Closure of an open cell in a CW complex












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Given a CW complex $X$ and any of its cells $e$, the closure $bar{e}$ in $X$ is covered by finitely many open cells by "C". Can we prove that $bar{e}$ is exactly a union of open cells, or, $(bar{e}-e)$ a union of open cells of strictly lower dimensions? If not, is there any counterexample?










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    Given a CW complex $X$ and any of its cells $e$, the closure $bar{e}$ in $X$ is covered by finitely many open cells by "C". Can we prove that $bar{e}$ is exactly a union of open cells, or, $(bar{e}-e)$ a union of open cells of strictly lower dimensions? If not, is there any counterexample?










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      Given a CW complex $X$ and any of its cells $e$, the closure $bar{e}$ in $X$ is covered by finitely many open cells by "C". Can we prove that $bar{e}$ is exactly a union of open cells, or, $(bar{e}-e)$ a union of open cells of strictly lower dimensions? If not, is there any counterexample?










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      Given a CW complex $X$ and any of its cells $e$, the closure $bar{e}$ in $X$ is covered by finitely many open cells by "C". Can we prove that $bar{e}$ is exactly a union of open cells, or, $(bar{e}-e)$ a union of open cells of strictly lower dimensions? If not, is there any counterexample?







      general-topology algebraic-topology






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      asked Dec 26 at 8:25









      Smart Yao

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          In general $overline e$ won't be a union of open cells.



          Simple example. One $0$-cell $e_0$, one $1$-cell $e_1$
          and one $2$-cell $e_2$. Attach $e_1$ to $e_0$ making an $S^1$. Now attach $e_2$
          to this $S_1$ by mapping the boundary of the unit disc to a point $P$ on $S^1$ that
          isn't $e_0$. Then $overline{e_2}=e_2cup{P}$ and that isn't a union of
          open cells.






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            In general $overline e$ won't be a union of open cells.



            Simple example. One $0$-cell $e_0$, one $1$-cell $e_1$
            and one $2$-cell $e_2$. Attach $e_1$ to $e_0$ making an $S^1$. Now attach $e_2$
            to this $S_1$ by mapping the boundary of the unit disc to a point $P$ on $S^1$ that
            isn't $e_0$. Then $overline{e_2}=e_2cup{P}$ and that isn't a union of
            open cells.






            share|cite|improve this answer




























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              In general $overline e$ won't be a union of open cells.



              Simple example. One $0$-cell $e_0$, one $1$-cell $e_1$
              and one $2$-cell $e_2$. Attach $e_1$ to $e_0$ making an $S^1$. Now attach $e_2$
              to this $S_1$ by mapping the boundary of the unit disc to a point $P$ on $S^1$ that
              isn't $e_0$. Then $overline{e_2}=e_2cup{P}$ and that isn't a union of
              open cells.






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                In general $overline e$ won't be a union of open cells.



                Simple example. One $0$-cell $e_0$, one $1$-cell $e_1$
                and one $2$-cell $e_2$. Attach $e_1$ to $e_0$ making an $S^1$. Now attach $e_2$
                to this $S_1$ by mapping the boundary of the unit disc to a point $P$ on $S^1$ that
                isn't $e_0$. Then $overline{e_2}=e_2cup{P}$ and that isn't a union of
                open cells.






                share|cite|improve this answer














                In general $overline e$ won't be a union of open cells.



                Simple example. One $0$-cell $e_0$, one $1$-cell $e_1$
                and one $2$-cell $e_2$. Attach $e_1$ to $e_0$ making an $S^1$. Now attach $e_2$
                to this $S_1$ by mapping the boundary of the unit disc to a point $P$ on $S^1$ that
                isn't $e_0$. Then $overline{e_2}=e_2cup{P}$ and that isn't a union of
                open cells.







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                edited Dec 26 at 8:42

























                answered Dec 26 at 8:30









                Lord Shark the Unknown

                101k958132




                101k958132






























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