Homeomorphism preserves regular openness

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Let $f: X to Y$ be a homeomorphism between two topological spaces. Let $U subset X$ be a regular open connected set, i.e., $textint(barU) = U$. Let $V = f(U)$. Then $f|_U: U to V$ is a homeomorphism on the two subspaces. Is it true $V$ is also regular open? I know $f(barU) = barV$ and $f(partial U) = partial V$ if $f$ is homeomorphism. But how do we conclude $V = barV setminus partial V$?




general-topology continuity




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    Let $f: X to Y$ be a homeomorphism between two topological spaces. Let $U subset X$ be a regular open connected set, i.e., $textint(barU) = U$. Let $V = f(U)$. Then $f|_U: U to V$ is a homeomorphism on the two subspaces. Is it true $V$ is also regular open? I know $f(barU) = barV$ and $f(partial U) = partial V$ if $f$ is homeomorphism. But how do we conclude $V = barV setminus partial V$?




    general-topology continuity




    share|cite|improve this question





















      up vote
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      favorite









      up vote
      1
      down vote

      favorite











      Let $f: X to Y$ be a homeomorphism between two topological spaces. Let $U subset X$ be a regular open connected set, i.e., $textint(barU) = U$. Let $V = f(U)$. Then $f|_U: U to V$ is a homeomorphism on the two subspaces. Is it true $V$ is also regular open? I know $f(barU) = barV$ and $f(partial U) = partial V$ if $f$ is homeomorphism. But how do we conclude $V = barV setminus partial V$?




      general-topology continuity




      share|cite|improve this question











      Let $f: X to Y$ be a homeomorphism between two topological spaces. Let $U subset X$ be a regular open connected set, i.e., $textint(barU) = U$. Let $V = f(U)$. Then $f|_U: U to V$ is a homeomorphism on the two subspaces. Is it true $V$ is also regular open? I know $f(barU) = barV$ and $f(partial U) = partial V$ if $f$ is homeomorphism. But how do we conclude $V = barV setminus partial V$?





      general-topology continuity





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      asked Aug 3 at 20:02









      user9527

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          Hint: Since $f$ is a bijection,
          $$f(U) = f(barU) backslash f(partial U).$$






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            up vote
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            Hint: Since $f$ is a bijection,
            $$f(U) = f(barU) backslash f(partial U).$$






            share|cite|improve this answer

























              up vote
              1
              down vote













              Hint: Since $f$ is a bijection,
              $$f(U) = f(barU) backslash f(partial U).$$






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Hint: Since $f$ is a bijection,
                $$f(U) = f(barU) backslash f(partial U).$$






                share|cite|improve this answer













                Hint: Since $f$ is a bijection,
                $$f(U) = f(barU) backslash f(partial U).$$







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Aug 3 at 21:19









                Zestylemonzi

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