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Is there a metric space such that for a sequence of nonempty subsets $K_1 supset K_2 …$ the intersection of all subsets is empty?
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Is there a metric space $(X,d)$ such that for a sequence of $emptyset neq K_j subset X$ with $K_1 supset K_2 ....$ such that the intersection of all $bigcap _jin mathbbNK_j$ is empty?
It seems very obvious that since all $K_j$ are not empty, there is an element in $K_1$ that is also in all $K_i$ with $i geq 1$ and that this element this is an element of the intersection, which thus is nonempty.
However in the task it was specified that the metric space is compact and the subsets $K_j$ are closed. I have used neither in the argument above and cannot figure out a counterexample where it would be necessary to have the conditions to the sets.
elementary-set-theory metric-spaces compactness
asked Jul 20 at 22:17
B.Swan
9701619
 |Â
show 1 more comment
up vote
2
down vote
favorite
Is there a metric space $(X,d)$ such that for a sequence of $emptyset neq K_j subset X$ with $K_1 supset K_2 ....$ such that the intersection of all $bigcap _jin mathbbNK_j$ is empty?
It seems very obvious that since all $K_j$ are not empty, there is an element in $K_1$ that is also in all $K_i$ with $i geq 1$ and that this element this is an element of the intersection, which thus is nonempty.
However in the task it was specified that the metric space is compact and the subsets $K_j$ are closed. I have used neither in the argument above and cannot figure out a counterexample where it would be necessary to have the conditions to the sets.
elementary-set-theory metric-spaces compactness
asked Jul 20 at 22:17
B.Swan
9701619
I think you may be getting the inclusions backwards. An element of $K_1$ does not have to be in all $K_i$ with $igeq 1$, since $K_1$ is a superset, not a subset of them.
– Eric Wofsey
Jul 20 at 22:35
@EricWofsey I did not claim that any element of $K_1$ is an element of any further subset. However if for example $K_1 supset K_2 neq emptyset$, then there exists an element in $K_1$ that is also an element in $K_2$
– B.Swan
Jul 20 at 22:39
That gets you an element of both $K_1$ and $K_2$. But what about an element of every single $K_i$ at once?
– Eric Wofsey
Jul 20 at 22:40
@EricWofsey You can use that argument iteratively for any pair $K_n supset K_n+1$ and since $K_1 supset K_n$ you get a point that is in all $K_i$
– B.Swan
Jul 20 at 22:43
I would suggest thinking about what happens when you try that argument on the examples given in the answers.
– Eric Wofsey
Jul 20 at 22:45
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is there a metric space $(X,d)$ such that for a sequence of $emptyset neq K_j subset X$ with $K_1 supset K_2 ....$ such that the intersection of all $bigcap _jin mathbbNK_j$ is empty?
It seems very obvious that since all $K_j$ are not empty, there is an element in $K_1$ that is also in all $K_i$ with $i geq 1$ and that this element this is an element of the intersection, which thus is nonempty.
However in the task it was specified that the metric space is compact and the subsets $K_j$ are closed. I have used neither in the argument above and cannot figure out a counterexample where it would be necessary to have the conditions to the sets.
elementary-set-theory metric-spaces compactness
asked Jul 20 at 22:17
B.Swan
9701619
Is there a metric space $(X,d)$ such that for a sequence of $emptyset neq K_j subset X$ with $K_1 supset K_2 ....$ such that the intersection of all $bigcap _jin mathbbNK_j$ is empty?
It seems very obvious that since all $K_j$ are not empty, there is an element in $K_1$ that is also in all $K_i$ with $i geq 1$ and that this element this is an element of the intersection, which thus is nonempty.
However in the task it was specified that the metric space is compact and the subsets $K_j$ are closed. I have used neither in the argument above and cannot figure out a counterexample where it would be necessary to have the conditions to the sets.
elementary-set-theory metric-spaces compactness
asked Jul 20 at 22:17
B.Swan
9701619
asked Jul 20 at 22:17
B.Swan
9701619
asked Jul 20 at 22:17
B.Swan
9701619
asked Jul 20 at 22:17
B.Swan
9701619
9701619
I think you may be getting the inclusions backwards. An element of $K_1$ does not have to be in all $K_i$ with $igeq 1$, since $K_1$ is a superset, not a subset of them.
