Prove $f(overline{U})) subset overline{f(U)}$












0












$begingroup$


I'm facing this problem: For $f : (X,mathcal{T}) to (Y,mathcal{T}')$ a continuous map between topological space, then for any subset $U subset X$ is $f(overline{U)) subset overline{f(U)}$.



I know the usual proof, but I´d like is this one is right:



Let be $x in f(overline{U})$. So for any open neighborhood $V$ of $x$ is $V cap f(U) neq emptyset$.



Since $f$ is continuous, $f^{-1}(V)$ is an open neighborhood of $f^{-1}(x)$.



Using $f^{-1}(x) in overline{U}$ is $f^{-1}(V) cap U neq emptyset$.



So $f(f^{-1}(V) cap U) = V cap f(U) neq emptyset$, then $x in overline{f(U)}$.



Is this right? I'm not sure.



And here another try:



Suppose exist $x in f(overline{U})$ with $x notin overline{f(U)}.$ Then exist an open neightborhood $V$ of $x$ such that $V cap f(U) = emptyset$.



For $z in overline{U}$ such that $f(z) = x$, for any $W$ open neighborhood of $z$ is $W cap A neq emptyset$.



Then exist $h in W cap A$ and $f(h) in f(W) cap f(A)$. So for the adequate $f(W) subset V$ is $f(W) cap U neq emptyset$ but $V cap f(U) = emptyset$. A contradiction.



Is this last one right?



Thanks!






general-topology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
    $endgroup$
    – o.h.
    Jan 22 at 19:25












  • $begingroup$
    In general is true $f(U cap V) = f(U) cap f(V)$?
    $endgroup$
    – LH8
    Jan 22 at 20:02










  • $begingroup$
    If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
    $endgroup$
    – Umberto P.
    Jan 22 at 20:05










  • $begingroup$
    And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
    $endgroup$
    – LH8
    Jan 22 at 20:19
















0












$begingroup$


I'm facing this problem: For $f : (X,mathcal{T}) to (Y,mathcal{T}')$ a continuous map between topological space, then for any subset $U subset X$ is $f(overline{U)) subset overline{f(U)}$.



I know the usual proof, but I´d like is this one is right:



Let be $x in f(overline{U})$. So for any open neighborhood $V$ of $x$ is $V cap f(U) neq emptyset$.



Since $f$ is continuous, $f^{-1}(V)$ is an open neighborhood of $f^{-1}(x)$.



Using $f^{-1}(x) in overline{U}$ is $f^{-1}(V) cap U neq emptyset$.



So $f(f^{-1}(V) cap U) = V cap f(U) neq emptyset$, then $x in overline{f(U)}$.



Is this right? I'm not sure.



And here another try:



Suppose exist $x in f(overline{U})$ with $x notin overline{f(U)}.$ Then exist an open neightborhood $V$ of $x$ such that $V cap f(U) = emptyset$.



For $z in overline{U}$ such that $f(z) = x$, for any $W$ open neighborhood of $z$ is $W cap A neq emptyset$.



Then exist $h in W cap A$ and $f(h) in f(W) cap f(A)$. So for the adequate $f(W) subset V$ is $f(W) cap U neq emptyset$ but $V cap f(U) = emptyset$. A contradiction.



Is this last one right?



Thanks!






general-topology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
    $endgroup$
    – o.h.
    Jan 22 at 19:25












  • $begingroup$
    In general is true $f(U cap V) = f(U) cap f(V)$?
    $endgroup$
    – LH8
    Jan 22 at 20:02










  • $begingroup$
    If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
    $endgroup$
    – Umberto P.
    Jan 22 at 20:05










  • $begingroup$
    And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
    $endgroup$
    – LH8
    Jan 22 at 20:19














0












0








0





$begingroup$


I'm facing this problem: For $f : (X,mathcal{T}) to (Y,mathcal{T}')$ a continuous map between topological space, then for any subset $U subset X$ is $f(overline{U)) subset overline{f(U)}$.



I know the usual proof, but I´d like is this one is right:



Let be $x in f(overline{U})$. So for any open neighborhood $V$ of $x$ is $V cap f(U) neq emptyset$.



Since $f$ is continuous, $f^{-1}(V)$ is an open neighborhood of $f^{-1}(x)$.



Using $f^{-1}(x) in overline{U}$ is $f^{-1}(V) cap U neq emptyset$.



So $f(f^{-1}(V) cap U) = V cap f(U) neq emptyset$, then $x in overline{f(U)}$.



Is this right? I'm not sure.



And here another try:



Suppose exist $x in f(overline{U})$ with $x notin overline{f(U)}.$ Then exist an open neightborhood $V$ of $x$ such that $V cap f(U) = emptyset$.



For $z in overline{U}$ such that $f(z) = x$, for any $W$ open neighborhood of $z$ is $W cap A neq emptyset$.



