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Math Help - Proof of Intersection of an open set

  1. #1
    Member thaopanda's Avatar
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    Proof of Intersection of an open set

    Prove that the intersection of an open set of R^2 with the x-axis is an open set of R \times {0} (where the latter is endowed with the distance d given by d(x, 0) := |x|, for x \in R).

    I don't understand what exactly this means.... any help is much appreciated
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  2. #2
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    Hi Panda,

    let A\subseteq \mathbb{R}^2 be the open set. We have to show that A \cap (\mathbb{R} \times \{0\}) is an open set of the metric space \mathbb{R} \times \{0\} (endowed with the metric p((x,0),(y,0)) = |x-y| which is the metric induced by the norm d(x,0) = |x|. Note that the statement of the problem says d is a distance but it is wrong, d such defined is a norm. We used this norm to induce the distance function (or metric) p ).

    This means that for every (x,0) \in A \cap (\mathbb{R} \times \{0\}) we must find an open ball in the space \mathbb{R} \times \{0\} with center at (x,0) such that the whole ball is a subset of A \cap (\mathbb{R} \times \{0\}).

    How do open balls of  \mathbb{R} \times \{0\} look like? Given some \varepsilon > 0 an open ball of  \mathbb{R} \times \{0\} with center at some point (y,0) \in \mathbb{R} \times \{0\} and radius \varepsilon is the set \{(z,0) \in \mathbb{R} \times \{0\};\, p((z,0),(y,0))<\varepsilon\} = \{(z,0) \in \mathbb{R} \times \{0\};\, |z-y|<\varepsilon\} . So they're open intervals.

    Since (x,0) \in A \cap (\mathbb{R} \times \{0\}) implies (x,0) \in A and A is an open set in \mathbb{R}^2, there must exist an open ball of \mathbb{R}^2 with center at (x,0) and some radius \varepsilon >0 such that this whole ball is a subset of A. This open ball is the set

    \{(z_1,z_2)\in \mathbb{R}^2;\, ||(z_1,z_2)-(x,0)||<\varepsilon\} = \{(z_1,z_2)\in \mathbb{R}^2;\, ||(z_1-x,z_2)||<\varepsilon\} =

    =\{(z_1,z_2)\in \mathbb{R}^2;\, \sqrt{(z_1-x)^2+z_2^2}<\varepsilon\}

    Now if we intersect this open ball with \mathbb{R} \times \{0\}, we get exactly an open ball of the space \mathbb{R} \times \{0\} centered at (x,0):

    \{(z_1,z_2)\in \mathbb{R}^2;\, \sqrt{(z_1-x)^2+z_2^2}<\varepsilon\} \cap (\mathbb{R} \times \{0\}) = \{(z_1,0)\in \mathbb{R}^2;\, \sqrt{(z_1-x)^2+0}<\varepsilon\}

     =\{(z_1,0)\in \mathbb{R}^2;\, |z_1-x|<\varepsilon\}

    This ball is obviously a subset of A \cap (\mathbb{R} \times \{0\}) (because the original ball is a subset of A).
    We conclude that the set A \cap (\mathbb{R} \times \{0\}) is an open set of the space \mathbb{R} \times \{0\}.
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