Mixing
An exercise on the quotient mapping $f:mathbb{R}tomathbb{Z}$
$begingroup$
Let $f:mathbb{R}tomathbb{Z}$ be a function such that each $xinmathbb{R}$ it associates the greatest integer less or equal to $x$. Suppose that $mathbb{R}$ is endowed with the eucledian topology such that $tau$ is the quotient topology in $mathbb{Z}$ defined by $f$.
a) Compute $f^{-1}(n)$ for $ninmathbb{Z}$
b)Prove $tau={emptyset,mathbb{Z}}cup{]-infty,n]capmathbb{Z}:ninmathbb{Z}}$
c) Determine if $(mathbb{Z},tau)$ is metrizable.
d)Determine if $(mathbb{Z},tau)$ is compact.
a)
If $f^{-1}(n)=[n,n+1)$ once $leqslant xleqslant n+1$
b)
If we have the infinite set (...,-n-1,...-3,-2,-1,0,1,2,3...n-1...)
Let $(mathbb{R},tau')$ be the topological space where $tau'$ is the Euclidean topology
$f^{-1}(mathbb{Z})=mathbb{R}in tau'impliesmathbb{Z}intau $
$f^{-1}(emptyset)=emptysetin tau'impliesemptysetintau $
If I pick up a set in $mathbb{Z}$ like }$(...,-n,...-3,-2,-1,0,1,2,3...n)$ then I prove it is open.
$f^{-1}((...,-n,...-3,-2,-1,0,1,2,3...n))=...cup [-n-1,-n)cup[-3,-2)cup[-2,-1)cup[-1,0)cup[1,2)cup[3,4)...cup[n-1,n)cup=(-infty,n)in tau'implies (infty,n)capmathbb{Z}intau$
Note the compliment of $mathbb{Z}setminus((infty,n)capmathbb{Z})=[n,infty)capmathbb{Z}$ which is closed hence I do need to try other sets.
This proves $tau={emptyset,mathbb{Z}}cup{]-infty,n]capmathbb{Z}:ninmathbb{Z}}$ is the topology $tau$.
c)
Proposition: Every metric space is Hausdorff.
Let $x,yinmathbb{Z}$ there is an open set $xin U$ and $yin V$ such that $U=(-infty,j]capmathbb{Z}$ and $V=(-infty,j']capmathbb{Z}$. It is trivial to notice $Ucap Vneq emptyset$. Then as $j$ and $j'$ are arbitrary elements of $mathbb{Z}$ it can be concluded that $mathbb{Z},tau$ is not Hausdorff hence it is not metrizable.
d)
Proposition: Every compact space is limited.
Since $mathbb{Z},tau$ is not limited hence it is not compact.
Question:
Is this answer right?
general-topology proof-verification
asked Jan 26 at 11:23
Pedro GomesPedro Gomes
1,9442721
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}tomathbb{Z}$ be a function such that each $xinmathbb{R}$ it associates the greatest integer less or equal to $x$. Suppose that $mathbb{R}$ is endowed with the eucledian topology such that $tau$ is the quotient topology in $mathbb{Z}$ defined by $f$.
a) Compute $f^{-1}(n)$ for $ninmathbb{Z}$
b)Prove $tau={emptyset,mathbb{Z}}cup{]-infty,n]capmathbb{Z}:ninmathbb{Z}}$
c) Determine if $(mathbb{Z},tau)$ is metrizable.
d)Determine if $(mathbb{Z},tau)$ is compact.
a)
If $f^{-1}(n)=[n,n+1)$ once $leqslant xleqslant n+1$
b)
If we have the infinite set (...,-n-1,...-3,-2,-1,0,1,2,3...n-1...)
Let $(mathbb{R},tau')$ be the topological space where $tau'$ is the Euclidean topology
$f^{-1}(mathbb{Z})=mathbb{R}in tau'impliesmathbb{Z}intau $
$f^{-1}(emptyset)=emptysetin tau'impliesemptysetintau $
If I pick up a set in $mathbb{Z}$ like }$(...,-n,...-3,-2,-1,0,1,2,3...n)$ then I prove it is open.
