If all slices of a subset of a product have covering dimension $n$, does the set have dimension $2n$?

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Let $Ssubset Atimes B$ be closed, where $A$ and $B$ are closed balls of dim $n$.



Assume that $forall bin B$, $Scap(Atimesb)$ has dimension $n$.



Then, does $S$ have dimension $2n$?



[Additional Question] If not, would there be any (not too strong) additional condition that makes this hold?




general-topology dimension-theory




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    up vote
    2
    down vote

    favorite












    Let $Ssubset Atimes B$ be closed, where $A$ and $B$ are closed balls of dim $n$.



    Assume that $forall bin B$, $Scap(Atimesb)$ has dimension $n$.



    Then, does $S$ have dimension $2n$?



    [Additional Question] If not, would there be any (not too strong) additional condition that makes this hold?




    general-topology dimension-theory




    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $Ssubset Atimes B$ be closed, where $A$ and $B$ are closed balls of dim $n$.



      Assume that $forall bin B$, $Scap(Atimesb)$ has dimension $n$.



      Then, does $S$ have dimension $2n$?



      [Additional Question] If not, would there be any (not too strong) additional condition that makes this hold?




      general-topology dimension-theory




      share|cite|improve this question













      Let $Ssubset Atimes B$ be closed, where $A$ and $B$ are closed balls of dim $n$.



      Assume that $forall bin B$, $Scap(Atimesb)$ has dimension $n$.



      Then, does $S$ have dimension $2n$?



      [Additional Question] If not, would there be any (not too strong) additional condition that makes this hold?





      general-topology dimension-theory





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 19 at 17:08









      user357151

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      asked Jul 19 at 15:52









      C.Park

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