Then exist $h in W cap A$ and $f(h) in f(W) cap f(A)$. So for the adequate $f(W) subset V$ is $f(W) cap U neq emptyset$ but $V cap f(U) = emptyset$. A contradiction.



Is this last one right?



Thanks!






general-topology







share|cite|improve this question











$endgroup$




I'm facing this problem: For $f : (X,mathcal{T}) to (Y,mathcal{T}')$ a continuous map between topological space, then for any subset $U subset X$ is $f(overline{U)) subset overline{f(U)}$.



I know the usual proof, but I´d like is this one is right:



Let be $x in f(overline{U})$. So for any open neighborhood $V$ of $x$ is $V cap f(U) neq emptyset$.



Since $f$ is continuous, $f^{-1}(V)$ is an open neighborhood of $f^{-1}(x)$.



Using $f^{-1}(x) in overline{U}$ is $f^{-1}(V) cap U neq emptyset$.



So $f(f^{-1}(V) cap U) = V cap f(U) neq emptyset$, then $x in overline{f(U)}$.



Is this right? I'm not sure.



And here another try:



Suppose exist $x in f(overline{U})$ with $x notin overline{f(U)}.$ Then exist an open neightborhood $V$ of $x$ such that $V cap f(U) = emptyset$.



For $z in overline{U}$ such that $f(z) = x$, for any $W$ open neighborhood of $z$ is $W cap A neq emptyset$.



Then exist $h in W cap A$ and $f(h) in f(W) cap f(A)$. So for the adequate $f(W) subset V$ is $f(W) cap U neq emptyset$ but $V cap f(U) = emptyset$. A contradiction.



Is this last one right?



Thanks!






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 19:20







LH8

















asked Jan 22 at 19:11









LH8LH8

1368




1368








  • 1




    $begingroup$
    Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
    $endgroup$
    – o.h.
    Jan 22 at 19:25












  • $begingroup$
    In general is true $f(U cap V) = f(U) cap f(V)$?
    $endgroup$
    – LH8
    Jan 22 at 20:02










  • $begingroup$
    If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
    $endgroup$
    – Umberto P.
    Jan 22 at 20:05










  • $begingroup$
    And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
    $endgroup$
    – LH8
    Jan 22 at 20:19














  • 1




    $begingroup$
    Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
    $endgroup$
    – o.h.
    Jan 22 at 19:25












  • $begingroup$
    In general is true $f(U cap V) = f(U) cap f(V)$?
    $endgroup$
    – LH8
    Jan 22 at 20:02










  • $begingroup$
    If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
    $endgroup$
    – Umberto P.
    Jan 22 at 20:05










  • $begingroup$
    And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
    $endgroup$
    – LH8
    Jan 22 at 20:19








1




1




$begingroup$
Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
$endgroup$
– o.h.
Jan 22 at 19:25






$begingroup$
Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
$endgroup$
– o.h.
Jan 22 at 19:25














$begingroup$
In general is true $f(U cap V) = f(U) cap f(V)$?
$endgroup$
– LH8
Jan 22 at 20:02




$begingroup$
In general is true $f(U cap V) = f(U) cap f(V)$?
$endgroup$
– LH8
Jan 22 at 20:02












$begingroup$
If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
$endgroup$
– Umberto P.
Jan 22 at 20:05




$begingroup$
If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
$endgroup$
– Umberto P.
Jan 22 at 20:05












$begingroup$
And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
$endgroup$
– LH8
Jan 22 at 20:19




$begingroup$
And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
$endgroup$
– LH8
Jan 22 at 20:19










2 Answers
2






active

oldest

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1












$begingroup$

Let $f$ be continuous and suppose $y in f[overline{U}]$. So $y=f(x)$ with $x in overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] cap U$ is non-empty (as $x in overline{U}$), say that $x' in f^{-1}[O] cap U$. But then $f(x') in f[U]$ and $f(x') in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y in overline{f[U]}$ and the inclusion has been shown.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    You shouldn't have to use points at all. Since $overline{f(U)}$ is closed the continuity of $f$ implies $f^{-1}(overline{f(U)})$ is closed too. But $$U subset f^{-1}(f(U)) subset f^{-1}(overline{f(U)}) implies overline U subset f^{-1}(overline{f(U)})$$ because the latter set is closed. Thus
    $$ f(overline U) subset f(f^{-1}(overline{f(U)})) = overline{f(U)}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
      $endgroup$
      – LH8
      Jan 22 at 19:48










    • $begingroup$
      Please, answer my question using elements. I knew that proof.
      $endgroup$
      – LH8
      Jan 22 at 20:47











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    2 Answers
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    2 Answers
    2