$f^{-1}((...,-n,...-3,-2,-1,0,1,2,3...n))=...cup [-n-1,-n)cup[-3,-2)cup[-2,-1)cup[-1,0)cup[1,2)cup[3,4)...cup[n-1,n)cup=(-infty,n)in tau'implies (infty,n)capmathbb{Z}intau$
Note the compliment of $mathbb{Z}setminus((infty,n)capmathbb{Z})=[n,infty)capmathbb{Z}$ which is closed hence I do need to try other sets.
This proves $tau={emptyset,mathbb{Z}}cup{]-infty,n]capmathbb{Z}:ninmathbb{Z}}$ is the topology $tau$.
c)
Proposition: Every metric space is Hausdorff.
Let $x,yinmathbb{Z}$ there is an open set $xin U$ and $yin V$ such that $U=(-infty,j]capmathbb{Z}$ and $V=(-infty,j']capmathbb{Z}$. It is trivial to notice $Ucap Vneq emptyset$. Then as $j$ and $j'$ are arbitrary elements of $mathbb{Z}$ it can be concluded that $mathbb{Z},tau$ is not Hausdorff hence it is not metrizable.
d)
Proposition: Every compact space is limited.
Since $mathbb{Z},tau$ is not limited hence it is not compact.
Question:
Is this answer right?
general-topology proof-verification
asked Jan 26 at 11:23
Pedro GomesPedro Gomes
1,9442721
$endgroup$
$begingroup$
What do you mean by "Every compact space is limited"? Do you mean that a compact metric space is bounded?
$endgroup$
– Servaes
Jan 26 at 11:37
$begingroup$
@Servaes Yes, I do
$endgroup$
– Pedro Gomes
Jan 26 at 11:37
$begingroup$
So then this proposition does no apply to $Bbb{Z}$ with the quotient topology, because you have just showed that it is not a metric space.
$endgroup$
– Servaes
Jan 26 at 11:41
$begingroup$
@Servaes But is there a more general result? I mean one proposition of this kind that applies to the topological spaces?
$endgroup$
– Pedro Gomes
Jan 26 at 11:45
add a comment |
$begingroup$
Let $f:mathbb{R}tomathbb{Z}$ be a function such that each $xinmathbb{R}$ it associates the greatest integer less or equal to $x$. Suppose that $mathbb{R}$ is endowed with the eucledian topology such that $tau$ is the quotient topology in $mathbb{Z}$ defined by $f$.
a) Compute $f^{-1}(n)$ for $ninmathbb{Z}$
b)Prove $tau={emptyset,mathbb{Z}}cup{]-infty,n]capmathbb{Z}:ninmathbb{Z}}$
c) Determine if $(mathbb{Z},tau)$ is metrizable.
d)Determine if $(mathbb{Z},tau)$ is compact.
a)
If $f^{-1}(n)=[n,n+1)$ once $leqslant xleqslant n+1$
b)
If we have the infinite set (...,-n-1,...-3,-2,-1,0,1,2,3...n-1...)
Let $(mathbb{R},tau')$ be the topological space where $tau'$ is the Euclidean topology
$f^{-1}(mathbb{Z})=mathbb{R}in tau'impliesmathbb{Z}intau $
$f^{-1}(emptyset)=emptysetin tau'impliesemptysetintau $
If I pick up a set in $mathbb{Z}$ like }$(...,-n,...-3,-2,-1,0,1,2,3...n)$ then I prove it is open.
$f^{-1}((...,-n,...-3,-2,-1,0,1,2,3...n))=...cup [-n-1,-n)cup[-3,-2)cup[-2,-1)cup[-1,0)cup[1,2)cup[3,4)...cup[n-1,n)cup=(-infty,n)in tau'implies (infty,n)capmathbb{Z}intau$
Note the compliment of $mathbb{Z}setminus((infty,n)capmathbb{Z})=[n,infty)capmathbb{Z}$ which is closed hence I do need to try other sets.
