Mixing
Proof: Every closed set of real numbers is an $F_sigma$ set
$begingroup$
I've been thinking about the proof. Although it seems very obvious that by the definition of $F_sigma$ sets, a set is an $F_sigma$ set if it is a countable union of closed sets, so intuitively, a single closed set is also a collection of '1' closed sets.... But this seems to be a little trivial, it might be that this is actually correct but i am really confused.
Another doubt that came to my mind is whether:' closed set of real numbers' means a collection of singletons?, if yes then would it be different from an 'open set of real numbers'? what is a 'closed set composed of in ANY topological space'?
Really new to the subject, trying to get the feel for it..
general-topology
asked Jan 26 at 11:42
CosmicCosmic
8810
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add a comment |
$begingroup$
I've been thinking about the proof. Although it seems very obvious that by the definition of $F_sigma$ sets, a set is an $F_sigma$ set if it is a countable union of closed sets, so intuitively, a single closed set is also a collection of '1' closed sets.... But this seems to be a little trivial, it might be that this is actually correct but i am really confused.
Another doubt that came to my mind is whether:' closed set of real numbers' means a collection of singletons?, if yes then would it be different from an 'open set of real numbers'? what is a 'closed set composed of in ANY topological space'?
Really new to the subject, trying to get the feel for it..
general-topology
asked Jan 26 at 11:42
CosmicCosmic
8810
$endgroup$
add a comment |
$begingroup$
I've been thinking about the proof. Although it seems very obvious that by the definition of $F_sigma$ sets, a set is an $F_sigma$ set if it is a countable union of closed sets, so intuitively, a single closed set is also a collection of '1' closed sets.... But this seems to be a little trivial, it might be that this is actually correct but i am really confused.
Another doubt that came to my mind is whether:' closed set of real numbers' means a collection of singletons?, if yes then would it be different from an 'open set of real numbers'? what is a 'closed set composed of in ANY topological space'?
Really new to the subject, trying to get the feel for it..
general-topology
asked Jan 26 at 11:42
CosmicCosmic
8810
$endgroup$
I've been thinking about the proof. Although it seems very obvious that by the definition of $F_sigma$ sets, a set is an $F_sigma$ set if it is a countable union of closed sets, so intuitively, a single closed set is also a collection of '1' closed sets.... But this seems to be a little trivial, it might be that this is actually correct but i am really confused.
Another doubt that came to my mind is whether:' closed set of real numbers' means a collection of singletons?, if yes then would it be different from an 'open set of real numbers'? what is a 'closed set composed of in ANY topological space'?
Really new to the subject, trying to get the feel for it..
general-topology
general-topology
asked Jan 26 at 11:42
CosmicCosmic
8810
asked Jan 26 at 11:42
CosmicCosmic
8810
asked Jan 26 at 11:42
CosmicCosmic
8810
asked Jan 26 at 11:42
CosmicCosmic
8810
asked Jan 26 at 11:42
CosmicCosmic
8810
8810
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is indeed as trivial as you think. A union of one closed set is a countable union of closed sets.
More interestingly: every open set (in $mathbb{R}$ or any metric space) is also the countable union of closed sets.
Also an open set is a $G_delta$ (a countable (1) intersection of open sets) and dually a closed set (in a metric space) also is a $G_delta$ (follows from the open set is an $F_sigma$ fact by taking complements and applying de Morgan).
answered Jan 26 at 11:55
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
Thanks, Could you help me with the second part of the question as well?
$endgroup$
– Cosmic
Jan 26 at 11:57
1
$begingroup$
@Cosmic a closed set of real numbers is a set that is the complement of an open set. Or equivalently a set that contains all its limit points. In defining a topology one defines the open sets and the closed sets are just their complements.
$endgroup$
– Henno Brandsma
Jan 26 at 12:11
add a comment |
$begingroup$
A closed set is, by definition, the complement of an open set. A closed set is indeed an $F_sigma$ set, as it is a countable union of closed sets, exactly as you said!
as for the part about singletons, well singletons are usually not open (for instance in $mathbb{R}$).
