In the infinite product topology, rigorously prove a given set is not open












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For example, how do we prove that $(-1, 1)^omega$ is not open in $mathbb{R}^omega$? I understand the general form of an open set in this topology, and it's difference with the box topology. The problem is that, while it seems intuitively evident, I'm not sure how to rigorously and formally prove that such set is not open in the product topology. Furthermore, is it necessary to us reductio ad absurdum to prove it?






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    $begingroup$


    For example, how do we prove that $(-1, 1)^omega$ is not open in $mathbb{R}^omega$? I understand the general form of an open set in this topology, and it's difference with the box topology. The problem is that, while it seems intuitively evident, I'm not sure how to rigorously and formally prove that such set is not open in the product topology. Furthermore, is it necessary to us reductio ad absurdum to prove it?






    general-topology







    share|cite|improve this question











    $endgroup$















      0












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      0





      $begingroup$


      For example, how do we prove that $(-1, 1)^omega$ is not open in $mathbb{R}^omega$? I understand the general form of an open set in this topology, and it's difference with the box topology. The problem is that, while it seems intuitively evident, I'm not sure how to rigorously and formally prove that such set is not open in the product topology. Furthermore, is it necessary to us reductio ad absurdum to prove it?






      general-topology







      share|cite|improve this question











      $endgroup$




      For example, how do we prove that $(-1, 1)^omega$ is not open in $mathbb{R}^omega$? I understand the general form of an open set in this topology, and it's difference with the box topology. The problem is that, while it seems intuitively evident, I'm not sure how to rigorously and formally prove that such set is not open in the product topology. Furthermore, is it necessary to us reductio ad absurdum to prove it?






      general-topology




      general-topology






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      edited Dec 19 '18 at 11:40









      Carl Mummert

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      67.8k7133252










      asked Dec 19 '18 at 1:54









      RyunaqRyunaq

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          3 Answers
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          $begingroup$

          You can directly show its complement is not closed. Let $x_n = (0,0,dots,0,0,2,2,2,2,2,dots)$ where there are $n$ $0$'s. Then each $x_n in mathbb{R}^omegasetminus (-1,1)^omega$ but $x_n to (0,0,0,dots) in (-1,1)^omega$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            In an infinite product like $mathbb{R}^omega$, every non-empty basic open subset $O$ has the property of "full projection":



            $$exists n in omega: pi_n[O] = mathbb{R}tag{1}$$



            As all non-empty open subsets are unions of non-empty basic open ones, and if $O$ has property (1), so has any superset of $O$, all non-empty open subsets of $mathbb{R}^omega$ have property (1).



            But $pi_n[(-1,1)^omega] = (-1,1) neq mathbb{R}$ for all $n$, so that set cannot be open.






            share|cite|improve this answer











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            • $begingroup$
              you also gave a proof that the empty set is not open
              $endgroup$
              – mathworker21
              Dec 19 '18 at 9:24












            • $begingroup$
              @mathworker21 OK, I'll add that detail.
              $endgroup$
              – Henno Brandsma
              Dec 19 '18 at 9:25



















            0












            $begingroup$

            It all depends on which definition of the product topology you would like to use. One proof I like is as follows:



            The sets of the form $prod_{i in Bbb N} X_i$, where each $X_i$ is open and $X_i = Bbb R$ for all but finitely many $i in Bbb N$, form a basis of the product topology on $Bbb R^omega$. However, the set $S = (-1,1)^omega$ contains no such subset. In particular, we note that $pi_i(S)$ (the image of the $i$th projection map, applied to $S$) fails to contain $Bbb R$ for any $i in Bbb N$. Since $S$ contains no basis element, $S$ cannot be open.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The set $S = emptyset$ contains no such subset. In particular, we note that $pi_i(S)$ fails to contain $mathbb{R}$ for any $i in mathbb{N}$. Since $S$ contains no basis element, $S$ cannot be open. :))))))
              $endgroup$
              – mathworker21
              Dec 19 '18 at 2:31










            • $begingroup$
              @mathworker21 if we take $X_1 = emptyset$ and $X_j = mathbb{R}$ for $j > 1$ then $prod X_i = emptyset$, so the empty set is of that form. The projection property does only apply to nonempty sets, as you noted.
              $endgroup$
              – Carl Mummert
              Dec 19 '18 at 11:43












