Equivalence of Baire Category Theorem for Complete Metric Spaces.
In my studies of functional analysis I have come across a statement of the Baire Category Theorem which states:
"The Baire Category Theorem says that if a complete
metric space $X$ is a countable union
of closed sets $S_n$ then at least one of the $S_n$ contains
an open ball".
And I am struggling with how to relate this to the more common form of the theorem for example from wikipedia:
A Baire space is a topological space with the following property: for each countable collection of open dense sets ${displaystyle {U_{n}}_{n=1}^{infty }}$, their intersection ${displaystyle textstyle bigcap _{n=1}^{infty }U_{n}}$ is dense. And Every complete metric space is Baire space.
Any insight to how these two are equivalent would be greatly appreciated.
general-topology functional-analysis metric-spaces complete-spaces baire-category
edited Nov 21 '18 at 14:42
Davide Giraudo
125k16150259
asked Nov 21 '18 at 12:32
Matthew
306112
add a comment |
In my studies of functional analysis I have come across a statement of the Baire Category Theorem which states:
"The Baire Category Theorem says that if a complete
metric space $X$ is a countable union
of closed sets $S_n$ then at least one of the $S_n$ contains
an open ball".
And I am struggling with how to relate this to the more common form of the theorem for example from wikipedia:
A Baire space is a topological space with the following property: for each countable collection of open dense sets ${displaystyle {U_{n}}_{n=1}^{infty }}$, their intersection ${displaystyle textstyle bigcap _{n=1}^{infty }U_{n}}$ is dense. And Every complete metric space is Baire space.
Any insight to how these two are equivalent would be greatly appreciated.
general-topology functional-analysis metric-spaces complete-spaces baire-category
edited Nov 21 '18 at 14:42
Davide Giraudo
125k16150259
asked Nov 21 '18 at 12:32
Matthew
306112
add a comment |
In my studies of functional analysis I have come across a statement of the Baire Category Theorem which states:
"The Baire Category Theorem says that if a complete
metric space $X$ is a countable union
of closed sets $S_n$ then at least one of the $S_n$ contains
an open ball".
And I am struggling with how to relate this to the more common form of the theorem for example from wikipedia:
A Baire space is a topological space with the following property: for each countable collection of open dense sets ${displaystyle {U_{n}}_{n=1}^{infty }}$, their intersection ${displaystyle textstyle bigcap _{n=1}^{infty }U_{n}}$ is dense. And Every complete metric space is Baire space.
Any insight to how these two are equivalent would be greatly appreciated.
general-topology functional-analysis metric-spaces complete-spaces baire-category
edited Nov 21 '18 at 14:42
Davide Giraudo
125k16150259
asked Nov 21 '18 at 12:32
Matthew
306112
In my studies of functional analysis I have come across a statement of the Baire Category Theorem which states:
"The Baire Category Theorem says that if a complete
metric space $X$ is a countable union
of closed sets $S_n$ then at least one of the $S_n$ contains
an open ball".
And I am struggling with how to relate this to the more common form of the theorem for example from wikipedia:
A Baire space is a topological space with the following property: for each countable collection of open dense sets ${displaystyle {U_{n}}_{n=1}^{infty }}$, their intersection ${displaystyle textstyle bigcap _{n=1}^{infty }U_{n}}$ is dense. And Every complete metric space is Baire space.
Any insight to how these two are equivalent would be greatly appreciated.
general-topology functional-analysis metric-spaces complete-spaces baire-category
general-topology functional-analysis metric-spaces complete-spaces baire-category
edited Nov 21 '18 at 14:42
Davide Giraudo
125k16150259
asked Nov 21 '18 at 12:32
Matthew
306112
edited Nov 21 '18 at 14:42
Davide Giraudo
125k16150259
asked Nov 21 '18 at 12:32
Matthew
306112
edited Nov 21 '18 at 14:42
Davide Giraudo
125k16150259
edited Nov 21 '18 at 14:42
Davide Giraudo
125k16150259
edited Nov 21 '18 at 14:42
Davide Giraudo
125k16150259
125k16150259
asked Nov 21 '18 at 12:32
Matthew
306112
asked Nov 21 '18 at 12:32
Matthew
306112
asked Nov 21 '18 at 12:32
Matthew
306112
306112
add a comment |
add a comment |
1 Answer
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Consider the assertion “every countable intersection of open dense sets is dense”. Passing to complements, it is seen to be equivalent to “every countable union of closed sets with empty interiors has empty interior”. But the interior of whole space $X$ is certainly not empty! Therefore, if it can be written has the union of a countable family of closed subsets, then at least one of these sets must have non-empty interior, which is equivalent to the assertion that it contains an open ball.
answered Nov 21 '18 at 12:39
José Carlos Santos
150k22122223
add a comment |
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1 Answer
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Consider the assertion “every countable intersection of open dense sets is dense”. Passing to complements, it is seen to be equivalent to “every countable union of closed sets with empty interiors has empty interior”. But the interior of whole space $X$ is certainly not empty! Therefore, if it can be written has the union of a countable family of closed subsets, then at least one of these sets must have non-empty interior, which is equivalent to the assertion that it contains an open ball.
answered Nov 21 '18 at 12:39
José Carlos Santos
150k22122223
add a comment |
Consider the assertion “every countable intersection of open dense sets is dense”. Passing to complements, it is seen to be equivalent to “every countable union of closed sets with empty interiors has empty interior”. But the interior of whole space $X$ is certainly not empty! Therefore, if it can be written has the union of a countable family of closed subsets, then at least one of these sets must have non-empty interior, which is equivalent to the assertion that it contains an open ball.
answered Nov 21 '18 at 12:39
José Carlos Santos
150k22122223
add a comment |
Consider the assertion “every countable intersection of open dense sets is dense”. Passing to complements, it is seen to be equivalent to “every countable union of closed sets with empty interiors has empty interior”. But the interior of whole space $X$ is certainly not empty! Therefore, if it can be written has the union of a countable family of closed subsets, then at least one of these sets must have non-empty interior, which is equivalent to the assertion that it contains an open ball.
answered Nov 21 '18 at 12:39
José Carlos Santos
150k22122223
Consider the assertion “every countable intersection of open dense sets is dense”. Passing to complements, it is seen to be equivalent to “every countable union of closed sets with empty interiors has empty interior”. But the interior of whole space $X$ is certainly not empty! Therefore, if it can be written has the union of a countable family of closed subsets, then at least one of these sets must have non-empty interior, which is equivalent to the assertion that it contains an open ball.
answered Nov 21 '18 at 12:39
José Carlos Santos
150k22122223
edited Nov 21 '18 at 14:44
answered Nov 21 '18 at 12:39
José Carlos Santos
150k22122223
answered Nov 21 '18 at 12:39
José Carlos Santos
150k22122223
answered Nov 21 '18 at 12:39
José Carlos Santos
150k22122223
150k22122223
add a comment |
add a comment |
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