billy2007 View Post
    The two sets V_{1}, V_{2} are subsets of the Euclidian space \mathbb{R}^n.
    You DO mean this to be that
    "The two sets V_{1}, V_{2} are open subsets of the Euclidian space \mathbb{R}^n." right?

    I think you need to be a bit more precise. By definition there exists an \epsilon _1 for V_1 and an \epsilon _2 for V_2. Can you use this to construct an \epsilon _3 for V_3 = V_1 \cup V_2?

    -Dan
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  3. #3
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    This known as a ball centered at x:  \varepsilon  > 0\quad  \Rightarrow \quad B_\varepsilon  (x) = \{ y:d(x,y) < \varepsilon \}.
    If V is open the for each point, x, in V some ball centered at x is a subset of V.
    So if V_1 \quad \& \quad V_2 are open and t \in \left( {V_1  \cap V_2 } \right) then \left[ {\exists B_\delta  (t) \subset V_1 } \right]\quad \& \quad \left[ {\exists B_\varepsilon  (t) \subset V_2 } \right].
    Let \alpha  = \min (\delta ,\varepsilon ) then B_\alpha  (t) \subset \left( {V_1  \cap V_2 } \right).
    Of course if \left( {V_1  \cap V_2 } \right) = \emptyset by definition the empty set is open.
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  4. #4
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    I'm going to start by nitpicking your definition. I would have said something like this instead:
    A subset V of \mathbb{R}^n is said to be open if there for every point x \in V exists a real number \epsilon > 0 such that every point y \in \mathbb{R}^n that satisfies |x - y| < \epsilon is a member of V.
    I realize that's what you meant, but it's important to be precise when defining stuff.

    I'll do 2 for you. You can probably handle 1 after that. I'm going to use the concept "open ball", so I'll define that first:

    An open ball of radius r around a point x is the set of points y such that |x-y|<r.

    Using this concept, we can state the definition of openness like this: A set V is open if, for every point x in V, there's an open ball around x that's a subset of V.

    OK, this is the solution of 2:

    Let x be a point in the intersection of V1 and V2. Then x is in V1 and x is in V2. Since x is in V1, there's an open ball around x that's a subset of V1. Since x is in V2, there's an open ball around x that's a subset of V2. The smaller of these open balls is a subset of the larger, and must therefore be a subset of both V1 and V2. That's equivalent to saying that the smaller of these two open balls around x is a subset of the intersection of V1 and V2, and that's exactly what we need to conclude that the intersection of V1 and V2 is open.

    Edit: I see that Plato beat me to it. His solution is exactly the same as mine.

    Or do I need to add something about the union of two open sets are closed? Bu if the union of two subsets are closed, then how possible can it be open?
    A union of open sets (even infinitely many) is always open. The intersection of a finite number of open sets is always open.
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  5. #5
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    Hi Guys,

    Thank You again very much for all Your help. I get it now.

    Best Regards and Enjoy whats left of the weekend

    Billy
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