– Eric Wofsey
Jul 20 at 22:35
@EricWofsey I did not claim that any element of $K_1$ is an element of any further subset. However if for example $K_1 supset K_2 neq emptyset$, then there exists an element in $K_1$ that is also an element in $K_2$
– B.Swan
Jul 20 at 22:39
That gets you an element of both $K_1$ and $K_2$. But what about an element of every single $K_i$ at once?
– Eric Wofsey
Jul 20 at 22:40
@EricWofsey You can use that argument iteratively for any pair $K_n supset K_n+1$ and since $K_1 supset K_n$ you get a point that is in all $K_i$
– B.Swan
Jul 20 at 22:43
I would suggest thinking about what happens when you try that argument on the examples given in the answers.
– Eric Wofsey
Jul 20 at 22:45
 |Â
show 1 more comment
I think you may be getting the inclusions backwards. An element of $K_1$ does not have to be in all $K_i$ with $igeq 1$, since $K_1$ is a superset, not a subset of them.
– Eric Wofsey
Jul 20 at 22:35
@EricWofsey I did not claim that any element of $K_1$ is an element of any further subset. However if for example $K_1 supset K_2 neq emptyset$, then there exists an element in $K_1$ that is also an element in $K_2$
– B.Swan
Jul 20 at 22:39
That gets you an element of both $K_1$ and $K_2$. But what about an element of every single $K_i$ at once?
– Eric Wofsey
Jul 20 at 22:40
@EricWofsey You can use that argument iteratively for any pair $K_n supset K_n+1$ and since $K_1 supset K_n$ you get a point that is in all $K_i$
– B.Swan
Jul 20 at 22:43
I would suggest thinking about what happens when you try that argument on the examples given in the answers.
– Eric Wofsey
Jul 20 at 22:45
I think you may be getting the inclusions backwards. An element of $K_1$ does not have to be in all $K_i$ with $igeq 1$, since $K_1$ is a superset, not a subset of them.
– Eric Wofsey
Jul 20 at 22:35
I think you may be getting the inclusions backwards. An element of $K_1$ does not have to be in all $K_i$ with $igeq 1$, since $K_1$ is a superset, not a subset of them.
– Eric Wofsey
Jul 20 at 22:35
@EricWofsey I did not claim that any element of $K_1$ is an element of any further subset. However if for example $K_1 supset K_2 neq emptyset$, then there exists an element in $K_1$ that is also an element in $K_2$
– B.Swan
Jul 20 at 22:39
@EricWofsey I did not claim that any element of $K_1$ is an element of any further subset. However if for example $K_1 supset K_2 neq emptyset$, then there exists an element in $K_1$ that is also an element in $K_2$
– B.Swan
Jul 20 at 22:39
That gets you an element of both $K_1$ and $K_2$. But what about an element of every single $K_i$ at once?
– Eric Wofsey
Jul 20 at 22:40
That gets you an element of both $K_1$ and $K_2$. But what about an element of every single $K_i$ at once?
– Eric Wofsey
Jul 20 at 22:40
@EricWofsey You can use that argument iteratively for any pair $K_n supset K_n+1$ and since $K_1 supset K_n$ you get a point that is in all $K_i$
– B.Swan
Jul 20 at 22:43
@EricWofsey You can use that argument iteratively for any pair $K_n supset K_n+1$ and since $K_1 supset K_n$ you get a point that is in all $K_i$
– B.Swan
Jul 20 at 22:43
I would suggest thinking about what happens when you try that argument on the examples given in the answers.
– Eric Wofsey
Jul 20 at 22:45
I would suggest thinking about what happens when you try that argument on the examples given in the answers.
– Eric Wofsey
Jul 20 at 22:45
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Both conditions are necessary for the theorem to hold. Here are two examples:
Sets are not closed:
Consider the metric space $[0,1]$ with the usual metric, and the sets $K_n = (0,frac1n]$. This space is compact, and the sets $K_n$ are nonempty and $K_n supset K_n+1$. However, the intersection of these sets is empty: remark that the sets $K_n$ are not closed.
Space is not compact:
Consider the metric space $(0,1]$ with the usual metric, and the sets $K_n = (0,frac1n]$. These sets are nonempty, we have $K_n supset K_n+1$, and importantly: the sets are closed in $(0,1]$ (you may have to convince yourself of this). However, again, the intersection of these sets is empty: remark that the space is not compact.
answered Jul 20 at 22:20
Sambo
1,2561427
add a comment |Â
up vote
5
down vote
Sure, the intersection can be empty: let $X =mathbbR$ in the standard metric
and $K_n = [n, infty)$ which are all closed. And if $x in mathbbR$, let $N$ be some integer above $x$, then $x notin K_N$ so $x notin bigcap_n K_n$, so that $bigcap_n K_n =emptyset$.
