Closure of a set in a metric space.
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I have a question with regards to the proof of the following:
Let $X = A cup A'$ be the closure of a set. $A'$ is the set of limit points of $A$. I wish to show that the closure is closed. I would very much like your opinions on the proof.
Let $ain A^ccap (A')^c$. Since $a$ is not a limit point of $A$ there exists a neighborhood of $a$ such that for every $q$ in the neighborhood of $a$, $q=a$ or $qnotin A$. This implies that $N_r(a)$ is contained within the complement of $A$. Now we would like to show that the neighborhood is contained within $(A')^c$, because then $N_r(a)$ $subset$ $A^c$ $cap$ $(A')^c$ which implies that the set is open, therefore the closure is closed. So assume that the neighborhood is not contained in $(A')^c$ this implies that $a$ $in$ $A'$ because $N_r(a)$ $subset$ $A'$, which is a contradiction.
real-analysis general-topology analysis proof-verification proof-writing
edited Jan 13 at 2:59
Janitha357
1,756516
asked Jan 13 at 2:39
mathsssislifemathsssislife
437
$endgroup$
add a comment |
$begingroup$
I have a question with regards to the proof of the following:
Let $X = A cup A'$ be the closure of a set. $A'$ is the set of limit points of $A$. I wish to show that the closure is closed. I would very much like your opinions on the proof.
Let $ain A^ccap (A')^c$. Since $a$ is not a limit point of $A$ there exists a neighborhood of $a$ such that for every $q$ in the neighborhood of $a$, $q=a$ or $qnotin A$. This implies that $N_r(a)$ is contained within the complement of $A$. Now we would like to show that the neighborhood is contained within $(A')^c$, because then $N_r(a)$ $subset$ $A^c$ $cap$ $(A')^c$ which implies that the set is open, therefore the closure is closed. So assume that the neighborhood is not contained in $(A')^c$ this implies that $a$ $in$ $A'$ because $N_r(a)$ $subset$ $A'$, which is a contradiction.
real-analysis general-topology analysis proof-verification proof-writing
edited Jan 13 at 2:59
Janitha357
1,756516
asked Jan 13 at 2:39
mathsssislifemathsssislife
437
$endgroup$
$begingroup$
You start by basically assuming a not in X and proceed to make contradicition; thusly proclaiming X is the whole space, which usually it isn't. For example when A is empty.
$endgroup$
– William Elliot
Jan 13 at 3:18
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You dont need the contradiction. Once you show that $N_r(a)$ exists and misses A, you've shown that X$^c$ is open. Thus X is closed.
$endgroup$
– Joel Pereira
Jan 13 at 3:19
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@WilliamElliot but if A is empty then X is closed.
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– mathsssislife
Jan 13 at 3:21
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@WilliamElliot what do you suggest I do differently?
$endgroup$
– mathsssislife
Jan 13 at 3:22
$begingroup$
Topologically limit points are nearly useless. It is weird that limit points play such a central role in metric space theory. They are clumbsy and their use is not helpful as an introduction to topology. A better alternative for analysis than limit points is punctured nhoods, to be used only as needed for those aspects of analysis that topology is less concerned about. @mathsssislife
$endgroup$
– William Elliot
Jan 13 at 9:11
add a comment |
$begingroup$
I have a question with regards to the proof of the following:
Let $X = A cup A'$ be the closure of a set. $A'$ is the set of limit points of $A$. I wish to show that the closure is closed. I would very much like your opinions on the proof.
Let $ain A^ccap (A')^c$. Since $a$ is not a limit point of $A$ there exists a neighborhood of $a$ such that for every $q$ in the neighborhood of $a$, $q=a$ or $qnotin A$. This implies that $N_r(a)$ is contained within the complement of $A$. Now we would like to show that the neighborhood is contained within $(A')^c$, because then $N_r(a)$ $subset$ $A^c$ $cap$ $(A')^c$ which implies that the set is open, therefore the closure is closed. So assume that the neighborhood is not contained in $(A')^c$ this implies that $a$ $in$ $A'$ because $N_r(a)$ $subset$ $A'$, which is a contradiction.
real-analysis general-topology analysis proof-verification proof-writing
edited Jan 13 at 2:59
Janitha357
1,756516
asked Jan 13 at 2:39
mathsssislifemathsssislife
437
$endgroup$
I have a question with regards to the proof of the following:
Let $X = A cup A'$ be the closure of a set. $A'$ is the set of limit points of $A$. I wish to show that the closure is closed. I would very much like your opinions on the proof.