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            3 Answers
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            3 Answers
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            $begingroup$

            You can directly show its complement is not closed. Let $x_n = (0,0,dots,0,0,2,2,2,2,2,dots)$ where there are $n$ $0$'s. Then each $x_n in mathbb{R}^omegasetminus (-1,1)^omega$ but $x_n to (0,0,0,dots) in (-1,1)^omega$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              You can directly show its complement is not closed. Let $x_n = (0,0,dots,0,0,2,2,2,2,2,dots)$ where there are $n$ $0$'s. Then each $x_n in mathbb{R}^omegasetminus (-1,1)^omega$ but $x_n to (0,0,0,dots) in (-1,1)^omega$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                You can directly show its complement is not closed. Let $x_n = (0,0,dots,0,0,2,2,2,2,2,dots)$ where there are $n$ $0$'s. Then each $x_n in mathbb{R}^omegasetminus (-1,1)^omega$ but $x_n to (0,0,0,dots) in (-1,1)^omega$.






                share|cite|improve this answer









                $endgroup$



                You can directly show its complement is not closed. Let $x_n = (0,0,dots,0,0,2,2,2,2,2,dots)$ where there are $n$ $0$'s. Then each $x_n in mathbb{R}^omegasetminus (-1,1)^omega$ but $x_n to (0,0,0,dots) in (-1,1)^omega$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 19 '18 at 1:59









                mathworker21mathworker21

                9,4711928




                9,4711928























                    1












                    $begingroup$

                    In an infinite product like $mathbb{R}^omega$, every non-empty basic open subset $O$ has the property of "full projection":



                    $$exists n in omega: pi_n[O] = mathbb{R}tag{1}$$



                    As all non-empty open subsets are unions of non-empty basic open ones, and if $O$ has property (1), so has any superset of $O$, all non-empty open subsets of $mathbb{R}^omega$ have property (1).



                    But $pi_n[(-1,1)^omega] = (-1,1) neq mathbb{R}$ for all $n$, so that set cannot be open.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      you also gave a proof that the empty set is not open
                      $endgroup$
                      – mathworker21
                      Dec 19 '18 at 9:24












                    • $begingroup$
                      @mathworker21 OK, I'll add that detail.
                      $endgroup$
                      – Henno Brandsma
                      Dec 19 '18 at 9:25
















                    1












                    $begingroup$

                    In an infinite product like $mathbb{R}^omega$, every non-empty basic open subset $O$ has the property of "full projection":



                    $$exists n in omega: pi_n[O] = mathbb{R}tag{1}$$



                    As all non-empty open subsets are unions of non-empty basic open ones, and if $O$ has property (1), so has any superset of $O$, all non-empty open subsets of $mathbb{R}^omega$ have property (1).



                    But $pi_n[(-1,1)^omega] = (-1,1) neq mathbb{R}$ for all $n$, so that set cannot be open.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      you also gave a proof that the empty set is not open
                      $endgroup$
                      – mathworker21
                      Dec 19 '18 at 9:24












                    • $begingroup$
                      @mathworker21 OK, I'll add that detail.
                      $endgroup$
                      – Henno Brandsma
                      Dec 19 '18 at 9:25














                    1












                    1








                    1





                    $begingroup$

                    In an infinite product like $mathbb{R}^omega$, every non-empty basic open subset $O$ has the property of "full projection":



                    $$exists n in omega: pi_n[O] = mathbb{R}tag{1}$$



                    As all non-empty open subsets are unions of non-empty basic open ones, and if $O$ has property (1), so has any superset of $O$, all non-empty open subsets of $mathbb{R}^omega$ have property (1).



                    But $pi_n[(-1,1)^omega] = (-1,1) neq mathbb{R}$ for all $n$, so that set cannot be open.






                    share|cite|improve this answer











                    $endgroup$



                    In an infinite product like $mathbb{R}^omega$, every non-empty basic open subset $O$ has the property of "full projection":



                    $$exists n in omega: pi_n[O] = mathbb{R}tag{1}$$



                    As all non-empty open subsets are unions of non-empty basic open ones, and if $O$ has property (1), so has any superset of $O$, all non-empty open subsets of $mathbb{R}^omega$ have property (1).