This proves $tau={emptyset,mathbb{Z}}cup{]-infty,n]capmathbb{Z}:ninmathbb{Z}}$ is the topology $tau$.
c)
Proposition: Every metric space is Hausdorff.
Let $x,yinmathbb{Z}$ there is an open set $xin U$ and $yin V$ such that $U=(-infty,j]capmathbb{Z}$ and $V=(-infty,j']capmathbb{Z}$. It is trivial to notice $Ucap Vneq emptyset$. Then as $j$ and $j'$ are arbitrary elements of $mathbb{Z}$ it can be concluded that $mathbb{Z},tau$ is not Hausdorff hence it is not metrizable.
d)
Proposition: Every compact space is limited.
Since $mathbb{Z},tau$ is not limited hence it is not compact.
Question:
Is this answer right?
general-topology proof-verification
asked Jan 26 at 11:23
Pedro GomesPedro Gomes
1,9442721
$endgroup$
Let $f:mathbb{R}tomathbb{Z}$ be a function such that each $xinmathbb{R}$ it associates the greatest integer less or equal to $x$. Suppose that $mathbb{R}$ is endowed with the eucledian topology such that $tau$ is the quotient topology in $mathbb{Z}$ defined by $f$.
a) Compute $f^{-1}(n)$ for $ninmathbb{Z}$
b)Prove $tau={emptyset,mathbb{Z}}cup{]-infty,n]capmathbb{Z}:ninmathbb{Z}}$
c) Determine if $(mathbb{Z},tau)$ is metrizable.
d)Determine if $(mathbb{Z},tau)$ is compact.
a)
If $f^{-1}(n)=[n,n+1)$ once $leqslant xleqslant n+1$
b)
If we have the infinite set (...,-n-1,...-3,-2,-1,0,1,2,3...n-1...)
Let $(mathbb{R},tau')$ be the topological space where $tau'$ is the Euclidean topology
$f^{-1}(mathbb{Z})=mathbb{R}in tau'impliesmathbb{Z}intau $
$f^{-1}(emptyset)=emptysetin tau'impliesemptysetintau $
If I pick up a set in $mathbb{Z}$ like }$(...,-n,...-3,-2,-1,0,1,2,3...n)$ then I prove it is open.
$f^{-1}((...,-n,...-3,-2,-1,0,1,2,3...n))=...cup [-n-1,-n)cup[-3,-2)cup[-2,-1)cup[-1,0)cup[1,2)cup[3,4)...cup[n-1,n)cup=(-infty,n)in tau'implies (infty,n)capmathbb{Z}intau$
Note the compliment of $mathbb{Z}setminus((infty,n)capmathbb{Z})=[n,infty)capmathbb{Z}$ which is closed hence I do need to try other sets.
This proves $tau={emptyset,mathbb{Z}}cup{]-infty,n]capmathbb{Z}:ninmathbb{Z}}$ is the topology $tau$.
c)
Proposition: Every metric space is Hausdorff.
Let $x,yinmathbb{Z}$ there is an open set $xin U$ and $yin V$ such that $U=(-infty,j]capmathbb{Z}$ and $V=(-infty,j']capmathbb{Z}$. It is trivial to notice $Ucap Vneq emptyset$. Then as $j$ and $j'$ are arbitrary elements of $mathbb{Z}$ it can be concluded that $mathbb{Z},tau$ is not Hausdorff hence it is not metrizable.
d)
Proposition: Every compact space is limited.
Since $mathbb{Z},tau$ is not limited hence it is not compact.
Question:
Is this answer right?
general-topology proof-verification
general-topology proof-verification
asked Jan 26 at 11:23
Pedro GomesPedro Gomes
1,9442721
asked Jan 26 at 11:23
Pedro GomesPedro Gomes
1,9442721
asked Jan 26 at 11:23
Pedro GomesPedro Gomes
1,9442721
asked Jan 26 at 11:23
Pedro GomesPedro Gomes
1,9442721
asked Jan 26 at 11:23
Pedro GomesPedro Gomes
1,9442721
1,9442721
$begingroup$
What do you mean by "Every compact space is limited"? Do you mean that a compact metric space is bounded?
