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Continuity of countable projection from non-first countable topological space



The Next CEO of Stack OverflowProof that product topology of subspace is same as induced product topologyEquivalence of continuous and sequential continuous implies first-countable?Compact topological space not having Countable Basis?Prove that $mathbb R/mathbb Z$ is sequential and not first countableCountable product of first countable Spaces is first countableIs Every Sequentially Compact, First Countable Space Also Compact?Sequential continuity implies continuity in the weak topology on a normed spaceWhat kind of a property implies (sequentially compact $iff$ compact)?Is this quotient space first countable?Locally Compact Hausdorff Topologial Group but not Sequential?Does a sequentially continuous function take its supremum on compacts?










0












$begingroup$


This might be trivial, but I just want to make sure I got this right:



Let $X$ be a metric space and $I$ an uncountable index set. Let us consider $X^I$ with the product topology (of course, the topology of $X$ is the one induced by the metric). Let $J subseteq I$ be an (infinite) countable subset of $I$. Let us also consider $X^J$ with the product topology and consider the map



$F: X^I to X^J$ given by $F((x_i)_i in I) := (x_j)_j in J$, i.e. $F$ is simply the projection from $I$ to $J$.



Note that $X^I$ is not a first countable topological space, so continuity of $F$ is not the same as sequential continuity of $F$. Now my question: Is the map $F$ continuous? It is trivially sequentially continuous and if $X^I$ was first countable, then the statement follows. However, I am not used to working with nets instead of sequences (which is necessary to check continuity of $F$, since its domain is not first countable) - so I wanted to make sure that my intuition that $F$ is continuous is correct. Intuitively, nothing can go wrong, because when I want to check continuity via nets, since I consider the same topology on both spaces, it seems to be straightforward.



Thankful for any answers on this!










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    This might be trivial, but I just want to make sure I got this right:



    Let $X$ be a metric space and $I$ an uncountable index set. Let us consider $X^I$ with the product topology (of course, the topology of $X$ is the one induced by the metric). Let $J subseteq I$ be an (infinite) countable subset of $I$. Let us also consider $X^J$ with the product topology and consider the map



    $F: X^I to X^J$ given by $F((x_i)_i in I) := (x_j)_j in J$, i.e. $F$ is simply the projection from $I$ to $J$.



    Note that $X^I$ is not a first countable topological space, so continuity of $F$ is not the same as sequential continuity of $F$. Now my question: Is the map $F$ continuous? It is trivially sequentially continuous and if $X^I$ was first countable, then the statement follows. However, I am not used to working with nets instead of sequences (which is necessary to check continuity of $F$, since its domain is not first countable) - so I wanted to make sure that my intuition that $F$ is continuous is correct. Intuitively, nothing can go wrong, because when I want to check continuity via nets, since I consider the same topology on both spaces, it seems to be straightforward.



    Thankful for any answers on this!










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      This might be trivial, but I just want to make sure I got this right:



      Let $X$ be a metric space and $I$ an uncountable index set. Let us consider $X^I$ with the product topology (of course, the topology of $X$ is the one induced by the metric). Let $J subseteq I$ be an (infinite) countable subset of $I$. Let us also consider $X^J$ with the product topology and consider the map



      $F: X^I to X^J$ given by $F((x_i)_i in I) := (x_j)_j in J$, i.e. $F$ is simply the projection from $I$ to $J$.



      Note that $X^I$ is not a first countable topological space, so continuity of $F$ is not the same as sequential continuity of $F$. Now my question: Is the map $F$ continuous? It is trivially sequentially continuous and if $X^I$ was first countable, then the statement follows. However, I am not used to working with nets instead of sequences (which is necessary to check continuity of $F$, since its domain is not first countable) - so I wanted to make sure that my intuition that $F$ is continuous is correct. Intuitively, nothing can go wrong, because when I want to check continuity via nets, since I consider the same topology on both spaces, it seems to be straightforward.



      Thankful for any answers on this!










      share|cite|improve this question









      $endgroup$




      This might be trivial, but I just want to make sure I got this right:



      Let $X$ be a metric space and $I$ an uncountable index set. Let us consider $X^I$ with the product topology (of course, the topology of $X$ is the one induced by the metric). Let $J subseteq I$ be an (infinite) countable subset of $I$. Let us also consider $X^J$ with the product topology and consider the map



      $F: X^I to X^J$ given by $F((x_i)_i in I) := (x_j)_j in J$, i.e. $F$ is simply the projection from $I$ to $J$.



      Note that $X^I$ is not a first countable topological space, so continuity of $F$ is not the same as sequential continuity of $F$. Now my question: Is the map $F$ continuous? It is trivially sequentially continuous and if $X^I$ was first countable, then the statement follows. However, I am not used to working with nets instead of sequences (which is necessary to check continuity of $F$, since its domain is not first countable) - so I wanted to make sure that my intuition that $F$ is continuous is correct. Intuitively, nothing can go wrong, because when I want to check continuity via nets, since I consider the same topology on both spaces, it seems to be straightforward.



      Thankful for any answers on this!







      general-topology product-space first-countable






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 27 at 18:43









      MarcoMarco

      607




      607




















          1 Answer
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          1












          $begingroup$

          Of course this map is continuous: if $pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $pi_i$ ($i$ in its index set), i.e. $ pi_i circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $pi_i circ F = pi_i$ for all $i in J$ trivially, so $F$ is continuous.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            You are right. Thanks a lot!
            $endgroup$
            – Marco
            2 days ago











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          active

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          active

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          1












          $begingroup$

          Of course this map is continuous: if $pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $pi_i$ ($i$ in its index set), i.e. $ pi_i circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $pi_i circ F = pi_i$ for all $i in J$ trivially, so $F$ is continuous.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            You are right. Thanks a lot!
            $endgroup$
            – Marco
            2 days ago















          1












          $begingroup$

          Of course this map is continuous: if $pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $pi_i$ ($i$ in its index set), i.e. $ pi_i circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $pi_i circ F = pi_i$ for all $i in J$ trivially, so $F$ is continuous.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            You are right. Thanks a lot!
            $endgroup$
            – Marco
            2 days ago













          1












          1








          1





          $begingroup$

          Of course this map is continuous: if $pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $pi_i$ ($i$ in its index set), i.e. $ pi_i circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $pi_i circ F = pi_i$ for all $i in J$ trivially, so $F$ is continuous.






          share|cite|improve this answer











          $endgroup$



          Of course this map is continuous: if $pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $pi_i$ ($i$ in its index set), i.e. $ pi_i circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $pi_i circ F = pi_i$ for all $i in J$ trivially, so $F$ is continuous.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Henno BrandsmaHenno Brandsma

          115k348124




          115k348124











          • $begingroup$
            You are right. Thanks a lot!
            $endgroup$
            – Marco
            2 days ago
















          • $begingroup$
            You are right. Thanks a lot!
            $endgroup$
            – Marco
            2 days ago















          $begingroup$
          You are right. Thanks a lot!
          $endgroup$
          – Marco
          2 days ago




          $begingroup$
          You are right. Thanks a lot!
          $endgroup$
          – Marco
          2 days ago

















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