                    But $pi_n[(-1,1)^omega] = (-1,1) neq mathbb{R}$ for all $n$, so that set cannot be open.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 19 '18 at 9:26

























                    answered Dec 19 '18 at 6:52









                    Henno BrandsmaHenno Brandsma

                    115k348124




                    115k348124












                    • $begingroup$
                      you also gave a proof that the empty set is not open
                      $endgroup$
                      – mathworker21
                      Dec 19 '18 at 9:24












                    • $begingroup$
                      @mathworker21 OK, I'll add that detail.
                      $endgroup$
                      – Henno Brandsma
                      Dec 19 '18 at 9:25


















                    • $begingroup$
                      you also gave a proof that the empty set is not open
                      $endgroup$
                      – mathworker21
                      Dec 19 '18 at 9:24












                    • $begingroup$
                      @mathworker21 OK, I'll add that detail.
                      $endgroup$
                      – Henno Brandsma
                      Dec 19 '18 at 9:25
















                    $begingroup$
                    you also gave a proof that the empty set is not open
                    $endgroup$
                    – mathworker21
                    Dec 19 '18 at 9:24






                    $begingroup$
                    you also gave a proof that the empty set is not open
                    $endgroup$
                    – mathworker21
                    Dec 19 '18 at 9:24














                    $begingroup$
                    @mathworker21 OK, I'll add that detail.
                    $endgroup$
                    – Henno Brandsma
                    Dec 19 '18 at 9:25




                    $begingroup$
                    @mathworker21 OK, I'll add that detail.
                    $endgroup$
                    – Henno Brandsma
                    Dec 19 '18 at 9:25











                    0












                    $begingroup$

                    It all depends on which definition of the product topology you would like to use. One proof I like is as follows:



                    The sets of the form $prod_{i in Bbb N} X_i$, where each $X_i$ is open and $X_i = Bbb R$ for all but finitely many $i in Bbb N$, form a basis of the product topology on $Bbb R^omega$. However, the set $S = (-1,1)^omega$ contains no such subset. In particular, we note that $pi_i(S)$ (the image of the $i$th projection map, applied to $S$) fails to contain $Bbb R$ for any $i in Bbb N$. Since $S$ contains no basis element, $S$ cannot be open.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      The set $S = emptyset$ contains no such subset. In particular, we note that $pi_i(S)$ fails to contain $mathbb{R}$ for any $i in mathbb{N}$. Since $S$ contains no basis element, $S$ cannot be open. :))))))
                      $endgroup$
                      – mathworker21
                      Dec 19 '18 at 2:31










                    • $begingroup$
                      @mathworker21 if we take $X_1 = emptyset$ and $X_j = mathbb{R}$ for $j > 1$ then $prod X_i = emptyset$, so the empty set is of that form. The projection property does only apply to nonempty sets, as you noted.
                      $endgroup$
                      – Carl Mummert
                      Dec 19 '18 at 11:43
















                    0












                    $begingroup$

                    It all depends on which definition of the product topology you would like to use. One proof I like is as follows:



                    The sets of the form $prod_{i in Bbb N} X_i$, where each $X_i$ is open and $X_i = Bbb R$ for all but finitely many $i in Bbb N$, form a basis of the product topology on $Bbb R^omega$. However, the set $S = (-1,1)^omega$ contains no such subset. In particular, we note that $pi_i(S)$ (the image of the $i$th projection map, applied to $S$) fails to contain $Bbb R$ for any $i in Bbb N$. Since $S$ contains no basis element, $S$ cannot be open.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      The set $S = emptyset$ contains no such subset. In particular, we note that $pi_i(S)$ fails to contain $mathbb{R}$ for any $i in mathbb{N}$. Since $S$ contains no basis element, $S$ cannot be open. :))))))
                      $endgroup$
                      – mathworker21
                      Dec 19 '18 at 2:31










                    • $begingroup$
                      @mathworker21 if we take $X_1 = emptyset$ and $X_j = mathbb{R}$ for $j > 1$ then $prod X_i = emptyset$, so the empty set is of that form. The projection property does only apply to nonempty sets, as you noted.
                      $endgroup$
                      – Carl Mummert
                      Dec 19 '18 at 11:43














