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

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

    How do open balls of $\displaystyle \mathbb{R} \times \{0\}$ look like? Given some $\displaystyle \varepsilon > 0$ an open ball of $\displaystyle \mathbb{R} \times \{0\}$ with center at some point $\displaystyle (y,0) \in \mathbb{R} \times \{0\}$ and radius $\displaystyle \varepsilon$ is the set $\displaystyle \{(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 $\displaystyle (x,0) \in A \cap (\mathbb{R} \times \{0\})$ implies $\displaystyle (x,0) \in A$ and $\displaystyle A$ is an open set in $\displaystyle \mathbb{R}^2$, there must exist an open ball of $\displaystyle \mathbb{R}^2$ with center at $\displaystyle (x,0)$ and some radius $\displaystyle \varepsilon >0$ such that this whole ball is a subset of $\displaystyle A$. This open ball is the set

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

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

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

    $\displaystyle \{(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\} $

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

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