$endgroup$
– Servaes
Jan 26 at 11:37
$begingroup$
@Servaes Yes, I do
$endgroup$
– Pedro Gomes
Jan 26 at 11:37
$begingroup$
So then this proposition does no apply to $Bbb{Z}$ with the quotient topology, because you have just showed that it is not a metric space.
$endgroup$
– Servaes
Jan 26 at 11:41
$begingroup$
@Servaes But is there a more general result? I mean one proposition of this kind that applies to the topological spaces?
$endgroup$
– Pedro Gomes
Jan 26 at 11:45
add a comment |
$begingroup$
What do you mean by "Every compact space is limited"? Do you mean that a compact metric space is bounded?
$endgroup$
– Servaes
Jan 26 at 11:37
$begingroup$
@Servaes Yes, I do
$endgroup$
– Pedro Gomes
Jan 26 at 11:37
$begingroup$
So then this proposition does no apply to $Bbb{Z}$ with the quotient topology, because you have just showed that it is not a metric space.
$endgroup$
– Servaes
Jan 26 at 11:41
$begingroup$
@Servaes But is there a more general result? I mean one proposition of this kind that applies to the topological spaces?
$endgroup$
– Pedro Gomes
Jan 26 at 11:45
$begingroup$
What do you mean by "Every compact space is limited"? Do you mean that a compact metric space is bounded?
$endgroup$
– Servaes
Jan 26 at 11:37
$begingroup$
What do you mean by "Every compact space is limited"? Do you mean that a compact metric space is bounded?
$endgroup$
– Servaes
Jan 26 at 11:37
$begingroup$
@Servaes Yes, I do
$endgroup$
– Pedro Gomes
Jan 26 at 11:37
$begingroup$
@Servaes Yes, I do
$endgroup$
– Pedro Gomes
Jan 26 at 11:37
$begingroup$
So then this proposition does no apply to $Bbb{Z}$ with the quotient topology, because you have just showed that it is not a metric space.
$endgroup$
– Servaes
Jan 26 at 11:41
$begingroup$
So then this proposition does no apply to $Bbb{Z}$ with the quotient topology, because you have just showed that it is not a metric space.
$endgroup$
– Servaes
Jan 26 at 11:41
$begingroup$
@Servaes But is there a more general result? I mean one proposition of this kind that applies to the topological spaces?
$endgroup$
– Pedro Gomes
Jan 26 at 11:45
$begingroup$
@Servaes But is there a more general result? I mean one proposition of this kind that applies to the topological spaces?
$endgroup$
– Pedro Gomes
Jan 26 at 11:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Indeed $f^{-1}[{n}]= [n, n+1)$. Note that this means that ${n}$ is not open in $tau$, as the quotient topology means that $O subseteq mathbb{Z}$ is open if and only if $f^{-1}[O]$ is open in $mathbb{R}$. And $[n,n+1)$ is not open.
Suppose $O$ is open in $mathbb{Z}$ under $tau$. This means $$f^{-1}[O]=bigcup {f^{-1}[{n}]: n in O} = bigcup_{n in O} [n,n+1)$$
is open in the Euclidean topology. If the integer $m$ is in $O$, so must $m-1$ be, as otherwise $m$ is not an interior point of $f^{-1}[O]$.
The only subsets $O$ of $mathbb{Z}$ that obey $forall M in O: m-1 in O$ are indeed $emptyset, mathbb{Z}$ and all sets of the form $(-infty,n]$ for $n in mathbb{Z}$ and these sets indeed have open pre-images, so these form exactly the quotient topolgoy $tau$.
As all non-empty open subsets of $tau$ must intersect (clear from the form of the open sets, or note that $n in O_1, m in O_2$ implies $min(n,m) in O_1 cap O_2$) $(mathbb{Z},tau)$ is not Hausdorff, so that idea checks out.