Let $ain A^ccap (A')^c$. Since $a$ is not a limit point of $A$ there exists a neighborhood of $a$ such that for every $q$ in the neighborhood of $a$, $q=a$ or $qnotin A$. This implies that $N_r(a)$ is contained within the complement of $A$. Now we would like to show that the neighborhood is contained within $(A')^c$, because then $N_r(a)$ $subset$ $A^c$ $cap$ $(A')^c$ which implies that the set is open, therefore the closure is closed. So assume that the neighborhood is not contained in $(A')^c$ this implies that $a$ $in$ $A'$ because $N_r(a)$ $subset$ $A'$, which is a contradiction.
real-analysis general-topology analysis proof-verification proof-writing
real-analysis general-topology analysis proof-verification proof-writing
edited Jan 13 at 2:59
Janitha357
1,756516
asked Jan 13 at 2:39
mathsssislifemathsssislife
437
edited Jan 13 at 2:59
Janitha357
1,756516
asked Jan 13 at 2:39
mathsssislifemathsssislife
437
edited Jan 13 at 2:59
Janitha357
1,756516
edited Jan 13 at 2:59
Janitha357
1,756516
edited Jan 13 at 2:59
Janitha357
1,756516
1,756516
asked Jan 13 at 2:39
mathsssislifemathsssislife
437
asked Jan 13 at 2:39
mathsssislifemathsssislife
437
asked Jan 13 at 2:39
mathsssislifemathsssislife
437
437
$begingroup$
You start by basically assuming a not in X and proceed to make contradicition; thusly proclaiming X is the whole space, which usually it isn't. For example when A is empty.
$endgroup$
– William Elliot
Jan 13 at 3:18
$begingroup$
You dont need the contradiction. Once you show that $N_r(a)$ exists and misses A, you've shown that X$^c$ is open. Thus X is closed.
$endgroup$
– Joel Pereira
Jan 13 at 3:19
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@WilliamElliot but if A is empty then X is closed.
$endgroup$
– mathsssislife
Jan 13 at 3:21
$begingroup$
@WilliamElliot what do you suggest I do differently?
$endgroup$
– mathsssislife
Jan 13 at 3:22
$begingroup$
Topologically limit points are nearly useless. It is weird that limit points play such a central role in metric space theory. They are clumbsy and their use is not helpful as an introduction to topology. A better alternative for analysis than limit points is punctured nhoods, to be used only as needed for those aspects of analysis that topology is less concerned about. @mathsssislife
$endgroup$
– William Elliot
Jan 13 at 9:11
add a comment |
$begingroup$
You start by basically assuming a not in X and proceed to make contradicition; thusly proclaiming X is the whole space, which usually it isn't. For example when A is empty.
$endgroup$
– William Elliot
Jan 13 at 3:18
$begingroup$
You dont need the contradiction. Once you show that $N_r(a)$ exists and misses A, you've shown that X$^c$ is open. Thus X is closed.
$endgroup$
– Joel Pereira
Jan 13 at 3:19
$begingroup$
@WilliamElliot but if A is empty then X is closed.
$endgroup$
– mathsssislife
Jan 13 at 3:21
$begingroup$
@WilliamElliot what do you suggest I do differently?
$endgroup$
– mathsssislife
Jan 13 at 3:22
$begingroup$
Topologically limit points are nearly useless. It is weird that limit points play such a central role in metric space theory. They are clumbsy and their use is not helpful as an introduction to topology. A better alternative for analysis than limit points is punctured nhoods, to be used only as needed for those aspects of analysis that topology is less concerned about. @mathsssislife
$endgroup$
– William Elliot
Jan 13 at 9:11
$begingroup$
You start by basically assuming a not in X and proceed to make contradicition; thusly proclaiming X is the whole space, which usually it isn't. For example when A is empty.
$endgroup$
– William Elliot
Jan 13 at 3:18
$begingroup$
You start by basically assuming a not in X and proceed to make contradicition; thusly proclaiming X is the whole space, which usually it isn't. For example when A is empty.