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    active

    oldest

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    1












    $begingroup$

    Let $f$ be continuous and suppose $y in f[overline{U}]$. So $y=f(x)$ with $x in overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] cap U$ is non-empty (as $x in overline{U}$), say that $x' in f^{-1}[O] cap U$. But then $f(x') in f[U]$ and $f(x') in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y in overline{f[U]}$ and the inclusion has been shown.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $f$ be continuous and suppose $y in f[overline{U}]$. So $y=f(x)$ with $x in overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] cap U$ is non-empty (as $x in overline{U}$), say that $x' in f^{-1}[O] cap U$. But then $f(x') in f[U]$ and $f(x') in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y in overline{f[U]}$ and the inclusion has been shown.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $f$ be continuous and suppose $y in f[overline{U}]$. So $y=f(x)$ with $x in overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] cap U$ is non-empty (as $x in overline{U}$), say that $x' in f^{-1}[O] cap U$. But then $f(x') in f[U]$ and $f(x') in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y in overline{f[U]}$ and the inclusion has been shown.






        share|cite|improve this answer









        $endgroup$



        Let $f$ be continuous and suppose $y in f[overline{U}]$. So $y=f(x)$ with $x in overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] cap U$ is non-empty (as $x in overline{U}$), say that $x' in f^{-1}[O] cap U$. But then $f(x') in f[U]$ and $f(x') in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y in overline{f[U]}$ and the inclusion has been shown.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 23 at 5:39









        Henno BrandsmaHenno Brandsma

        111k348118




        111k348118























            2












            $begingroup$

            You shouldn't have to use points at all. Since $overline{f(U)}$ is closed the continuity of $f$ implies $f^{-1}(overline{f(U)})$ is closed too. But $$U subset f^{-1}(f(U)) subset f^{-1}(overline{f(U)}) implies overline U subset f^{-1}(overline{f(U)})$$ because the latter set is closed. Thus
            $$ f(overline U) subset f(f^{-1}(overline{f(U)})) = overline{f(U)}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
              $endgroup$
              – LH8
              Jan 22 at 19:48










            • $begingroup$
              Please, answer my question using elements. I knew that proof.
              $endgroup$
              – LH8
              Jan 22 at 20:47
















            2












            $begingroup$

            You shouldn't have to use points at all. Since $overline{f(U)}$ is closed the continuity of $f$ implies $f^{-1}(overline{f(U)})$ is closed too. But $$U subset f^{-1}(f(U)) subset f^{-1}(overline{f(U)}) implies overline U subset f^{-1}(overline{f(U)})$$ because the latter set is closed. Thus
            $$ f(overline U) subset f(f^{-1}(overline{f(U)})) = overline{f(U)}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
              $endgroup$
              – LH8
              Jan 22 at 19:48










            • $begingroup$
              Please, answer my question using elements. I knew that proof.
              $endgroup$
              – LH8
              Jan 22 at 20:47














            2












            2








            2





            $begingroup$

            You shouldn't have to use points at all. Since $overline{f(U)}$ is closed the continuity of $f$ implies $f^{-1}(overline{f(U)})$ is closed too. But $$U subset f^{-1}(f(U)) subset f^{-1}(overline{f(U)}) implies overline U subset f^{-1}(overline{f(U)})$$ because the latter set is closed. Thus
            $$ f(overline U) subset f(f^{-1}(overline{f(U)})) = overline{f(U)}.$$






            share|cite|improve this answer









            $endgroup$



            You shouldn't have to use points at all. Since $overline{f(U)}$ is closed the continuity of $f$ implies $f^{-1}(overline{f(U)})$ is closed too. But $$U subset f^{-1}(f(U)) subset f^{-1}(overline{f(U)}) implies overline U subset f^{-1}(overline{f(U)})$$ because the latter set is closed. Thus
            $$ f(overline U) subset f(f^{-1}(overline{f(U)})) = overline{f(U)}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 22 at 19:26









            Umberto P.Umberto P.

            39.6k13267




            39.6k13267












            • $begingroup$
              I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
              $endgroup$
              – LH8
              Jan 22 at 19:48










            • $begingroup$
              Please, answer my question using elements. I knew that proof.
              $endgroup$
              – LH8
              Jan 22 at 20:47


















            • $begingroup$
              I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
              $endgroup$
              – LH8
              Jan 22 at 19:48










            • $begingroup$
              Please, answer my question using elements. I knew that proof.
              $endgroup$
              – LH8
              Jan 22 at 20:47
















            $begingroup$
            I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
            $endgroup$
            – LH8
            Jan 22 at 19:48




            $begingroup$
            I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
            $endgroup$
            – LH8
            Jan 22 at 19:48












            $begingroup$
            Please, answer my question using elements. I knew that proof.
            $endgroup$
            – LH8
            Jan 22 at 20:47




            $begingroup$
            Please, answer my question using elements. I knew that proof.
            $endgroup$
            – LH8
            Jan 22 at 20:47


















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            What does “Dominus providebit” mean?

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            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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