Mixing
Prove the metric space is Complete.
$begingroup$
I am trying to solve the following question:
Question: Let $(X,d)$ be a metric space such that $d(A,B)>0$ for every pair of disjoint closed subsets $A$ and $B$. Show that $(X,d)$ is complete.
My Attempt :
Let $(x_n)_n$ be a Cauchy sequence with distinct terms in $X$ which is not convergent in $X$. Now I am trying to arrive a contradiction…and then our result will be done…!!!
To do this I consider the sets
$A={x_{2k}| k in mathbb{N} }$ and $B={ x_{2k+1}|k in mathbb{N} }$
then Cauchyness of $(x_n)_n$ imply $d(A, B)=0$ and also $A$ and $B$ are disjoint.
Now If A and B are closed sets then we are done…!!!!
But I cannot prove these sets to be closed in $X$.
I know the range set of a sequence is not closed, in general, in a metric space.
Am I on the right track to prove the theorem? If $A$ and $B$ are closed, how can I prove it here…? If this procedure is not in the right way, suggest me any other solution.
And sorry if my question is so classic....!!
Thank you.!!
metric-spaces complete-spaces
asked May 26 '17 at 9:10
Indrajit GhoshIndrajit Ghosh
1,0971718
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following question:
Question: Let $(X,d)$ be a metric space such that $d(A,B)>0$ for every pair of disjoint closed subsets $A$ and $B$. Show that $(X,d)$ is complete.
My Attempt :
Let $(x_n)_n$ be a Cauchy sequence with distinct terms in $X$ which is not convergent in $X$. Now I am trying to arrive a contradiction…and then our result will be done…!!!
To do this I consider the sets
$A={x_{2k}| k in mathbb{N} }$ and $B={ x_{2k+1}|k in mathbb{N} }$
then Cauchyness of $(x_n)_n$ imply $d(A, B)=0$ and also $A$ and $B$ are disjoint.
Now If A and B are closed sets then we are done…!!!!
But I cannot prove these sets to be closed in $X$.
I know the range set of a sequence is not closed, in general, in a metric space.
Am I on the right track to prove the theorem? If $A$ and $B$ are closed, how can I prove it here…? If this procedure is not in the right way, suggest me any other solution.
And sorry if my question is so classic....!!
Thank you.!!
metric-spaces complete-spaces
asked May 26 '17 at 9:10
Indrajit GhoshIndrajit Ghosh
1,0971718
$endgroup$
1
$begingroup$
Yes, you are on the right track. Use the fact that a Cauchy sequence converges if and only if it has a convergent subsequence. And, yes, in this case the sets $A$ and $B$ are closed.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:18
1
$begingroup$
@ Jose Carlos Santos......Oho yes....So this sequence has no convergent subsequence. this means A has no limit points outsides A and similar for B....Am I right sir..?
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:23
$begingroup$
@Indrajit Ghosh Indeed you are.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:25
$begingroup$
@ J C Santos...thank you so much sir..
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:26
add a comment |
$begingroup$
I am trying to solve the following question:
Question: Let $(X,d)$ be a metric space such that $d(A,B)>0$ for every pair of disjoint closed subsets $A$ and $B$. Show that $(X,d)$ is complete.
My Attempt :
Let $(x_n)_n$ be a Cauchy sequence with distinct terms in $X$ which is not convergent in $X$. Now I am trying to arrive a contradiction…and then our result will be done…!!!
To do this I consider the sets
$A={x_{2k}| k in mathbb{N} }$ and $B={ x_{2k+1}|k in mathbb{N} }$
then Cauchyness of $(x_n)_n$ imply $d(A, B)=0$ and also $A$ and $B$ are disjoint.
Now If A and B are closed sets then we are done…!!!!
But I cannot prove these sets to be closed in $X$.
I know the range set of a sequence is not closed, in general, in a metric space.
Am I on the right track to prove the theorem? If $A$ and $B$ are closed, how can I prove it here…? If this procedure is not in the right way, suggest me any other solution.
And sorry if my question is so classic....!!
