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Math Help - open closed cartesian product

  1. #1
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    open closed cartesian product

    I am considering the space \mathbb{Q}x \mathbb{Q} in \mathbb{R}^2 .

    I know that \mathbb{Q} is neither open nor closed in \mathbb{R}, so can I say that the cartesian product \mathbb{Q}x \mathbb{Q} is neither open nor closed in in \mathbb{R}^2 ?
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  2. #2
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    Because \mathbb{Q} \times \mathbb{Q} \subseteq \mathbb{R}^2 we use the metric topology of \mathbb{R}^2.
    Every point of \mathbb{Q} \times \mathbb{Q} is a limit point of \mathbb{R}^2.
    And every point of \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right) is a limit point of \mathbb{Q} \times \mathbb{Q}.

    Can you answer your questions now?
    BTW: The set \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right) is pathwise connected. It is actually easy to exhibit the path between two points.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Because \mathbb{Q} \times \mathbb{Q} \subseteq \mathbb{R}^2 we use the metric topology of \mathbb{R}^2.
    Every point of \mathbb{Q} \times \mathbb{Q} is a limit point of \mathbb{R}^2.
    And every point of \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right) is a limit point of \mathbb{Q} \times \mathbb{Q}.

    Can you answer your questions now?
    BTW: The set \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right) is pathwise connected. It is actually easy to exhibit the path between two points.
    I don't understand why \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right) is path connected; can you explain your reasoning?
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  5. #5
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    The set \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right) is the set of all ordered pairs in \mathbb{R}^2 that have at least one irrational coordinate. The is a point \left( {\alpha ,\beta } \right) \in \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right) which has both coordinates irrational.
    Suppose that \left( {p,q} \right) \in \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right) is any other point. One of its coordinates is irrational, say itís q.
    Using the three points \left( {\alpha ,\beta } \right),\;\left( {\alpha ,q} \right)\,\& \,\left( {p,q} \right) construct two line segments: l_1 (t):\left( {t\left[ {\alpha  - p} \right] + p,q} \right)\,\& \,l_2 (t):\left( {\alpha ,t\left[ {\beta  - q} \right] + q} \right)\;\; 0\le t \le 1.
    Both line segments are subsets of \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right) because at least on coordinate of each point is irrational.
    The first segment connects \left( {\alpha ,q} \right)\,\& \,\left( {p,q} \right) and the second connects \left( {\alpha ,q} \right)\,\& \,\left( {\alpha ,\beta } \right).
    Therefore the set is clearly pathwise connected.
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