You have the following rules:
- any union of open set is open
- any finite intersection of open sets is open
Hence, you get for closed sets that :
- any intersection of closed sets is closed
- any finite union of closed sets is closed
answered Jan 26 at 11:48
J.FJ.F
36713
$endgroup$
add a comment |
$begingroup$
Let $X$ be a closed subset of reals.
Because $X$ is closed,
$$X={xinmathbb{R}:(forall varepsilon>0)[ Xcap(x-varepsilon,x+varepsilon)nevarnothing]}=:mathrm{closure}(X).$$
That is to say, the set $X$ contains its limit points. So, for example, the interval $(0,1)$ is not closed because it is missing $0$ and $1$---limit points.
Observe $X=Xcupvarnothingcupvarnothingcupcdots$ where the union is countable. Therefore $X$ is an $F_sigma$-set.
Assume $X=[0,1]$ the closed interval from $0$ to $1$. Then it is true that $X=bigcup_{xin X}{x}$ however the union here is not countable. Therefore this is not an argument for why $X$ is indeed an $F_sigma$-set. See the above paragraph for the true argument.
answered Jan 26 at 11:48
Alberto TakaseAlberto Takase
2,380719
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is indeed as trivial as you think. A union of one closed set is a countable union of closed sets.
More interestingly: every open set (in $mathbb{R}$ or any metric space) is also the countable union of closed sets.
Also an open set is a $G_delta$ (a countable (1) intersection of open sets) and dually a closed set (in a metric space) also is a $G_delta$ (follows from the open set is an $F_sigma$ fact by taking complements and applying de Morgan).
answered Jan 26 at 11:55
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
Thanks, Could you help me with the second part of the question as well?
$endgroup$
– Cosmic
Jan 26 at 11:57
1
$begingroup$
@Cosmic a closed set of real numbers is a set that is the complement of an open set. Or equivalently a set that contains all its limit points. In defining a topology one defines the open sets and the closed sets are just their complements.
$endgroup$
– Henno Brandsma
Jan 26 at 12:11
add a comment |
$begingroup$
It is indeed as trivial as you think. A union of one closed set is a countable union of closed sets.
More interestingly: every open set (in $mathbb{R}$ or any metric space) is also the countable union of closed sets.
Also an open set is a $G_delta$ (a countable (1) intersection of open sets) and dually a closed set (in a metric space) also is a $G_delta$ (follows from the open set is an $F_sigma$ fact by taking complements and applying de Morgan).
answered Jan 26 at 11:55
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
Thanks, Could you help me with the second part of the question as well?
$endgroup$
– Cosmic
Jan 26 at 11:57
1
$begingroup$
@Cosmic a closed set of real numbers is a set that is the complement of an open set. Or equivalently a set that contains all its limit points. In defining a topology one defines the open sets and the closed sets are just their complements.
$endgroup$
– Henno Brandsma
Jan 26 at 12:11
add a comment |
$begingroup$
It is indeed as trivial as you think. A union of one closed set is a countable union of closed sets.
More interestingly: every open set (in $mathbb{R}$ or any metric space) is also the countable union of closed sets.
Also an open set is a $G_delta$ (a countable (1) intersection of open sets) and dually a closed set (in a metric space) also is a $G_delta$ (follows from the open set is an $F_sigma$ fact by taking complements and applying de Morgan).
answered Jan 26 at 11:55
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
It is indeed as trivial as you think. A union of one closed set is a countable union of closed sets.
More interestingly: every open set (in $mathbb{R}$ or any metric space) is also the countable union of closed sets.
Also an open set is a $G_delta$ (a countable (1) intersection of open sets) and dually a closed set (in a metric space) also is a $G_delta$ (follows from the open set is an $F_sigma$ fact by taking complements and applying de Morgan).
answered Jan 26 at 11:55
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 26 at 11:55
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 26 at 11:55
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 26 at 11:55
Henno BrandsmaHenno Brandsma
113k348123
113k348123
$begingroup$
Thanks, Could you help me with the second part of the question as well?