Thank you.!!
metric-spaces complete-spaces
asked May 26 '17 at 9:10
Indrajit GhoshIndrajit Ghosh
1,0971718
$endgroup$
I am trying to solve the following question:
Question: Let $(X,d)$ be a metric space such that $d(A,B)>0$ for every pair of disjoint closed subsets $A$ and $B$. Show that $(X,d)$ is complete.
My Attempt :
Let $(x_n)_n$ be a Cauchy sequence with distinct terms in $X$ which is not convergent in $X$. Now I am trying to arrive a contradiction…and then our result will be done…!!!
To do this I consider the sets
$A={x_{2k}| k in mathbb{N} }$ and $B={ x_{2k+1}|k in mathbb{N} }$
then Cauchyness of $(x_n)_n$ imply $d(A, B)=0$ and also $A$ and $B$ are disjoint.
Now If A and B are closed sets then we are done…!!!!
But I cannot prove these sets to be closed in $X$.
I know the range set of a sequence is not closed, in general, in a metric space.
Am I on the right track to prove the theorem? If $A$ and $B$ are closed, how can I prove it here…? If this procedure is not in the right way, suggest me any other solution.
And sorry if my question is so classic....!!
Thank you.!!
metric-spaces complete-spaces
metric-spaces complete-spaces
asked May 26 '17 at 9:10
Indrajit GhoshIndrajit Ghosh
1,0971718
asked May 26 '17 at 9:10
Indrajit GhoshIndrajit Ghosh
1,0971718
asked May 26 '17 at 9:10
Indrajit GhoshIndrajit Ghosh
1,0971718
asked May 26 '17 at 9:10
Indrajit GhoshIndrajit Ghosh
1,0971718
asked May 26 '17 at 9:10
Indrajit GhoshIndrajit Ghosh
1,0971718
1,0971718
1
$begingroup$
Yes, you are on the right track. Use the fact that a Cauchy sequence converges if and only if it has a convergent subsequence. And, yes, in this case the sets $A$ and $B$ are closed.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:18
1
$begingroup$
@ Jose Carlos Santos......Oho yes....So this sequence has no convergent subsequence. this means A has no limit points outsides A and similar for B....Am I right sir..?
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:23
$begingroup$
@Indrajit Ghosh Indeed you are.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:25
$begingroup$
@ J C Santos...thank you so much sir..
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:26
add a comment |
1
$begingroup$
Yes, you are on the right track. Use the fact that a Cauchy sequence converges if and only if it has a convergent subsequence. And, yes, in this case the sets $A$ and $B$ are closed.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:18
1
$begingroup$
@ Jose Carlos Santos......Oho yes....So this sequence has no convergent subsequence. this means A has no limit points outsides A and similar for B....Am I right sir..?
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:23
$begingroup$
@Indrajit Ghosh Indeed you are.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:25
$begingroup$
@ J C Santos...thank you so much sir..
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:26
1
1
$begingroup$
Yes, you are on the right track. Use the fact that a Cauchy sequence converges if and only if it has a convergent subsequence. And, yes, in this case the sets $A$ and $B$ are closed.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:18
$begingroup$
Yes, you are on the right track. Use the fact that a Cauchy sequence converges if and only if it has a convergent subsequence. And, yes, in this case the sets $A$ and $B$ are closed.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:18
1
1
$begingroup$
@ Jose Carlos Santos......Oho yes....So this sequence has no convergent subsequence. this means A has no limit points outsides A and similar for B....Am I right sir..?
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:23
$begingroup$
@ Jose Carlos Santos......Oho yes....So this sequence has no convergent subsequence. this means A has no limit points outsides A and similar for B....Am I right sir..?
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:23
$begingroup$
@Indrajit Ghosh Indeed you are.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:25
$begingroup$
@Indrajit Ghosh Indeed you are.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:25
$begingroup$
@ J C Santos...thank you so much sir..
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:26
$begingroup$
@ J C Santos...thank you so much sir..
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:26
add a comment |
1 Answer
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$begingroup$
Cauchy sequences are convergent iff it has a convergent subsequence.
Now the two sets you defined, has no limit points, for if $A$ has limit point, say $y$, there is a subsequence of $A$ that converges to $y$, which is also a convergent subsequence of $(x_n)$, but since $(x_n)$ is a Cauchy sequence which is not convergent in $X$, then no subsequence of it can converge, hence a contradiction.