The nice thing is that this cannot happen if $X$ or some $K_n$ is compact.
answered Jul 20 at 22:21
Henno Brandsma
91.5k342100
Henno.Very nice answer, thanks.
– Peter Szilas
Jul 21 at 8:18
add a comment |Â
up vote
3
down vote
Yes. Take the reals (with the usual topology) and $K_n=[n,+infty)$.
answered Jul 20 at 22:22
José Carlos Santos
114k1698177
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Both conditions are necessary for the theorem to hold. Here are two examples:
Sets are not closed:
Consider the metric space $[0,1]$ with the usual metric, and the sets $K_n = (0,frac1n]$. This space is compact, and the sets $K_n$ are nonempty and $K_n supset K_n+1$. However, the intersection of these sets is empty: remark that the sets $K_n$ are not closed.
Space is not compact:
Consider the metric space $(0,1]$ with the usual metric, and the sets $K_n = (0,frac1n]$. These sets are nonempty, we have $K_n supset K_n+1$, and importantly: the sets are closed in $(0,1]$ (you may have to convince yourself of this). However, again, the intersection of these sets is empty: remark that the space is not compact.
answered Jul 20 at 22:20
Sambo
1,2561427
add a comment |Â
up vote
4
down vote
accepted
Both conditions are necessary for the theorem to hold. Here are two examples:
Sets are not closed:
Consider the metric space $[0,1]$ with the usual metric, and the sets $K_n = (0,frac1n]$. This space is compact, and the sets $K_n$ are nonempty and $K_n supset K_n+1$. However, the intersection of these sets is empty: remark that the sets $K_n$ are not closed.
Space is not compact:
Consider the metric space $(0,1]$ with the usual metric, and the sets $K_n = (0,frac1n]$. These sets are nonempty, we have $K_n supset K_n+1$, and importantly: the sets are closed in $(0,1]$ (you may have to convince yourself of this). However, again, the intersection of these sets is empty: remark that the space is not compact.
answered Jul 20 at 22:20
Sambo
1,2561427
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Both conditions are necessary for the theorem to hold. Here are two examples:
Sets are not closed:
Consider the metric space $[0,1]$ with the usual metric, and the sets $K_n = (0,frac1n]$. This space is compact, and the sets $K_n$ are nonempty and $K_n supset K_n+1$. However, the intersection of these sets is empty: remark that the sets $K_n$ are not closed.
Space is not compact:
Consider the metric space $(0,1]$ with the usual metric, and the sets $K_n = (0,frac1n]$. These sets are nonempty, we have $K_n supset K_n+1$, and importantly: the sets are closed in $(0,1]$ (you may have to convince yourself of this). However, again, the intersection of these sets is empty: remark that the space is not compact.
answered Jul 20 at 22:20
Sambo
1,2561427
Both conditions are necessary for the theorem to hold. Here are two examples:
Sets are not closed:
Consider the metric space $[0,1]$ with the usual metric, and the sets $K_n = (0,frac1n]$. This space is compact, and the sets $K_n$ are nonempty and $K_n supset K_n+1$. However, the intersection of these sets is empty: remark that the sets $K_n$ are not closed.
Space is not compact:
Consider the metric space $(0,1]$ with the usual metric, and the sets $K_n = (0,frac1n]$. These sets are nonempty, we have $K_n supset K_n+1$, and importantly: the sets are closed in $(0,1]$ (you may have to convince yourself of this). However, again, the intersection of these sets is empty: remark that the space is not compact.
answered Jul 20 at 22:20
Sambo
1,2561427
edited Jul 20 at 23:55
answered Jul 20 at 22:20
Sambo
1,2561427
answered Jul 20 at 22:20
Sambo
1,2561427
answered Jul 20 at 22:20
Sambo
1,2561427
1,2561427
add a comment |Â
add a comment |Â
up vote
5
down vote
Sure, the intersection can be empty: let $X =mathbbR$ in the standard metric
and $K_n = [n, infty)$ which are all closed. And if $x in mathbbR$, let $N$ be some integer above $x$, then $x notin K_N$ so $x notin bigcap_n K_n$, so that $bigcap_n K_n =emptyset$.
