Mixing
Convergence sequence between discrete and cocountable topology [closed]
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How to prove a convergence sequence into discrete and
co-countable topological space is eventually constant sequence.
My attempt :
We know that any open set in discrete topology is x.
If we consider a sequence Xn then by defination of
convergence in topology for n>N, Xn=x.??
Then how to proceed??
And how to proceed in co-countable topology?
Please help! Thank you.
general-topology
asked Jul 16 at 14:20
Golam biswas
399
closed as off-topic by user21820, John Ma, José Carlos Santos, amWhy, Xander Henderson Aug 1 at 2:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser21820, John Ma, Jos̩ Carlos Santos, amWhy, Xander Henderson
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up vote
0
down vote
favorite
How to prove a convergence sequence into discrete and
co-countable topological space is eventually constant sequence.
My attempt :
We know that any open set in discrete topology is x.
If we consider a sequence Xn then by defination of
convergence in topology for n>N, Xn=x.??
Then how to proceed??
And how to proceed in co-countable topology?
Please help! Thank you.
general-topology
asked Jul 16 at 14:20
Golam biswas
399
closed as off-topic by user21820, John Ma, José Carlos Santos, amWhy, Xander Henderson Aug 1 at 2:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser21820, John Ma, Jos̩ Carlos Santos, amWhy, Xander Henderson
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to prove a convergence sequence into discrete and
co-countable topological space is eventually constant sequence.
My attempt :
We know that any open set in discrete topology is x.
If we consider a sequence Xn then by defination of
convergence in topology for n>N, Xn=x.??
Then how to proceed??
And how to proceed in co-countable topology?
Please help! Thank you.
general-topology
asked Jul 16 at 14:20
Golam biswas
399
How to prove a convergence sequence into discrete and
co-countable topological space is eventually constant sequence.
My attempt :
We know that any open set in discrete topology is x.
If we consider a sequence Xn then by defination of
convergence in topology for n>N, Xn=x.??
Then how to proceed??
And how to proceed in co-countable topology?
Please help! Thank you.
general-topology
asked Jul 16 at 14:20
Golam biswas
399
asked Jul 16 at 14:20
Golam biswas
399
asked Jul 16 at 14:20
Golam biswas
399
asked Jul 16 at 14:20
Golam biswas
399
399
closed as off-topic by user21820, John Ma, José Carlos Santos, amWhy, Xander Henderson Aug 1 at 2:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser21820, John Ma, Jos̩ Carlos Santos, amWhy, Xander Henderson
closed as off-topic by user21820, John Ma, José Carlos Santos, amWhy, Xander Henderson Aug 1 at 2:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser21820, John Ma, Jos̩ Carlos Santos, amWhy, Xander Henderson
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1 Answer
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Let $(x_n)_nin mathbbN subseteq E$ be a convergent sequence and let $lin E$ such that $x_n rightarrow l$. Define $V:=Esetminus x_n : x_n neq l$. Clearly $V$ has countable complement and is therefore open in our topology. As $x_n rightarrow l$ there exists $Nin mathbbN$ such that for all $ngeq N$ holds $x_n in V$. By definition of $V$ this means $x_n = l$ for all $ngeq N$, i.e. $(x_n)_nin mathbbN$ is eventually constant.
Let $(x_n)_nin mathbbN subseteq E$ be eventually constant. Then there exists $lin E$ and $N in mathbbN$ such that for all $ngeq N$ holds $x_n=l$. Let $Vsubseteq E$ be an open neighborhood of $l$. By definition of neighborhood $lin V$. Hence, for all $ngeq N$ holds $x_nin V$. As $V$ was an arbitrary neighborhood of $l$ we conclude $x_n rightarrow l$. Note that we didn't use any information about the topology, i.e. this holds in every topological space.
answered Jul 16 at 14:37
Simone Masiero
326113
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $(x_n)_nin mathbbN subseteq E$ be a convergent sequence and let $lin E$ such that $x_n rightarrow l$. Define $V:=Esetminus x_n : x_n neq l$. Clearly $V$ has countable complement and is therefore open in our topology. As $x_n rightarrow l$ there exists $Nin mathbbN$ such that for all $ngeq N$ holds $x_n in V$. By definition of $V$ this means $x_n = l$ for all $ngeq N$, i.e. $(x_n)_nin mathbbN$ is eventually constant.