So $A$ and $B$ have no limit points, so the closure of $A$ is $A$ and closure of $B$ is $B$, and closures are closed sets, and hence we got two disjoint closed sets whose distance is $0$.
answered May 26 '17 at 9:24
Arpan1729Arpan1729
2,7971320
$endgroup$
add a comment |
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$begingroup$
Cauchy sequences are convergent iff it has a convergent subsequence.
Now the two sets you defined, has no limit points, for if $A$ has limit point, say $y$, there is a subsequence of $A$ that converges to $y$, which is also a convergent subsequence of $(x_n)$, but since $(x_n)$ is a Cauchy sequence which is not convergent in $X$, then no subsequence of it can converge, hence a contradiction.
So $A$ and $B$ have no limit points, so the closure of $A$ is $A$ and closure of $B$ is $B$, and closures are closed sets, and hence we got two disjoint closed sets whose distance is $0$.
answered May 26 '17 at 9:24
Arpan1729Arpan1729
2,7971320
$endgroup$
add a comment |
$begingroup$
Cauchy sequences are convergent iff it has a convergent subsequence.
Now the two sets you defined, has no limit points, for if $A$ has limit point, say $y$, there is a subsequence of $A$ that converges to $y$, which is also a convergent subsequence of $(x_n)$, but since $(x_n)$ is a Cauchy sequence which is not convergent in $X$, then no subsequence of it can converge, hence a contradiction.
So $A$ and $B$ have no limit points, so the closure of $A$ is $A$ and closure of $B$ is $B$, and closures are closed sets, and hence we got two disjoint closed sets whose distance is $0$.
answered May 26 '17 at 9:24
Arpan1729Arpan1729
2,7971320
$endgroup$
add a comment |
$begingroup$
Cauchy sequences are convergent iff it has a convergent subsequence.
Now the two sets you defined, has no limit points, for if $A$ has limit point, say $y$, there is a subsequence of $A$ that converges to $y$, which is also a convergent subsequence of $(x_n)$, but since $(x_n)$ is a Cauchy sequence which is not convergent in $X$, then no subsequence of it can converge, hence a contradiction.
So $A$ and $B$ have no limit points, so the closure of $A$ is $A$ and closure of $B$ is $B$, and closures are closed sets, and hence we got two disjoint closed sets whose distance is $0$.
answered May 26 '17 at 9:24
Arpan1729Arpan1729
2,7971320
$endgroup$
Cauchy sequences are convergent iff it has a convergent subsequence.
Now the two sets you defined, has no limit points, for if $A$ has limit point, say $y$, there is a subsequence of $A$ that converges to $y$, which is also a convergent subsequence of $(x_n)$, but since $(x_n)$ is a Cauchy sequence which is not convergent in $X$, then no subsequence of it can converge, hence a contradiction.
So $A$ and $B$ have no limit points, so the closure of $A$ is $A$ and closure of $B$ is $B$, and closures are closed sets, and hence we got two disjoint closed sets whose distance is $0$.
answered May 26 '17 at 9:24
Arpan1729Arpan1729
2,7971320
edited May 26 '17 at 9:28
answered May 26 '17 at 9:24
Arpan1729Arpan1729
2,7971320
answered May 26 '17 at 9:24
Arpan1729Arpan1729
2,7971320
answered May 26 '17 at 9:24
Arpan1729Arpan1729
2,7971320
2,7971320
add a comment |
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$begingroup$
Yes, you are on the right track. Use the fact that a Cauchy sequence converges if and only if it has a convergent subsequence. And, yes, in this case the sets $A$ and $B$ are closed.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:18
1
$begingroup$
@ Jose Carlos Santos......Oho yes....So this sequence has no convergent subsequence. this means A has no limit points outsides A and similar for B....Am I right sir..?
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:23
$begingroup$
@Indrajit Ghosh Indeed you are.
$endgroup$
– José Carlos Santos
May 26 '17 at 9:25
$begingroup$
@ J C Santos...thank you so much sir..
$endgroup$
– Indrajit Ghosh
May 26 '17 at 9:26