$endgroup$
– William Elliot
Jan 13 at 3:18
$begingroup$
You dont need the contradiction. Once you show that $N_r(a)$ exists and misses A, you've shown that X$^c$ is open. Thus X is closed.
$endgroup$
– Joel Pereira
Jan 13 at 3:19
$begingroup$
You dont need the contradiction. Once you show that $N_r(a)$ exists and misses A, you've shown that X$^c$ is open. Thus X is closed.
$endgroup$
– Joel Pereira
Jan 13 at 3:19
$begingroup$
@WilliamElliot but if A is empty then X is closed.
$endgroup$
– mathsssislife
Jan 13 at 3:21
$begingroup$
@WilliamElliot but if A is empty then X is closed.
$endgroup$
– mathsssislife
Jan 13 at 3:21
$begingroup$
@WilliamElliot what do you suggest I do differently?
$endgroup$
– mathsssislife
Jan 13 at 3:22
$begingroup$
@WilliamElliot what do you suggest I do differently?
$endgroup$
– mathsssislife
Jan 13 at 3:22
$begingroup$
Topologically limit points are nearly useless. It is weird that limit points play such a central role in metric space theory. They are clumbsy and their use is not helpful as an introduction to topology. A better alternative for analysis than limit points is punctured nhoods, to be used only as needed for those aspects of analysis that topology is less concerned about. @mathsssislife
$endgroup$
– William Elliot
Jan 13 at 9:11
$begingroup$
Topologically limit points are nearly useless. It is weird that limit points play such a central role in metric space theory. They are clumbsy and their use is not helpful as an introduction to topology. A better alternative for analysis than limit points is punctured nhoods, to be used only as needed for those aspects of analysis that topology is less concerned about. @mathsssislife
$endgroup$
– William Elliot
Jan 13 at 9:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The idea in the first half of the proof showing $N_r(a)subseteq A^c$ is correct. But in my opinion when showing that $N_r(a)$ is not contained in $(A')^c$ your reasoning lacks explanations. I would do it as follows.
Assume $N_r(a)nsubseteq (A')^c.$ Then $N_r(a)cap A'neqemptyset.$ So let $bin N_r(a)cap A'.$ Then there exists $t>0$ such that $N_t(a)subseteq N_r(a)$ (why?). Therefore $N_t(b)cap Aneq emptyset$ (why?). So $N_r(a)cap Aneqemptyset,$ which is a contradiction. Hence $N_r(a)subseteq (A')^c.$
answered Jan 13 at 3:46
Janitha357Janitha357
1,756516
$endgroup$
$begingroup$
Would this be better: Let p$in$ $(E cup E′)^c$ which implies that p $notin$ E and p $notin$ E'. Since p$notin$E', $exists$ $N_r(p)$ : $forall$ q$in$ Nr(p) , q $=$ p or q$notin$ E which is equivalent to Nr(p) $ cap$ E $=$ $emptyset$ , therefore Nr(p) ⊂ $E^c$ . Now we would like to show that $N_r(p)$ $subset$ $E′^c$, so assume that is not the case this implies that $N_r(p)$ $subset$ E' $implies$ p $in$ E' which is a contradiction, therefore $N_r(p)$ $subset$ $E^c$ $cup$ E' which means that it is open, therefore the closure is closed.
$endgroup$
– mathsssislife
Jan 13 at 23:46
add a comment |
$begingroup$
So I assume that your definition of "$A$ is closed" is "$A$ contains all its limit points", or equivalently $A' subseteq A$.
If we then define $overline{A}=A cup A'$ then indeed this set is closed:
$$overline{A}' = (A cup A')' = A' cup A'' subseteq A' subseteq overline{A}$$
using that in a $T_1$ space (thus certainly in a metric space) we always have $B'' subseteq B'$ for all subsets $B$ and also $(C cup D)' = C' cup D'$ for all subsets $C,D$. Thus $B = overline{A}$ obeys $B' subseteq B$ so is closed.
Moreover, if $C$ is closed and $A subseteq C $ then $A' subseteq C' subseteq C$, so $overline{A}=A cup A' subseteq C$, so it follows that $overline{A}$ is the smallest closed set that contains $A$ as a subset, another way the closure could have been defined.
answered Jan 13 at 15:45
Henno BrandsmaHenno Brandsma
107k347114
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add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
The idea in the first half of the proof showing $N_r(a)subseteq A^c$ is correct. But in my opinion when showing that $N_r(a)$ is not contained in $(A')^c$ your reasoning lacks explanations. I would do it as follows.