The open cover $(-infty,1], (-infty,2], ldots$ has no finite subcover, so $(mathbb{Z}, tau)$ is not compact.
answered Jan 26 at 12:30
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
add a comment |
$begingroup$
a) Correct. Though I don't know what you mean by "once $nleq xleq n+1$".
b) You have proved that $tau$ contains the given topology, i.e. that it is at least as fine as the given topology. You have not shown equality; there could be other sets in $tau$. I would suggest that you prove the implication
$$nin UquadRightarrowquad n-1in U,$$
for all open sets $UsubsetBbb{Z}$.
c) Correct, though poorly worded. I would suggest a proof by contradiction; suppose $(Bbb{Z},tau)$ is Hausdorff...
d) This is wrong, the proposition as stated is false, or at best very unclear. Perhaps you had in mind
Proposition: Every compact metric space is bounded.
Unfortunately you have just shown in part c that $(Bbb{Z},tau)$ is not a metric space, so the proposition does not apply. But it is not hard to give an explicit open cover that does not have a finite subcover; any infinite open cover that doesn't contain $Bbb{Z}intau$ will do, in fact.
answered Jan 26 at 11:48
ServaesServaes
28.5k34099
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Indeed $f^{-1}[{n}]= [n, n+1)$. Note that this means that ${n}$ is not open in $tau$, as the quotient topology means that $O subseteq mathbb{Z}$ is open if and only if $f^{-1}[O]$ is open in $mathbb{R}$. And $[n,n+1)$ is not open.
Suppose $O$ is open in $mathbb{Z}$ under $tau$. This means $$f^{-1}[O]=bigcup {f^{-1}[{n}]: n in O} = bigcup_{n in O} [n,n+1)$$
is open in the Euclidean topology. If the integer $m$ is in $O$, so must $m-1$ be, as otherwise $m$ is not an interior point of $f^{-1}[O]$.
The only subsets $O$ of $mathbb{Z}$ that obey $forall M in O: m-1 in O$ are indeed $emptyset, mathbb{Z}$ and all sets of the form $(-infty,n]$ for $n in mathbb{Z}$ and these sets indeed have open pre-images, so these form exactly the quotient topolgoy $tau$.
As all non-empty open subsets of $tau$ must intersect (clear from the form of the open sets, or note that $n in O_1, m in O_2$ implies $min(n,m) in O_1 cap O_2$) $(mathbb{Z},tau)$ is not Hausdorff, so that idea checks out.
The open cover $(-infty,1], (-infty,2], ldots$ has no finite subcover, so $(mathbb{Z}, tau)$ is not compact.
answered Jan 26 at 12:30
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
add a comment |
$begingroup$
Indeed $f^{-1}[{n}]= [n, n+1)$. Note that this means that ${n}$ is not open in $tau$, as the quotient topology means that $O subseteq mathbb{Z}$ is open if and only if $f^{-1}[O]$ is open in $mathbb{R}$. And $[n,n+1)$ is not open.
Suppose $O$ is open in $mathbb{Z}$ under $tau$. This means $$f^{-1}[O]=bigcup {f^{-1}[{n}]: n in O} = bigcup_{n in O} [n,n+1)$$
is open in the Euclidean topology. If the integer $m$ is in $O$, so must $m-1$ be, as otherwise $m$ is not an interior point of $f^{-1}[O]$.
The only subsets $O$ of $mathbb{Z}$ that obey $forall M in O: m-1 in O$ are indeed $emptyset, mathbb{Z}$ and all sets of the form $(-infty,n]$ for $n in mathbb{Z}$ and these sets indeed have open pre-images, so these form exactly the quotient topolgoy $tau$.
As all non-empty open subsets of $tau$ must intersect (clear from the form of the open sets, or note that $n in O_1, m in O_2$ implies $min(n,m) in O_1 cap O_2$) $(mathbb{Z},tau)$ is not Hausdorff, so that idea checks out.