Let $(x_n)_nin mathbbN subseteq E$ be eventually constant. Then there exists $lin E$ and $N in mathbbN$ such that for all $ngeq N$ holds $x_n=l$. Let $Vsubseteq E$ be an open neighborhood of $l$. By definition of neighborhood $lin V$. Hence, for all $ngeq N$ holds $x_nin V$. As $V$ was an arbitrary neighborhood of $l$ we conclude $x_n rightarrow l$. Note that we didn't use any information about the topology, i.e. this holds in every topological space.
answered Jul 16 at 14:37
Simone Masiero
326113
add a comment |Â
up vote
2
down vote
accepted
Let $(x_n)_nin mathbbN subseteq E$ be a convergent sequence and let $lin E$ such that $x_n rightarrow l$. Define $V:=Esetminus x_n : x_n neq l$. Clearly $V$ has countable complement and is therefore open in our topology. As $x_n rightarrow l$ there exists $Nin mathbbN$ such that for all $ngeq N$ holds $x_n in V$. By definition of $V$ this means $x_n = l$ for all $ngeq N$, i.e. $(x_n)_nin mathbbN$ is eventually constant.
Let $(x_n)_nin mathbbN subseteq E$ be eventually constant. Then there exists $lin E$ and $N in mathbbN$ such that for all $ngeq N$ holds $x_n=l$. Let $Vsubseteq E$ be an open neighborhood of $l$. By definition of neighborhood $lin V$. Hence, for all $ngeq N$ holds $x_nin V$. As $V$ was an arbitrary neighborhood of $l$ we conclude $x_n rightarrow l$. Note that we didn't use any information about the topology, i.e. this holds in every topological space.
answered Jul 16 at 14:37
Simone Masiero
326113
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $(x_n)_nin mathbbN subseteq E$ be a convergent sequence and let $lin E$ such that $x_n rightarrow l$. Define $V:=Esetminus x_n : x_n neq l$. Clearly $V$ has countable complement and is therefore open in our topology. As $x_n rightarrow l$ there exists $Nin mathbbN$ such that for all $ngeq N$ holds $x_n in V$. By definition of $V$ this means $x_n = l$ for all $ngeq N$, i.e. $(x_n)_nin mathbbN$ is eventually constant.
Let $(x_n)_nin mathbbN subseteq E$ be eventually constant. Then there exists $lin E$ and $N in mathbbN$ such that for all $ngeq N$ holds $x_n=l$. Let $Vsubseteq E$ be an open neighborhood of $l$. By definition of neighborhood $lin V$. Hence, for all $ngeq N$ holds $x_nin V$. As $V$ was an arbitrary neighborhood of $l$ we conclude $x_n rightarrow l$. Note that we didn't use any information about the topology, i.e. this holds in every topological space.
answered Jul 16 at 14:37
Simone Masiero
326113
Let $(x_n)_nin mathbbN subseteq E$ be a convergent sequence and let $lin E$ such that $x_n rightarrow l$. Define $V:=Esetminus x_n : x_n neq l$. Clearly $V$ has countable complement and is therefore open in our topology. As $x_n rightarrow l$ there exists $Nin mathbbN$ such that for all $ngeq N$ holds $x_n in V$. By definition of $V$ this means $x_n = l$ for all $ngeq N$, i.e. $(x_n)_nin mathbbN$ is eventually constant.
Let $(x_n)_nin mathbbN subseteq E$ be eventually constant. Then there exists $lin E$ and $N in mathbbN$ such that for all $ngeq N$ holds $x_n=l$. Let $Vsubseteq E$ be an open neighborhood of $l$. By definition of neighborhood $lin V$. Hence, for all $ngeq N$ holds $x_nin V$. As $V$ was an arbitrary neighborhood of $l$ we conclude $x_n rightarrow l$. Note that we didn't use any information about the topology, i.e. this holds in every topological space.
answered Jul 16 at 14:37
Simone Masiero
326113
answered Jul 16 at 14:37
Simone Masiero
326113
answered Jul 16 at 14:37
Simone Masiero
326113
answered Jul 16 at 14:37
Simone Masiero
326113
326113
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