Assume $N_r(a)nsubseteq (A')^c.$ Then $N_r(a)cap A'neqemptyset.$ So let $bin N_r(a)cap A'.$ Then there exists $t>0$ such that $N_t(a)subseteq N_r(a)$ (why?). Therefore $N_t(b)cap Aneq emptyset$ (why?). So $N_r(a)cap Aneqemptyset,$ which is a contradiction. Hence $N_r(a)subseteq (A')^c.$
answered Jan 13 at 3:46
Janitha357Janitha357
1,756516
$endgroup$
$begingroup$
Would this be better: Let p$in$ $(E cup E′)^c$ which implies that p $notin$ E and p $notin$ E'. Since p$notin$E', $exists$ $N_r(p)$ : $forall$ q$in$ Nr(p) , q $=$ p or q$notin$ E which is equivalent to Nr(p) $ cap$ E $=$ $emptyset$ , therefore Nr(p) ⊂ $E^c$ . Now we would like to show that $N_r(p)$ $subset$ $E′^c$, so assume that is not the case this implies that $N_r(p)$ $subset$ E' $implies$ p $in$ E' which is a contradiction, therefore $N_r(p)$ $subset$ $E^c$ $cup$ E' which means that it is open, therefore the closure is closed.
$endgroup$
– mathsssislife
Jan 13 at 23:46
add a comment |
$begingroup$
The idea in the first half of the proof showing $N_r(a)subseteq A^c$ is correct. But in my opinion when showing that $N_r(a)$ is not contained in $(A')^c$ your reasoning lacks explanations. I would do it as follows.
Assume $N_r(a)nsubseteq (A')^c.$ Then $N_r(a)cap A'neqemptyset.$ So let $bin N_r(a)cap A'.$ Then there exists $t>0$ such that $N_t(a)subseteq N_r(a)$ (why?). Therefore $N_t(b)cap Aneq emptyset$ (why?). So $N_r(a)cap Aneqemptyset,$ which is a contradiction. Hence $N_r(a)subseteq (A')^c.$
answered Jan 13 at 3:46
Janitha357Janitha357
1,756516
$endgroup$
$begingroup$
Would this be better: Let p$in$ $(E cup E′)^c$ which implies that p $notin$ E and p $notin$ E'. Since p$notin$E', $exists$ $N_r(p)$ : $forall$ q$in$ Nr(p) , q $=$ p or q$notin$ E which is equivalent to Nr(p) $ cap$ E $=$ $emptyset$ , therefore Nr(p) ⊂ $E^c$ . Now we would like to show that $N_r(p)$ $subset$ $E′^c$, so assume that is not the case this implies that $N_r(p)$ $subset$ E' $implies$ p $in$ E' which is a contradiction, therefore $N_r(p)$ $subset$ $E^c$ $cup$ E' which means that it is open, therefore the closure is closed.
$endgroup$
– mathsssislife
Jan 13 at 23:46
add a comment |
$begingroup$
The idea in the first half of the proof showing $N_r(a)subseteq A^c$ is correct. But in my opinion when showing that $N_r(a)$ is not contained in $(A')^c$ your reasoning lacks explanations. I would do it as follows.
Assume $N_r(a)nsubseteq (A')^c.$ Then $N_r(a)cap A'neqemptyset.$ So let $bin N_r(a)cap A'.$ Then there exists $t>0$ such that $N_t(a)subseteq N_r(a)$ (why?). Therefore $N_t(b)cap Aneq emptyset$ (why?). So $N_r(a)cap Aneqemptyset,$ which is a contradiction. Hence $N_r(a)subseteq (A')^c.$
answered Jan 13 at 3:46
Janitha357Janitha357
1,756516
$endgroup$
The idea in the first half of the proof showing $N_r(a)subseteq A^c$ is correct. But in my opinion when showing that $N_r(a)$ is not contained in $(A')^c$ your reasoning lacks explanations. I would do it as follows.
