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Show that a closed subset is first category.
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I have the following problem:
Suppose $A$ is a closed subset of a complete metric space $X$. Show that $A$ is of the first category if and only if $A$ has empty interior.
My attemp: if $A$ is a closed subset of $X$ and has empty interior, i.e., $int(A)=emptyset$, we can conlude that $A=overlineAimplies int(A)=int(overlineA)=emptyset$. So, $A$ is nowhere dense.
Now I define the countable sequence $U_n$ equal to $A$ if $n=1$ and $emptyset$ if $n>1$. Therefore, $A=displaystylecup_n=1^infty U_n$ is the first category.
Now, if $A$ is a closed subset of $X$ and is of the first category, my idea was prove this by contradiction, suppose that a closed ball $Y$ is contained in $A$. Since $Ysubset X$, and $X$ is a complete metric space, we can consider $Y$ as a complete metric space, by Baire's theorem, $Y$ is the second category, and I don't know how continuous this. I need some idea, thanks!
real-analysis baire-category
asked Jul 29 at 3:22
julios
564
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favorite
I have the following problem:
Suppose $A$ is a closed subset of a complete metric space $X$. Show that $A$ is of the first category if and only if $A$ has empty interior.
My attemp: if $A$ is a closed subset of $X$ and has empty interior, i.e., $int(A)=emptyset$, we can conlude that $A=overlineAimplies int(A)=int(overlineA)=emptyset$. So, $A$ is nowhere dense.
Now I define the countable sequence $U_n$ equal to $A$ if $n=1$ and $emptyset$ if $n>1$. Therefore, $A=displaystylecup_n=1^infty U_n$ is the first category.
Now, if $A$ is a closed subset of $X$ and is of the first category, my idea was prove this by contradiction, suppose that a closed ball $Y$ is contained in $A$. Since $Ysubset X$, and $X$ is a complete metric space, we can consider $Y$ as a complete metric space, by Baire's theorem, $Y$ is the second category, and I don't know how continuous this. I need some idea, thanks!
real-analysis baire-category
asked Jul 29 at 3:22
julios
564
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following problem:
Suppose $A$ is a closed subset of a complete metric space $X$. Show that $A$ is of the first category if and only if $A$ has empty interior.
My attemp: if $A$ is a closed subset of $X$ and has empty interior, i.e., $int(A)=emptyset$, we can conlude that $A=overlineAimplies int(A)=int(overlineA)=emptyset$. So, $A$ is nowhere dense.
Now I define the countable sequence $U_n$ equal to $A$ if $n=1$ and $emptyset$ if $n>1$. Therefore, $A=displaystylecup_n=1^infty U_n$ is the first category.
Now, if $A$ is a closed subset of $X$ and is of the first category, my idea was prove this by contradiction, suppose that a closed ball $Y$ is contained in $A$. Since $Ysubset X$, and $X$ is a complete metric space, we can consider $Y$ as a complete metric space, by Baire's theorem, $Y$ is the second category, and I don't know how continuous this. I need some idea, thanks!
real-analysis baire-category
asked Jul 29 at 3:22
julios
564
I have the following problem:
Suppose $A$ is a closed subset of a complete metric space $X$. Show that $A$ is of the first category if and only if $A$ has empty interior.
My attemp: if $A$ is a closed subset of $X$ and has empty interior, i.e., $int(A)=emptyset$, we can conlude that $A=overlineAimplies int(A)=int(overlineA)=emptyset$. So, $A$ is nowhere dense.
Now I define the countable sequence $U_n$ equal to $A$ if $n=1$ and $emptyset$ if $n>1$. Therefore, $A=displaystylecup_n=1^infty U_n$ is the first category.
Now, if $A$ is a closed subset of $X$ and is of the first category, my idea was prove this by contradiction, suppose that a closed ball $Y$ is contained in $A$. Since $Ysubset X$, and $X$ is a complete metric space, we can consider $Y$ as a complete metric space, by Baire's theorem, $Y$ is the second category, and I don't know how continuous this. I need some idea, thanks!
real-analysis baire-category
asked Jul 29 at 3:22
julios
564
asked Jul 29 at 3:22
julios
564
asked Jul 29 at 3:22
julios
564
asked Jul 29 at 3:22
julios
564
564
add a comment |Â
add a comment |Â
1 Answer
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If $A$ has non-empty interior, then $B[x; r]$ is contained in $A$, for some $x in X$ and $r > 0$. So, if we can express $A$ as a countable union of nowhere dense sets $bigcup_n in mathbbN C_n$, then $B[x; r]$ is covered by this countable union of nowhere dense sets. Consider $D_n = C_n cap B[x; r]$, and you have that $B[x; r]$ is itself the countable union of nowhere dense sets $D_n$.