$endgroup$
– Cosmic
Jan 26 at 11:57
1
$begingroup$
@Cosmic a closed set of real numbers is a set that is the complement of an open set. Or equivalently a set that contains all its limit points. In defining a topology one defines the open sets and the closed sets are just their complements.
$endgroup$
– Henno Brandsma
Jan 26 at 12:11
add a comment |
$begingroup$
Thanks, Could you help me with the second part of the question as well?
$endgroup$
– Cosmic
Jan 26 at 11:57
1
$begingroup$
@Cosmic a closed set of real numbers is a set that is the complement of an open set. Or equivalently a set that contains all its limit points. In defining a topology one defines the open sets and the closed sets are just their complements.
$endgroup$
– Henno Brandsma
Jan 26 at 12:11
$begingroup$
Thanks, Could you help me with the second part of the question as well?
$endgroup$
– Cosmic
Jan 26 at 11:57
$begingroup$
Thanks, Could you help me with the second part of the question as well?
$endgroup$
– Cosmic
Jan 26 at 11:57
1
1
$begingroup$
@Cosmic a closed set of real numbers is a set that is the complement of an open set. Or equivalently a set that contains all its limit points. In defining a topology one defines the open sets and the closed sets are just their complements.
$endgroup$
– Henno Brandsma
Jan 26 at 12:11
$begingroup$
@Cosmic a closed set of real numbers is a set that is the complement of an open set. Or equivalently a set that contains all its limit points. In defining a topology one defines the open sets and the closed sets are just their complements.
$endgroup$
– Henno Brandsma
Jan 26 at 12:11
add a comment |
$begingroup$
A closed set is, by definition, the complement of an open set. A closed set is indeed an $F_sigma$ set, as it is a countable union of closed sets, exactly as you said!
as for the part about singletons, well singletons are usually not open (for instance in $mathbb{R}$).
You have the following rules:
- any union of open set is open
- any finite intersection of open sets is open
Hence, you get for closed sets that :
- any intersection of closed sets is closed
- any finite union of closed sets is closed
answered Jan 26 at 11:48
J.FJ.F
36713
$endgroup$
add a comment |
$begingroup$
A closed set is, by definition, the complement of an open set. A closed set is indeed an $F_sigma$ set, as it is a countable union of closed sets, exactly as you said!
as for the part about singletons, well singletons are usually not open (for instance in $mathbb{R}$).
You have the following rules:
- any union of open set is open
- any finite intersection of open sets is open
Hence, you get for closed sets that :
- any intersection of closed sets is closed
- any finite union of closed sets is closed
answered Jan 26 at 11:48
J.FJ.F
36713
$endgroup$
add a comment |
$begingroup$
A closed set is, by definition, the complement of an open set. A closed set is indeed an $F_sigma$ set, as it is a countable union of closed sets, exactly as you said!
as for the part about singletons, well singletons are usually not open (for instance in $mathbb{R}$).
You have the following rules:
- any union of open set is open
- any finite intersection of open sets is open
Hence, you get for closed sets that :
- any intersection of closed sets is closed
- any finite union of closed sets is closed
answered Jan 26 at 11:48
J.FJ.F
36713
$endgroup$
A closed set is, by definition, the complement of an open set. A closed set is indeed an $F_sigma$ set, as it is a countable union of closed sets, exactly as you said!
as for the part about singletons, well singletons are usually not open (for instance in $mathbb{R}$).
You have the following rules:
- any union of open set is open
- any finite intersection of open sets is open
Hence, you get for closed sets that :
- any intersection of closed sets is closed
- any finite union of closed sets is closed
answered Jan 26 at 11:48
J.FJ.F
36713
answered Jan 26 at 11:48
J.FJ.F
36713
answered Jan 26 at 11:48
J.FJ.F
36713
answered Jan 26 at 11:48
J.FJ.F
36713
36713
add a comment |
add a comment |
$begingroup$
Let $X$ be a closed subset of reals.