The open cover $(-infty,1], (-infty,2], ldots$ has no finite subcover, so $(mathbb{Z}, tau)$ is not compact.
answered Jan 26 at 12:30
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
add a comment |
$begingroup$
Indeed $f^{-1}[{n}]= [n, n+1)$. Note that this means that ${n}$ is not open in $tau$, as the quotient topology means that $O subseteq mathbb{Z}$ is open if and only if $f^{-1}[O]$ is open in $mathbb{R}$. And $[n,n+1)$ is not open.
Suppose $O$ is open in $mathbb{Z}$ under $tau$. This means $$f^{-1}[O]=bigcup {f^{-1}[{n}]: n in O} = bigcup_{n in O} [n,n+1)$$
is open in the Euclidean topology. If the integer $m$ is in $O$, so must $m-1$ be, as otherwise $m$ is not an interior point of $f^{-1}[O]$.
The only subsets $O$ of $mathbb{Z}$ that obey $forall M in O: m-1 in O$ are indeed $emptyset, mathbb{Z}$ and all sets of the form $(-infty,n]$ for $n in mathbb{Z}$ and these sets indeed have open pre-images, so these form exactly the quotient topolgoy $tau$.
As all non-empty open subsets of $tau$ must intersect (clear from the form of the open sets, or note that $n in O_1, m in O_2$ implies $min(n,m) in O_1 cap O_2$) $(mathbb{Z},tau)$ is not Hausdorff, so that idea checks out.
The open cover $(-infty,1], (-infty,2], ldots$ has no finite subcover, so $(mathbb{Z}, tau)$ is not compact.
answered Jan 26 at 12:30
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
Indeed $f^{-1}[{n}]= [n, n+1)$. Note that this means that ${n}$ is not open in $tau$, as the quotient topology means that $O subseteq mathbb{Z}$ is open if and only if $f^{-1}[O]$ is open in $mathbb{R}$. And $[n,n+1)$ is not open.
Suppose $O$ is open in $mathbb{Z}$ under $tau$. This means $$f^{-1}[O]=bigcup {f^{-1}[{n}]: n in O} = bigcup_{n in O} [n,n+1)$$
is open in the Euclidean topology. If the integer $m$ is in $O$, so must $m-1$ be, as otherwise $m$ is not an interior point of $f^{-1}[O]$.
The only subsets $O$ of $mathbb{Z}$ that obey $forall M in O: m-1 in O$ are indeed $emptyset, mathbb{Z}$ and all sets of the form $(-infty,n]$ for $n in mathbb{Z}$ and these sets indeed have open pre-images, so these form exactly the quotient topolgoy $tau$.
As all non-empty open subsets of $tau$ must intersect (clear from the form of the open sets, or note that $n in O_1, m in O_2$ implies $min(n,m) in O_1 cap O_2$) $(mathbb{Z},tau)$ is not Hausdorff, so that idea checks out.
The open cover $(-infty,1], (-infty,2], ldots$ has no finite subcover, so $(mathbb{Z}, tau)$ is not compact.
answered Jan 26 at 12:30
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 26 at 12:30
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 26 at 12:30
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 26 at 12:30
Henno BrandsmaHenno Brandsma
113k348123
113k348123
add a comment |
add a comment |
$begingroup$
a) Correct. Though I don't know what you mean by "once $nleq xleq n+1$".
b) You have proved that $tau$ contains the given topology, i.e. that it is at least as fine as the given topology. You have not shown equality; there could be other sets in $tau$. I would suggest that you prove the implication
$$nin UquadRightarrowquad n-1in U,$$
for all open sets $UsubsetBbb{Z}$.
c) Correct, though poorly worded. I would suggest a proof by contradiction; suppose $(Bbb{Z},tau)$ is Hausdorff...
d) This is wrong, the proposition as stated is false, or at best very unclear. Perhaps you had in mind
Proposition: Every compact metric space is bounded.
