Closure of an open cell in a CW complex
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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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
add a comment |
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
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
general-topology algebraic-topology
asked Dec 26 at 8:25
Smart Yao
1168
1168
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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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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |
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.
add a comment |
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.
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.
edited Dec 26 at 8:42
answered Dec 26 at 8:30
Lord Shark the Unknown
101k958132
101k958132
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