The Domain of a continuous function is an open set












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I have started studying the book "Differentiable manifolds; An Introduction" by Brickell & Clark. But I have encountered the following paragraph on page 4.



image



Why is the domain of a continuous functions nacessarily an open set?



Thanks for any help.






geometry metric-spaces differential-topology







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  • $begingroup$
    If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
    $endgroup$
    – Kavi Rama Murthy
    Jan 18 at 8:10
















0












$begingroup$


I have started studying the book "Differentiable manifolds; An Introduction" by Brickell & Clark. But I have encountered the following paragraph on page 4.



image



Why is the domain of a continuous functions nacessarily an open set?



Thanks for any help.






geometry metric-spaces differential-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
    $endgroup$
    – Kavi Rama Murthy
    Jan 18 at 8:10














0












0








0





$begingroup$


I have started studying the book "Differentiable manifolds; An Introduction" by Brickell & Clark. But I have encountered the following paragraph on page 4.



image



Why is the domain of a continuous functions nacessarily an open set?



Thanks for any help.






geometry metric-spaces differential-topology







share|cite|improve this question











$endgroup$




I have started studying the book "Differentiable manifolds; An Introduction" by Brickell & Clark. But I have encountered the following paragraph on page 4.



image



Why is the domain of a continuous functions nacessarily an open set?



Thanks for any help.






geometry metric-spaces differential-topology




geometry metric-spaces differential-topology






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share|cite|improve this question













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edited Jan 18 at 10:11







user

















asked Jan 18 at 7:59









useruser

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  • $begingroup$
    If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
    $endgroup$
    – Kavi Rama Murthy
    Jan 18 at 8:10


















  • $begingroup$
    If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
    $endgroup$
    – Kavi Rama Murthy
    Jan 18 at 8:10
















$begingroup$
If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:10




$begingroup$
If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:10










2 Answers
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Let $X,Y$ be topological spaces. Then by topological definition $f:X rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.



Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.






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    $S$ and $T$ are topological spaces and the domain of $f:S to T$ is the set $S$ and $S$ is open.






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      2 Answers
      2






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      2 Answers
      2






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      $begingroup$

      Let $X,Y$ be topological spaces. Then by topological definition $f:X rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.



      Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $X,Y$ be topological spaces. Then by topological definition $f:X rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.



        Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $X,Y$ be topological spaces. Then by topological definition $f:X rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.



          Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.






          share|cite|improve this answer









          $endgroup$



          Let $X,Y$ be topological spaces. Then by topological definition $f:X rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.



          Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 at 8:25









          ZeeklessZeekless

          577111




          577111























              0












              $begingroup$

              $S$ and $T$ are topological spaces and the domain of $f:S to T$ is the set $S$ and $S$ is open.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $S$ and $T$ are topological spaces and the domain of $f:S to T$ is the set $S$ and $S$ is open.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $S$ and $T$ are topological spaces and the domain of $f:S to T$ is the set $S$ and $S$ is open.






                  share|cite|improve this answer









                  $endgroup$



                  $S$ and $T$ are topological spaces and the domain of $f:S to T$ is the set $S$ and $S$ is open.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 8:09









                  FredFred

                  46.9k1848




                  46.9k1848






























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