Unfortunately you have just shown in part c that $(Bbb{Z},tau)$ is not a metric space, so the proposition does not apply. But it is not hard to give an explicit open cover that does not have a finite subcover; any infinite open cover that doesn't contain $Bbb{Z}intau$ will do, in fact.
answered Jan 26 at 11:48
ServaesServaes
28.5k34099
$endgroup$
add a comment |
$begingroup$
a) Correct. Though I don't know what you mean by "once $nleq xleq n+1$".
b) You have proved that $tau$ contains the given topology, i.e. that it is at least as fine as the given topology. You have not shown equality; there could be other sets in $tau$. I would suggest that you prove the implication
$$nin UquadRightarrowquad n-1in U,$$
for all open sets $UsubsetBbb{Z}$.
c) Correct, though poorly worded. I would suggest a proof by contradiction; suppose $(Bbb{Z},tau)$ is Hausdorff...
d) This is wrong, the proposition as stated is false, or at best very unclear. Perhaps you had in mind
Proposition: Every compact metric space is bounded.
Unfortunately you have just shown in part c that $(Bbb{Z},tau)$ is not a metric space, so the proposition does not apply. But it is not hard to give an explicit open cover that does not have a finite subcover; any infinite open cover that doesn't contain $Bbb{Z}intau$ will do, in fact.
answered Jan 26 at 11:48
ServaesServaes
28.5k34099
$endgroup$
add a comment |
$begingroup$
a) Correct. Though I don't know what you mean by "once $nleq xleq n+1$".
b) You have proved that $tau$ contains the given topology, i.e. that it is at least as fine as the given topology. You have not shown equality; there could be other sets in $tau$. I would suggest that you prove the implication
$$nin UquadRightarrowquad n-1in U,$$
for all open sets $UsubsetBbb{Z}$.
c) Correct, though poorly worded. I would suggest a proof by contradiction; suppose $(Bbb{Z},tau)$ is Hausdorff...
d) This is wrong, the proposition as stated is false, or at best very unclear. Perhaps you had in mind
Proposition: Every compact metric space is bounded.
Unfortunately you have just shown in part c that $(Bbb{Z},tau)$ is not a metric space, so the proposition does not apply. But it is not hard to give an explicit open cover that does not have a finite subcover; any infinite open cover that doesn't contain $Bbb{Z}intau$ will do, in fact.
answered Jan 26 at 11:48
ServaesServaes
28.5k34099
$endgroup$
a) Correct. Though I don't know what you mean by "once $nleq xleq n+1$".
b) You have proved that $tau$ contains the given topology, i.e. that it is at least as fine as the given topology. You have not shown equality; there could be other sets in $tau$. I would suggest that you prove the implication
$$nin UquadRightarrowquad n-1in U,$$
for all open sets $UsubsetBbb{Z}$.
c) Correct, though poorly worded. I would suggest a proof by contradiction; suppose $(Bbb{Z},tau)$ is Hausdorff...
d) This is wrong, the proposition as stated is false, or at best very unclear. Perhaps you had in mind
Proposition: Every compact metric space is bounded.
Unfortunately you have just shown in part c that $(Bbb{Z},tau)$ is not a metric space, so the proposition does not apply. But it is not hard to give an explicit open cover that does not have a finite subcover; any infinite open cover that doesn't contain $Bbb{Z}intau$ will do, in fact.
answered Jan 26 at 11:48
ServaesServaes
28.5k34099
answered Jan 26 at 11:48
ServaesServaes
28.5k34099
answered Jan 26 at 11:48
ServaesServaes
28.5k34099
answered Jan 26 at 11:48
ServaesServaes
28.5k34099
28.5k34099
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$begingroup$
What do you mean by "Every compact space is limited"? Do you mean that a compact metric space is bounded?
$endgroup$
– Servaes
Jan 26 at 11:37
$begingroup$
@Servaes Yes, I do
$endgroup$
– Pedro Gomes
Jan 26 at 11:37
$begingroup$
So then this proposition does no apply to $Bbb{Z}$ with the quotient topology, because you have just showed that it is not a metric space.
$endgroup$
– Servaes
Jan 26 at 11:41
$begingroup$
@Servaes But is there a more general result? I mean one proposition of this kind that applies to the topological spaces?
$endgroup$
– Pedro Gomes
Jan 26 at 11:45