Of course, $B[x; r] setminus overlineD_n$ is an open dense subset of $B[x; r]$, whose intersection is empty. This contradicts the BCT on $B[x; r]$.
answered Jul 29 at 3:36
Theo Bendit
11.8k1843
Thanks a lot, I don't saw that $C_ncap B=D_n$.
– julios
Jul 29 at 3:49
Wouldn't the converse of Baire CT imply that if $A$ is a first category then $A$ is empty? (Since $A$ is a complete metric space)
– Akababa
Aug 1 at 22:35
@Akababa I might be misunderstanding your question, but a single point in a complete metric space is closed, first category, and non-empty.
– Theo Bendit
Aug 2 at 0:03
Doesn't BCT say that a nonempty complete metric space is a second category?
– Akababa
Aug 2 at 13:55
@Akababa Right, but $A$ is first category in the context of $X$, not as a subset of itself. In, say, the real numbers, every singleton $lbrace x rbrace$ is nowhere dense, but considered as a subset of itself, $$lbrace x rbrace = B[x; 1],$$ which implies it's not nowhere dense.
– Theo Bendit
Aug 2 at 14:07
 |Â
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $A$ has non-empty interior, then $B[x; r]$ is contained in $A$, for some $x in X$ and $r > 0$. So, if we can express $A$ as a countable union of nowhere dense sets $bigcup_n in mathbbN C_n$, then $B[x; r]$ is covered by this countable union of nowhere dense sets. Consider $D_n = C_n cap B[x; r]$, and you have that $B[x; r]$ is itself the countable union of nowhere dense sets $D_n$.
Of course, $B[x; r] setminus overlineD_n$ is an open dense subset of $B[x; r]$, whose intersection is empty. This contradicts the BCT on $B[x; r]$.
answered Jul 29 at 3:36
Theo Bendit
11.8k1843
Thanks a lot, I don't saw that $C_ncap B=D_n$.
– julios
Jul 29 at 3:49
Wouldn't the converse of Baire CT imply that if $A$ is a first category then $A$ is empty? (Since $A$ is a complete metric space)
– Akababa
Aug 1 at 22:35
@Akababa I might be misunderstanding your question, but a single point in a complete metric space is closed, first category, and non-empty.
– Theo Bendit
Aug 2 at 0:03
Doesn't BCT say that a nonempty complete metric space is a second category?
– Akababa
Aug 2 at 13:55
@Akababa Right, but $A$ is first category in the context of $X$, not as a subset of itself. In, say, the real numbers, every singleton $lbrace x rbrace$ is nowhere dense, but considered as a subset of itself, $$lbrace x rbrace = B[x; 1],$$ which implies it's not nowhere dense.
– Theo Bendit
Aug 2 at 14:07
 |Â
show 1 more comment
up vote
0
down vote
If $A$ has non-empty interior, then $B[x; r]$ is contained in $A$, for some $x in X$ and $r > 0$. So, if we can express $A$ as a countable union of nowhere dense sets $bigcup_n in mathbbN C_n$, then $B[x; r]$ is covered by this countable union of nowhere dense sets. Consider $D_n = C_n cap B[x; r]$, and you have that $B[x; r]$ is itself the countable union of nowhere dense sets $D_n$.
Of course, $B[x; r] setminus overlineD_n$ is an open dense subset of $B[x; r]$, whose intersection is empty. This contradicts the BCT on $B[x; r]$.
answered Jul 29 at 3:36
Theo Bendit
11.8k1843
Thanks a lot, I don't saw that $C_ncap B=D_n$.
– julios
Jul 29 at 3:49
Wouldn't the converse of Baire CT imply that if $A$ is a first category then $A$ is empty? (Since $A$ is a complete metric space)
– Akababa
Aug 1 at 22:35
@Akababa I might be misunderstanding your question, but a single point in a complete metric space is closed, first category, and non-empty.
– Theo Bendit
Aug 2 at 0:03
Doesn't BCT say that a nonempty complete metric space is a second category?
– Akababa
Aug 2 at 13:55
@Akababa Right, but $A$ is first category in the context of $X$, not as a subset of itself. In, say, the real numbers, every singleton $lbrace x rbrace$ is nowhere dense, but considered as a subset of itself, $$lbrace x rbrace = B[x; 1],$$ which implies it's not nowhere dense.
– Theo Bendit
Aug 2 at 14:07
 |Â
show 1 more comment
up vote
0
down vote
up vote
0
down vote
If $A$ has non-empty interior, then $B[x; r]$ is contained in $A$, for some $x in X$ and $r > 0$. So, if we can express $A$ as a countable union of nowhere dense sets $bigcup_n in mathbbN C_n$, then $B[x; r]$ is covered by this countable union of nowhere dense sets. Consider $D_n = C_n cap B[x; r]$, and you have that $B[x; r]$ is itself the countable union of nowhere dense sets $D_n$.