Because $X$ is closed,
$$X={xinmathbb{R}:(forall varepsilon>0)[ Xcap(x-varepsilon,x+varepsilon)nevarnothing]}=:mathrm{closure}(X).$$
That is to say, the set $X$ contains its limit points. So, for example, the interval $(0,1)$ is not closed because it is missing $0$ and $1$---limit points.
Observe $X=Xcupvarnothingcupvarnothingcupcdots$ where the union is countable. Therefore $X$ is an $F_sigma$-set.
Assume $X=[0,1]$ the closed interval from $0$ to $1$. Then it is true that $X=bigcup_{xin X}{x}$ however the union here is not countable. Therefore this is not an argument for why $X$ is indeed an $F_sigma$-set. See the above paragraph for the true argument.
answered Jan 26 at 11:48
Alberto TakaseAlberto Takase
2,380719
$endgroup$
add a comment |
$begingroup$
Let $X$ be a closed subset of reals.
Because $X$ is closed,
$$X={xinmathbb{R}:(forall varepsilon>0)[ Xcap(x-varepsilon,x+varepsilon)nevarnothing]}=:mathrm{closure}(X).$$
That is to say, the set $X$ contains its limit points. So, for example, the interval $(0,1)$ is not closed because it is missing $0$ and $1$---limit points.
Observe $X=Xcupvarnothingcupvarnothingcupcdots$ where the union is countable. Therefore $X$ is an $F_sigma$-set.
Assume $X=[0,1]$ the closed interval from $0$ to $1$. Then it is true that $X=bigcup_{xin X}{x}$ however the union here is not countable. Therefore this is not an argument for why $X$ is indeed an $F_sigma$-set. See the above paragraph for the true argument.
answered Jan 26 at 11:48
Alberto TakaseAlberto Takase
2,380719
$endgroup$
add a comment |
$begingroup$
Let $X$ be a closed subset of reals.
Because $X$ is closed,
$$X={xinmathbb{R}:(forall varepsilon>0)[ Xcap(x-varepsilon,x+varepsilon)nevarnothing]}=:mathrm{closure}(X).$$
That is to say, the set $X$ contains its limit points. So, for example, the interval $(0,1)$ is not closed because it is missing $0$ and $1$---limit points.
Observe $X=Xcupvarnothingcupvarnothingcupcdots$ where the union is countable. Therefore $X$ is an $F_sigma$-set.
Assume $X=[0,1]$ the closed interval from $0$ to $1$. Then it is true that $X=bigcup_{xin X}{x}$ however the union here is not countable. Therefore this is not an argument for why $X$ is indeed an $F_sigma$-set. See the above paragraph for the true argument.
answered Jan 26 at 11:48
Alberto TakaseAlberto Takase
2,380719
$endgroup$
Let $X$ be a closed subset of reals.
Because $X$ is closed,
$$X={xinmathbb{R}:(forall varepsilon>0)[ Xcap(x-varepsilon,x+varepsilon)nevarnothing]}=:mathrm{closure}(X).$$
That is to say, the set $X$ contains its limit points. So, for example, the interval $(0,1)$ is not closed because it is missing $0$ and $1$---limit points.
Observe $X=Xcupvarnothingcupvarnothingcupcdots$ where the union is countable. Therefore $X$ is an $F_sigma$-set.
Assume $X=[0,1]$ the closed interval from $0$ to $1$. Then it is true that $X=bigcup_{xin X}{x}$ however the union here is not countable. Therefore this is not an argument for why $X$ is indeed an $F_sigma$-set. See the above paragraph for the true argument.
answered Jan 26 at 11:48
Alberto TakaseAlberto Takase
2,380719
edited Jan 26 at 11:54
answered Jan 26 at 11:48
Alberto TakaseAlberto Takase
2,380719
answered Jan 26 at 11:48
Alberto TakaseAlberto Takase
2,380719
answered Jan 26 at 11:48
Alberto TakaseAlberto Takase
2,380719
2,380719
add a comment |
add a comment |
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