Assume $N_r(a)nsubseteq (A')^c.$ Then $N_r(a)cap A'neqemptyset.$ So let $bin N_r(a)cap A'.$ Then there exists $t>0$ such that $N_t(a)subseteq N_r(a)$ (why?). Therefore $N_t(b)cap Aneq emptyset$ (why?). So $N_r(a)cap Aneqemptyset,$ which is a contradiction. Hence $N_r(a)subseteq (A')^c.$
answered Jan 13 at 3:46
Janitha357Janitha357
1,756516
answered Jan 13 at 3:46
Janitha357Janitha357
1,756516
answered Jan 13 at 3:46
Janitha357Janitha357
1,756516
answered Jan 13 at 3:46
Janitha357Janitha357
1,756516
1,756516
$begingroup$
Would this be better: Let p$in$ $(E cup E′)^c$ which implies that p $notin$ E and p $notin$ E'. Since p$notin$E', $exists$ $N_r(p)$ : $forall$ q$in$ Nr(p) , q $=$ p or q$notin$ E which is equivalent to Nr(p) $ cap$ E $=$ $emptyset$ , therefore Nr(p) ⊂ $E^c$ . Now we would like to show that $N_r(p)$ $subset$ $E′^c$, so assume that is not the case this implies that $N_r(p)$ $subset$ E' $implies$ p $in$ E' which is a contradiction, therefore $N_r(p)$ $subset$ $E^c$ $cup$ E' which means that it is open, therefore the closure is closed.
$endgroup$
– mathsssislife
Jan 13 at 23:46
add a comment |
$begingroup$
Would this be better: Let p$in$ $(E cup E′)^c$ which implies that p $notin$ E and p $notin$ E'. Since p$notin$E', $exists$ $N_r(p)$ : $forall$ q$in$ Nr(p) , q $=$ p or q$notin$ E which is equivalent to Nr(p) $ cap$ E $=$ $emptyset$ , therefore Nr(p) ⊂ $E^c$ . Now we would like to show that $N_r(p)$ $subset$ $E′^c$, so assume that is not the case this implies that $N_r(p)$ $subset$ E' $implies$ p $in$ E' which is a contradiction, therefore $N_r(p)$ $subset$ $E^c$ $cup$ E' which means that it is open, therefore the closure is closed.
$endgroup$
– mathsssislife
Jan 13 at 23:46
$begingroup$
Would this be better: Let p$in$ $(E cup E′)^c$ which implies that p $notin$ E and p $notin$ E'. Since p$notin$E', $exists$ $N_r(p)$ : $forall$ q$in$ Nr(p) , q $=$ p or q$notin$ E which is equivalent to Nr(p) $ cap$ E $=$ $emptyset$ , therefore Nr(p) ⊂ $E^c$ . Now we would like to show that $N_r(p)$ $subset$ $E′^c$, so assume that is not the case this implies that $N_r(p)$ $subset$ E' $implies$ p $in$ E' which is a contradiction, therefore $N_r(p)$ $subset$ $E^c$ $cup$ E' which means that it is open, therefore the closure is closed.
$endgroup$
– mathsssislife
Jan 13 at 23:46
$begingroup$
Would this be better: Let p$in$ $(E cup E′)^c$ which implies that p $notin$ E and p $notin$ E'. Since p$notin$E', $exists$ $N_r(p)$ : $forall$ q$in$ Nr(p) , q $=$ p or q$notin$ E which is equivalent to Nr(p) $ cap$ E $=$ $emptyset$ , therefore Nr(p) ⊂ $E^c$ . Now we would like to show that $N_r(p)$ $subset$ $E′^c$, so assume that is not the case this implies that $N_r(p)$ $subset$ E' $implies$ p $in$ E' which is a contradiction, therefore $N_r(p)$ $subset$ $E^c$ $cup$ E' which means that it is open, therefore the closure is closed.
$endgroup$
– mathsssislife
Jan 13 at 23:46
add a comment |
$begingroup$
So I assume that your definition of "$A$ is closed" is "$A$ contains all its limit points", or equivalently $A' subseteq A$.
If we then define $overline{A}=A cup A'$ then indeed this set is closed:
$$overline{A}' = (A cup A')' = A' cup A'' subseteq A' subseteq overline{A}$$
using that in a $T_1$ space (thus certainly in a metric space) we always have $B'' subseteq B'$ for all subsets $B$ and also $(C cup D)' = C' cup D'$ for all subsets $C,D$. Thus $B = overline{A}$ obeys $B' subseteq B$ so is closed.
