How can i prove that? Connected components and continuous function












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$X_1$, $X_2$ are closed and open connected components of the сompact $X = sin (1/x)$. ( It is compact $X subseteq mathbb{R}^2$ - the union of a vertical segment $X_1 = {(0,y) : −1 leq y leq 1}$ and graph $X_2$ of function $y = sin(1/x)$, $0 < x leq 1$)



Prove that if we have a continuous function $f:Xrightarrow X$, then exist point $x in X_2$: if $f(x)in X_1$, then $f(X) subseteq X_1$.



Please, help me understand how i can do it. And please excuse my English.






general-topology algebraic-topology







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    0












    $begingroup$


    $X_1$, $X_2$ are closed and open connected components of the сompact $X = sin (1/x)$. ( It is compact $X subseteq mathbb{R}^2$ - the union of a vertical segment $X_1 = {(0,y) : −1 leq y leq 1}$ and graph $X_2$ of function $y = sin(1/x)$, $0 < x leq 1$)



    Prove that if we have a continuous function $f:Xrightarrow X$, then exist point $x in X_2$: if $f(x)in X_1$, then $f(X) subseteq X_1$.



    Please, help me understand how i can do it. And please excuse my English.






    general-topology algebraic-topology







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $X_1$, $X_2$ are closed and open connected components of the сompact $X = sin (1/x)$. ( It is compact $X subseteq mathbb{R}^2$ - the union of a vertical segment $X_1 = {(0,y) : −1 leq y leq 1}$ and graph $X_2$ of function $y = sin(1/x)$, $0 < x leq 1$)



      Prove that if we have a continuous function $f:Xrightarrow X$, then exist point $x in X_2$: if $f(x)in X_1$, then $f(X) subseteq X_1$.



      Please, help me understand how i can do it. And please excuse my English.






      general-topology algebraic-topology







      share|cite|improve this question











      $endgroup$




      $X_1$, $X_2$ are closed and open connected components of the сompact $X = sin (1/x)$. ( It is compact $X subseteq mathbb{R}^2$ - the union of a vertical segment $X_1 = {(0,y) : −1 leq y leq 1}$ and graph $X_2$ of function $y = sin(1/x)$, $0 < x leq 1$)



      Prove that if we have a continuous function $f:Xrightarrow X$, then exist point $x in X_2$: if $f(x)in X_1$, then $f(X) subseteq X_1$.



      Please, help me understand how i can do it. And please excuse my English.






      general-topology algebraic-topology




      general-topology algebraic-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 16 at 14:08









      Tyrone

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      asked Jan 16 at 13:34









      pentarrapentarra

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      31






















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          Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $overline{X}_2 = X$ and $overline{X}_1 = X_1$.



          Given a continuous $f : X to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.



          Under the assumption in your question we infer that $f(X_2) subset X_1$. But then continuity implies $f(X) = f(overline{X}_2) subset overline{f(X_2)} subset overline{X}_1 = X_1$.






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            $begingroup$

            Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $overline{X}_2 = X$ and $overline{X}_1 = X_1$.



            Given a continuous $f : X to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.



            Under the assumption in your question we infer that $f(X_2) subset X_1$. But then continuity implies $f(X) = f(overline{X}_2) subset overline{f(X_2)} subset overline{X}_1 = X_1$.






            share|cite|improve this answer









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              0












              $begingroup$

              Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $overline{X}_2 = X$ and $overline{X}_1 = X_1$.



              Given a continuous $f : X to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.



              Under the assumption in your question we infer that $f(X_2) subset X_1$. But then continuity implies $f(X) = f(overline{X}_2) subset overline{f(X_2)} subset overline{X}_1 = X_1$.






              share|cite|improve this answer









              $endgroup$
















                0












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                0





                $begingroup$

                Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $overline{X}_2 = X$ and $overline{X}_1 = X_1$.



                Given a continuous $f : X to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.



                Under the assumption in your question we infer that $f(X_2) subset X_1$. But then continuity implies $f(X) = f(overline{X}_2) subset overline{f(X_2)} subset overline{X}_1 = X_1$.






                share|cite|improve this answer









                $endgroup$



                Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $overline{X}_2 = X$ and $overline{X}_1 = X_1$.



                Given a continuous $f : X to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.



                Under the assumption in your question we infer that $f(X_2) subset X_1$. But then continuity implies $f(X) = f(overline{X}_2) subset overline{f(X_2)} subset overline{X}_1 = X_1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 16 at 22:19









                Paul FrostPaul Frost

                11.1k3934




                11.1k3934






























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