Of course, $B[x; r] setminus overlineD_n$ is an open dense subset of $B[x; r]$, whose intersection is empty. This contradicts the BCT on $B[x; r]$.
answered Jul 29 at 3:36
Theo Bendit
11.8k1843
If $A$ has non-empty interior, then $B[x; r]$ is contained in $A$, for some $x in X$ and $r > 0$. So, if we can express $A$ as a countable union of nowhere dense sets $bigcup_n in mathbbN C_n$, then $B[x; r]$ is covered by this countable union of nowhere dense sets. Consider $D_n = C_n cap B[x; r]$, and you have that $B[x; r]$ is itself the countable union of nowhere dense sets $D_n$.
Of course, $B[x; r] setminus overlineD_n$ is an open dense subset of $B[x; r]$, whose intersection is empty. This contradicts the BCT on $B[x; r]$.
answered Jul 29 at 3:36
Theo Bendit
11.8k1843
answered Jul 29 at 3:36
Theo Bendit
11.8k1843
answered Jul 29 at 3:36
Theo Bendit
11.8k1843
answered Jul 29 at 3:36
Theo Bendit
11.8k1843
11.8k1843
Thanks a lot, I don't saw that $C_ncap B=D_n$.
– julios
Jul 29 at 3:49
Wouldn't the converse of Baire CT imply that if $A$ is a first category then $A$ is empty? (Since $A$ is a complete metric space)
– Akababa
Aug 1 at 22:35
@Akababa I might be misunderstanding your question, but a single point in a complete metric space is closed, first category, and non-empty.
– Theo Bendit
Aug 2 at 0:03
Doesn't BCT say that a nonempty complete metric space is a second category?
– Akababa
Aug 2 at 13:55
@Akababa Right, but $A$ is first category in the context of $X$, not as a subset of itself. In, say, the real numbers, every singleton $lbrace x rbrace$ is nowhere dense, but considered as a subset of itself, $$lbrace x rbrace = B[x; 1],$$ which implies it's not nowhere dense.
– Theo Bendit
Aug 2 at 14:07
 |Â
show 1 more comment
Thanks a lot, I don't saw that $C_ncap B=D_n$.
– julios
Jul 29 at 3:49
Wouldn't the converse of Baire CT imply that if $A$ is a first category then $A$ is empty? (Since $A$ is a complete metric space)
– Akababa
Aug 1 at 22:35
@Akababa I might be misunderstanding your question, but a single point in a complete metric space is closed, first category, and non-empty.
– Theo Bendit
Aug 2 at 0:03
Doesn't BCT say that a nonempty complete metric space is a second category?
– Akababa
Aug 2 at 13:55
@Akababa Right, but $A$ is first category in the context of $X$, not as a subset of itself. In, say, the real numbers, every singleton $lbrace x rbrace$ is nowhere dense, but considered as a subset of itself, $$lbrace x rbrace = B[x; 1],$$ which implies it's not nowhere dense.
– Theo Bendit
Aug 2 at 14:07
Thanks a lot, I don't saw that $C_ncap B=D_n$.
– julios
Jul 29 at 3:49
Thanks a lot, I don't saw that $C_ncap B=D_n$.
– julios
Jul 29 at 3:49
Wouldn't the converse of Baire CT imply that if $A$ is a first category then $A$ is empty? (Since $A$ is a complete metric space)
– Akababa
Aug 1 at 22:35
Wouldn't the converse of Baire CT imply that if $A$ is a first category then $A$ is empty? (Since $A$ is a complete metric space)
– Akababa
Aug 1 at 22:35
@Akababa I might be misunderstanding your question, but a single point in a complete metric space is closed, first category, and non-empty.
– Theo Bendit
Aug 2 at 0:03
@Akababa I might be misunderstanding your question, but a single point in a complete metric space is closed, first category, and non-empty.
– Theo Bendit
Aug 2 at 0:03
Doesn't BCT say that a nonempty complete metric space is a second category?
– Akababa
Aug 2 at 13:55
Doesn't BCT say that a nonempty complete metric space is a second category?
– Akababa
Aug 2 at 13:55
@Akababa Right, but $A$ is first category in the context of $X$, not as a subset of itself. In, say, the real numbers, every singleton $lbrace x rbrace$ is nowhere dense, but considered as a subset of itself, $$lbrace x rbrace = B[x; 1],$$ which implies it's not nowhere dense.
– Theo Bendit
Aug 2 at 14:07
@Akababa Right, but $A$ is first category in the context of $X$, not as a subset of itself. In, say, the real numbers, every singleton $lbrace x rbrace$ is nowhere dense, but considered as a subset of itself, $$lbrace x rbrace = B[x; 1],$$ which implies it's not nowhere dense.
– Theo Bendit
Aug 2 at 14:07
 |Â
show 1 more comment
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