                    0












                    0








                    0





                    $begingroup$

                    It all depends on which definition of the product topology you would like to use. One proof I like is as follows:



                    The sets of the form $prod_{i in Bbb N} X_i$, where each $X_i$ is open and $X_i = Bbb R$ for all but finitely many $i in Bbb N$, form a basis of the product topology on $Bbb R^omega$. However, the set $S = (-1,1)^omega$ contains no such subset. In particular, we note that $pi_i(S)$ (the image of the $i$th projection map, applied to $S$) fails to contain $Bbb R$ for any $i in Bbb N$. Since $S$ contains no basis element, $S$ cannot be open.






                    share|cite|improve this answer









                    $endgroup$



                    It all depends on which definition of the product topology you would like to use. One proof I like is as follows:



                    The sets of the form $prod_{i in Bbb N} X_i$, where each $X_i$ is open and $X_i = Bbb R$ for all but finitely many $i in Bbb N$, form a basis of the product topology on $Bbb R^omega$. However, the set $S = (-1,1)^omega$ contains no such subset. In particular, we note that $pi_i(S)$ (the image of the $i$th projection map, applied to $S$) fails to contain $Bbb R$ for any $i in Bbb N$. Since $S$ contains no basis element, $S$ cannot be open.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 19 '18 at 2:26









                    OmnomnomnomOmnomnomnom

                    129k792186




                    129k792186












                    • $begingroup$
                      The set $S = emptyset$ contains no such subset. In particular, we note that $pi_i(S)$ fails to contain $mathbb{R}$ for any $i in mathbb{N}$. Since $S$ contains no basis element, $S$ cannot be open. :))))))
                      $endgroup$
                      – mathworker21
                      Dec 19 '18 at 2:31










                    • $begingroup$
                      @mathworker21 if we take $X_1 = emptyset$ and $X_j = mathbb{R}$ for $j > 1$ then $prod X_i = emptyset$, so the empty set is of that form. The projection property does only apply to nonempty sets, as you noted.
                      $endgroup$
                      – Carl Mummert
                      Dec 19 '18 at 11:43


















                    • $begingroup$
                      The set $S = emptyset$ contains no such subset. In particular, we note that $pi_i(S)$ fails to contain $mathbb{R}$ for any $i in mathbb{N}$. Since $S$ contains no basis element, $S$ cannot be open. :))))))
                      $endgroup$
                      – mathworker21
                      Dec 19 '18 at 2:31










                    • $begingroup$
                      @mathworker21 if we take $X_1 = emptyset$ and $X_j = mathbb{R}$ for $j > 1$ then $prod X_i = emptyset$, so the empty set is of that form. The projection property does only apply to nonempty sets, as you noted.
                      $endgroup$
                      – Carl Mummert
                      Dec 19 '18 at 11:43
















                    $begingroup$
                    The set $S = emptyset$ contains no such subset. In particular, we note that $pi_i(S)$ fails to contain $mathbb{R}$ for any $i in mathbb{N}$. Since $S$ contains no basis element, $S$ cannot be open. :))))))
                    $endgroup$
                    – mathworker21
                    Dec 19 '18 at 2:31




                    $begingroup$
                    The set $S = emptyset$ contains no such subset. In particular, we note that $pi_i(S)$ fails to contain $mathbb{R}$ for any $i in mathbb{N}$. Since $S$ contains no basis element, $S$ cannot be open. :))))))
                    $endgroup$
                    – mathworker21
                    Dec 19 '18 at 2:31












                    $begingroup$
                    @mathworker21 if we take $X_1 = emptyset$ and $X_j = mathbb{R}$ for $j > 1$ then $prod X_i = emptyset$, so the empty set is of that form. The projection property does only apply to nonempty sets, as you noted.
                    $endgroup$
                    – Carl Mummert
                    Dec 19 '18 at 11:43




                    $begingroup$
                    @mathworker21 if we take $X_1 = emptyset$ and $X_j = mathbb{R}$ for $j > 1$ then $prod X_i = emptyset$, so the empty set is of that form. The projection property does only apply to nonempty sets, as you noted.
                    $endgroup$
                    – Carl Mummert
                    Dec 19 '18 at 11:43


















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