Moreover, if $C$ is closed and $A subseteq C $ then $A' subseteq C' subseteq C$, so $overline{A}=A cup A' subseteq C$, so it follows that $overline{A}$ is the smallest closed set that contains $A$ as a subset, another way the closure could have been defined.
answered Jan 13 at 15:45
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
$begingroup$
So I assume that your definition of "$A$ is closed" is "$A$ contains all its limit points", or equivalently $A' subseteq A$.
If we then define $overline{A}=A cup A'$ then indeed this set is closed:
$$overline{A}' = (A cup A')' = A' cup A'' subseteq A' subseteq overline{A}$$
using that in a $T_1$ space (thus certainly in a metric space) we always have $B'' subseteq B'$ for all subsets $B$ and also $(C cup D)' = C' cup D'$ for all subsets $C,D$. Thus $B = overline{A}$ obeys $B' subseteq B$ so is closed.
Moreover, if $C$ is closed and $A subseteq C $ then $A' subseteq C' subseteq C$, so $overline{A}=A cup A' subseteq C$, so it follows that $overline{A}$ is the smallest closed set that contains $A$ as a subset, another way the closure could have been defined.
answered Jan 13 at 15:45
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
$begingroup$
So I assume that your definition of "$A$ is closed" is "$A$ contains all its limit points", or equivalently $A' subseteq A$.
If we then define $overline{A}=A cup A'$ then indeed this set is closed:
$$overline{A}' = (A cup A')' = A' cup A'' subseteq A' subseteq overline{A}$$
using that in a $T_1$ space (thus certainly in a metric space) we always have $B'' subseteq B'$ for all subsets $B$ and also $(C cup D)' = C' cup D'$ for all subsets $C,D$. Thus $B = overline{A}$ obeys $B' subseteq B$ so is closed.
Moreover, if $C$ is closed and $A subseteq C $ then $A' subseteq C' subseteq C$, so $overline{A}=A cup A' subseteq C$, so it follows that $overline{A}$ is the smallest closed set that contains $A$ as a subset, another way the closure could have been defined.
answered Jan 13 at 15:45
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
So I assume that your definition of "$A$ is closed" is "$A$ contains all its limit points", or equivalently $A' subseteq A$.
If we then define $overline{A}=A cup A'$ then indeed this set is closed:
$$overline{A}' = (A cup A')' = A' cup A'' subseteq A' subseteq overline{A}$$
using that in a $T_1$ space (thus certainly in a metric space) we always have $B'' subseteq B'$ for all subsets $B$ and also $(C cup D)' = C' cup D'$ for all subsets $C,D$. Thus $B = overline{A}$ obeys $B' subseteq B$ so is closed.
Moreover, if $C$ is closed and $A subseteq C $ then $A' subseteq C' subseteq C$, so $overline{A}=A cup A' subseteq C$, so it follows that $overline{A}$ is the smallest closed set that contains $A$ as a subset, another way the closure could have been defined.
answered Jan 13 at 15:45
Henno BrandsmaHenno Brandsma
107k347114
answered Jan 13 at 15:45
Henno BrandsmaHenno Brandsma
107k347114
answered Jan 13 at 15:45
Henno BrandsmaHenno Brandsma
107k347114
answered Jan 13 at 15:45
Henno BrandsmaHenno Brandsma
107k347114
107k347114
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$begingroup$
You start by basically assuming a not in X and proceed to make contradicition; thusly proclaiming X is the whole space, which usually it isn't. For example when A is empty.
$endgroup$
– William Elliot
Jan 13 at 3:18
$begingroup$
You dont need the contradiction. Once you show that $N_r(a)$ exists and misses A, you've shown that X$^c$ is open. Thus X is closed.
$endgroup$
– Joel Pereira
Jan 13 at 3:19
$begingroup$
@WilliamElliot but if A is empty then X is closed.
$endgroup$
– mathsssislife
Jan 13 at 3:21
$begingroup$
@WilliamElliot what do you suggest I do differently?
$endgroup$
– mathsssislife
Jan 13 at 3:22
$begingroup$
Topologically limit points are nearly useless. It is weird that limit points play such a central role in metric space theory. They are clumbsy and their use is not helpful as an introduction to topology. A better alternative for analysis than limit points is punctured nhoods, to be used only as needed for those aspects of analysis that topology is less concerned about. @mathsssislife
$endgroup$
– William Elliot
Jan 13 at 9:11