Does ${x_n:nge1}cup{y_n:nge1}$ has no accumulation point in $X?$
Let $(x_n),(y_n)$ be two sequences in a metrizable topological space $X$ such that none of $(x_n),(y_n)$ has cluster point in $X.$ Can we then conclude the set ${x_n:nge1}cup{y_n:nge1}$ has no accumulation point in $X?$
I think that should be correct yet I cannot get sure.
Please help.
general-topology
asked Nov 28 '18 at 2:29
Jave
458114
add a comment |
Let $(x_n),(y_n)$ be two sequences in a metrizable topological space $X$ such that none of $(x_n),(y_n)$ has cluster point in $X.$ Can we then conclude the set ${x_n:nge1}cup{y_n:nge1}$ has no accumulation point in $X?$
I think that should be correct yet I cannot get sure.
Please help.
general-topology
asked Nov 28 '18 at 2:29
Jave
458114
add a comment |
Let $(x_n),(y_n)$ be two sequences in a metrizable topological space $X$ such that none of $(x_n),(y_n)$ has cluster point in $X.$ Can we then conclude the set ${x_n:nge1}cup{y_n:nge1}$ has no accumulation point in $X?$
I think that should be correct yet I cannot get sure.
Please help.
general-topology
asked Nov 28 '18 at 2:29
Jave
458114
Let $(x_n),(y_n)$ be two sequences in a metrizable topological space $X$ such that none of $(x_n),(y_n)$ has cluster point in $X.$ Can we then conclude the set ${x_n:nge1}cup{y_n:nge1}$ has no accumulation point in $X?$
I think that should be correct yet I cannot get sure.
Please help.
general-topology
general-topology
asked Nov 28 '18 at 2:29
Jave
458114
asked Nov 28 '18 at 2:29
Jave
458114
asked Nov 28 '18 at 2:29
Jave
458114
asked Nov 28 '18 at 2:29
Jave
458114
asked Nov 28 '18 at 2:29
Jave
458114
458114
add a comment |
add a comment |
1 Answer
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Yes. Because if $z_0$ is an accumulation point for
$A={x_n:nge1}cup{y_n:nge1}$ then there is a subsequence of
$A$ like $(z_n)_{nin mathbb{N}} $ such that $z_nto z_0$. But either the set ${ z_n|nin mathbb{N}} cap {x_n:nge 1} $ have infinte elements or ${ z_n|nin mathbb{N} }cap {y_n:nge1}$ have infinite elements. So there is a subsequence of $(z_n)$ like $(z_{n_k})_{kin mathbb{N}} $ such that $(z_{n_k}) subseteq {x_n:nge1}$(or $subseteq {y_n:nge1} $). But $z_{n_k}to z_0$. So $z_0 $ is an accumulation point for ${x_n:nge1}$(or for ${y_n:nge1}$) and this is a contradiction.
answered Nov 28 '18 at 3:36
Darmad
441112
add a comment |
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Yes. Because if $z_0$ is an accumulation point for
$A={x_n:nge1}cup{y_n:nge1}$ then there is a subsequence of
$A$ like $(z_n)_{nin mathbb{N}} $ such that $z_nto z_0$. But either the set ${ z_n|nin mathbb{N}} cap {x_n:nge 1} $ have infinte elements or ${ z_n|nin mathbb{N} }cap {y_n:nge1}$ have infinite elements. So there is a subsequence of $(z_n)$ like $(z_{n_k})_{kin mathbb{N}} $ such that $(z_{n_k}) subseteq {x_n:nge1}$(or $subseteq {y_n:nge1} $). But $z_{n_k}to z_0$. So $z_0 $ is an accumulation point for ${x_n:nge1}$(or for ${y_n:nge1}$) and this is a contradiction.
answered Nov 28 '18 at 3:36
Darmad
441112
add a comment |
Yes. Because if $z_0$ is an accumulation point for
$A={x_n:nge1}cup{y_n:nge1}$ then there is a subsequence of
$A$ like $(z_n)_{nin mathbb{N}} $ such that $z_nto z_0$. But either the set ${ z_n|nin mathbb{N}} cap {x_n:nge 1} $ have infinte elements or ${ z_n|nin mathbb{N} }cap {y_n:nge1}$ have infinite elements. So there is a subsequence of $(z_n)$ like $(z_{n_k})_{kin mathbb{N}} $ such that $(z_{n_k}) subseteq {x_n:nge1}$(or $subseteq {y_n:nge1} $). But $z_{n_k}to z_0$. So $z_0 $ is an accumulation point for ${x_n:nge1}$(or for ${y_n:nge1}$) and this is a contradiction.
answered Nov 28 '18 at 3:36
Darmad
441112
add a comment |
Yes. Because if $z_0$ is an accumulation point for
$A={x_n:nge1}cup{y_n:nge1}$ then there is a subsequence of
$A$ like $(z_n)_{nin mathbb{N}} $ such that $z_nto z_0$. But either the set ${ z_n|nin mathbb{N}} cap {x_n:nge 1} $ have infinte elements or ${ z_n|nin mathbb{N} }cap {y_n:nge1}$ have infinite elements. So there is a subsequence of $(z_n)$ like $(z_{n_k})_{kin mathbb{N}} $ such that $(z_{n_k}) subseteq {x_n:nge1}$(or $subseteq {y_n:nge1} $). But $z_{n_k}to z_0$. So $z_0 $ is an accumulation point for ${x_n:nge1}$(or for ${y_n:nge1}$) and this is a contradiction.
answered Nov 28 '18 at 3:36
Darmad
441112
Yes. Because if $z_0$ is an accumulation point for
$A={x_n:nge1}cup{y_n:nge1}$ then there is a subsequence of
$A$ like $(z_n)_{nin mathbb{N}} $ such that $z_nto z_0$. But either the set ${ z_n|nin mathbb{N}} cap {x_n:nge 1} $ have infinte elements or ${ z_n|nin mathbb{N} }cap {y_n:nge1}$ have infinite elements. So there is a subsequence of $(z_n)$ like $(z_{n_k})_{kin mathbb{N}} $ such that $(z_{n_k}) subseteq {x_n:nge1}$(or $subseteq {y_n:nge1} $). But $z_{n_k}to z_0$. So $z_0 $ is an accumulation point for ${x_n:nge1}$(or for ${y_n:nge1}$) and this is a contradiction.
answered Nov 28 '18 at 3:36
Darmad
441112
answered Nov 28 '18 at 3:36
Darmad
441112
answered Nov 28 '18 at 3:36
Darmad
441112
answered Nov 28 '18 at 3:36
Darmad
441112
441112
add a comment |
add a comment |
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