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Math Help - subspace, closed map

  1. #1
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    subspace, closed map

    Let \pi_1:\mathbb{R} \times \mathbb{R}, (x, y) \mapsto \pi_1(x, y)=x, let X be the subspace (\{0\}\times \mathbb{R})\cup (\mathbb{R}\times \{0\}) \subseteq \mathbb{R} \times \mathbb{R}

    Prove f=\pi_1|_X is a closed map but not an open map.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by GenoaTopologist View Post
    Let \pi_1:\mathbb{R} \times \mathbb{R}, (x, y) \mapsto \pi_1(x, y)=x, let X be the subspace (\{0\}\times \mathbb{R})\cup (\mathbb{R}\times \{0\}) \subseteq \mathbb{R} \times \mathbb{R}

    Prove f=\pi_1|_X is a closed map but not an open map.
    Hello, could you maybe clarify your notation? I am not exactly sure what this means, maybe some other more senior and knowledgable member recognizes this notation.
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  3. #3
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    Quote Originally Posted by GenoaTopologist View Post
    Let \pi_1:\mathbb{R} \times \mathbb{R}, (x, y) \mapsto \pi_1(x, y)=x, let X be the subspace (\{0\}\times \mathbb{R})\cup (\mathbb{R}\times \{0\}) \subseteq \mathbb{R} \times \mathbb{R}

    Prove f=\pi_1|_X is a closed map but not an open map.
    Quote Originally Posted by Mathstud28 View Post
    Hello, could you maybe clarify your notation? I am not exactly sure what this means, maybe some other more senior and knowledgable member recognizes this notation.
    Though I am not used to this notation either, I am guessing it is to prove that
    the function f (which behaviors like \pi_1 but restricted in the subspace X) maps closed sets to closed sets but not open sets to open sets in X.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chabmgph View Post
    Though I am not used to this notation either, I am guessing it is to prove that
    the function f (which behaviors like \pi_1 but restricted in the subspace X) maps closed sets to closed sets but not open sets to open sets in X.
    My main question is what does (x,y)\mapsto \pi_1(x,y)=x mean? I have not seen that notation.
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    My main question is what does (x,y)\mapsto \pi_1(x,y)=x mean? I have not seen that notation.
    It just means that to (x,y), you give the value \pi_1(x,y) (which is very logic since it's the image of (x,y) by the function pi_1 !)
    And that you define \pi_1(x,y) as being x.

    It's a correct notation And I'm sure you know what this function is
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    Quote Originally Posted by GenoaTopologist View Post
    Let \pi_1:\mathbb{R} \times \mathbb{R}, (x, y) \mapsto \pi_1(x, y)=x, let X be the subspace (\{0\}\times \mathbb{R})\cup (\mathbb{R}\times \{0\}) \subseteq \mathbb{R} \times \mathbb{R}

    Prove f=\pi_1|_X is a closed map but not an open map.
    The space X is just the union of the two coordinate axes. The mapping f (projection onto the first coordinate) acts as the identity on the x-axis, and takes everything on the y-axis to the origin. You should be able to see from that that the image under f of a closed set will always be closed (think of the set as the union of two parts, the part on the x-axis and the part on the y-axis: the image under f of each of these two parts is fairly obviously closed).

    The map is not open, because for example the image of the open subset \{0\}\times(0,1) of X is a single point and therefore not open.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    It just means that to (x,y), you give the value \pi_1(x,y) (which is very logic since it's the image of (x,y) by the function pi_1 !)
    And that you define \pi_1(x,y) as being x.

    It's a correct notation And I'm sure you know what this function is
    Thank you Moo, I need to be careful using only one book since it has a different set of notation. My book is not the only book.
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