The nice thing is that this cannot happen if $X$ or some $K_n$ is compact.
answered Jul 20 at 22:21
Henno Brandsma
91.5k342100
Henno.Very nice answer, thanks.
– Peter Szilas
Jul 21 at 8:18
add a comment |Â
up vote
5
down vote
Sure, the intersection can be empty: let $X =mathbbR$ in the standard metric
and $K_n = [n, infty)$ which are all closed. And if $x in mathbbR$, let $N$ be some integer above $x$, then $x notin K_N$ so $x notin bigcap_n K_n$, so that $bigcap_n K_n =emptyset$.
The nice thing is that this cannot happen if $X$ or some $K_n$ is compact.
answered Jul 20 at 22:21
Henno Brandsma
91.5k342100
Henno.Very nice answer, thanks.
– Peter Szilas
Jul 21 at 8:18
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Sure, the intersection can be empty: let $X =mathbbR$ in the standard metric
and $K_n = [n, infty)$ which are all closed. And if $x in mathbbR$, let $N$ be some integer above $x$, then $x notin K_N$ so $x notin bigcap_n K_n$, so that $bigcap_n K_n =emptyset$.
The nice thing is that this cannot happen if $X$ or some $K_n$ is compact.
answered Jul 20 at 22:21
Henno Brandsma
91.5k342100
Sure, the intersection can be empty: let $X =mathbbR$ in the standard metric
and $K_n = [n, infty)$ which are all closed. And if $x in mathbbR$, let $N$ be some integer above $x$, then $x notin K_N$ so $x notin bigcap_n K_n$, so that $bigcap_n K_n =emptyset$.
The nice thing is that this cannot happen if $X$ or some $K_n$ is compact.
answered Jul 20 at 22:21
Henno Brandsma
91.5k342100
answered Jul 20 at 22:21
Henno Brandsma
91.5k342100
answered Jul 20 at 22:21
Henno Brandsma
91.5k342100
answered Jul 20 at 22:21
Henno Brandsma
91.5k342100
91.5k342100
Henno.Very nice answer, thanks.
– Peter Szilas
Jul 21 at 8:18
add a comment |Â
Henno.Very nice answer, thanks.
– Peter Szilas
Jul 21 at 8:18
Henno.Very nice answer, thanks.
– Peter Szilas
Jul 21 at 8:18
Henno.Very nice answer, thanks.
– Peter Szilas
Jul 21 at 8:18
add a comment |Â
up vote
3
down vote
Yes. Take the reals (with the usual topology) and $K_n=[n,+infty)$.
answered Jul 20 at 22:22
José Carlos Santos
114k1698177
add a comment |Â
up vote
3
down vote
Yes. Take the reals (with the usual topology) and $K_n=[n,+infty)$.
answered Jul 20 at 22:22
José Carlos Santos
114k1698177
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Yes. Take the reals (with the usual topology) and $K_n=[n,+infty)$.
answered Jul 20 at 22:22
José Carlos Santos
114k1698177
Yes. Take the reals (with the usual topology) and $K_n=[n,+infty)$.
answered Jul 20 at 22:22
José Carlos Santos
114k1698177
answered Jul 20 at 22:22
José Carlos Santos
114k1698177
answered Jul 20 at 22:22
José Carlos Santos
114k1698177
answered Jul 20 at 22:22
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
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I think you may be getting the inclusions backwards. An element of $K_1$ does not have to be in all $K_i$ with $igeq 1$, since $K_1$ is a superset, not a subset of them.
– Eric Wofsey
Jul 20 at 22:35
@EricWofsey I did not claim that any element of $K_1$ is an element of any further subset. However if for example $K_1 supset K_2 neq emptyset$, then there exists an element in $K_1$ that is also an element in $K_2$
– B.Swan
Jul 20 at 22:39
That gets you an element of both $K_1$ and $K_2$. But what about an element of every single $K_i$ at once?
– Eric Wofsey
Jul 20 at 22:40
@EricWofsey You can use that argument iteratively for any pair $K_n supset K_n+1$ and since $K_1 supset K_n$ you get a point that is in all $K_i$
– B.Swan
Jul 20 at 22:43
I would suggest thinking about what happens when you try that argument on the examples given in the answers.
– Eric Wofsey
Jul 20 at 22:45