Mixing
Proof verification: Restriction of f to two closed subsets of X covering X continuous implies f on X is...
Let $A,B subseteq X$ be closed in $X$ with $A cup B = X$.
Suppose there exists a function, $f: X rightarrow Y$, with both
$f {restriction_{A}}: A rightarrow Y$ and $ f {restriction_{B}}: B rightarrow Y$ continuous.
Prove that $f$ is continuous.
Proposed proof:
It is sufficient to show that if $F subseteq Y$ is closed in $Y$, then $f^{-1}(F)$ is closed in $X$.
Let $F subseteq Y$ be closed in $Y$.
By continuity of $f{restriction_{A}}$ and $f{restriction_{B}}$, one has it that $f^{-1}{restriction_{A}}(F)$ and $f^{-1}{restriction_{B}}(F)$ are closed in their subspace topologies.
I.e. $f^{-1}{restriction_{A}}(F) = A backslash (bigcup_limits{{i in I}}U_i cap A) $ and similarly, $f^{-1}{restriction_{B}}(F) = B backslash (bigcup_limits{{j in J}}U_j cap B) $
Also being restrictions of $f$ we have that:
$$f^{-1}{restriction_{A}}(F) = f^{-1}(F) cap A$$
$$f^{-1}{restriction_{B}}(F) = f^{-1}(F) cap B$$
$Rightarrow f^{-1}{restriction_{A}}(F) cup f^{-1}{restriction_{B}}(F) = (f^{-1}(F) cap B) cup (f^{-1}(F) cap A) = f^{-1}(F) cap (A cup B) = f^{-1}(F)$
Since $A cup B = X$.
It now suffices to show that $f^{-1}{restriction_{A}}(F) = A backslash (bigcup_limits{{i in I}}U_i cap A) $ is closed in $X$ and similarly for $f^{-1}{restriction_{B}}(F)$, as the union of two closed sets is closed.
$$X backslash f^{-1}{restriction_{A}}(F) = X backslash (A backslash (bigcup_limits{{i in I}}U_i cap A)) $$
$$= X backslash ((A backslash bigcup_limits{{i in I}}U_i) cup (A backslash A) $$
$$= (X backslash A) cup bigcup_limits{{i in I}}U_i $$
Which is just the union of open sets in X (since A is closed by assumption).
Hence $X backslash f^{-1}{restriction_{A}}(F)$ is open $Rightarrow f^{-1}{restriction_{A}}(F)$ is closed.
One can show $f^{-1}{restriction_{B}}(F)$ is closed similarly, thereby completing the proof.
general-topology proof-verification
asked Nov 20 '18 at 12:52
Eugaurie
61
add a comment |
Let $A,B subseteq X$ be closed in $X$ with $A cup B = X$.
Suppose there exists a function, $f: X rightarrow Y$, with both
$f {restriction_{A}}: A rightarrow Y$ and $ f {restriction_{B}}: B rightarrow Y$ continuous.
Prove that $f$ is continuous.
Proposed proof:
It is sufficient to show that if $F subseteq Y$ is closed in $Y$, then $f^{-1}(F)$ is closed in $X$.
Let $F subseteq Y$ be closed in $Y$.
By continuity of $f{restriction_{A}}$ and $f{restriction_{B}}$, one has it that $f^{-1}{restriction_{A}}(F)$ and $f^{-1}{restriction_{B}}(F)$ are closed in their subspace topologies.
I.e. $f^{-1}{restriction_{A}}(F) = A backslash (bigcup_limits{{i in I}}U_i cap A) $ and similarly, $f^{-1}{restriction_{B}}(F) = B backslash (bigcup_limits{{j in J}}U_j cap B) $
Also being restrictions of $f$ we have that:
$$f^{-1}{restriction_{A}}(F) = f^{-1}(F) cap A$$
$$f^{-1}{restriction_{B}}(F) = f^{-1}(F) cap B$$
$Rightarrow f^{-1}{restriction_{A}}(F) cup f^{-1}{restriction_{B}}(F) = (f^{-1}(F) cap B) cup (f^{-1}(F) cap A) = f^{-1}(F) cap (A cup B) = f^{-1}(F)$
Since $A cup B = X$.
It now suffices to show that $f^{-1}{restriction_{A}}(F) = A backslash (bigcup_limits{{i in I}}U_i cap A) $ is closed in $X$ and similarly for $f^{-1}{restriction_{B}}(F)$, as the union of two closed sets is closed.
$$X backslash f^{-1}{restriction_{A}}(F) = X backslash (A backslash (bigcup_limits{{i in I}}U_i cap A)) $$
$$= X backslash ((A backslash bigcup_limits{{i in I}}U_i) cup (A backslash A) $$
$$= (X backslash A) cup bigcup_limits{{i in I}}U_i $$
Which is just the union of open sets in X (since A is closed by assumption).
Hence $X backslash f^{-1}{restriction_{A}}(F)$ is open $Rightarrow f^{-1}{restriction_{A}}(F)$ is closed.
One can show $f^{-1}{restriction_{B}}(F)$ is closed similarly, thereby completing the proof.
general-topology proof-verification
asked Nov 20 '18 at 12:52
Eugaurie
61
add a comment |
Let $A,B subseteq X$ be closed in $X$ with $A cup B = X$.
Suppose there exists a function, $f: X rightarrow Y$, with both
$f {restriction_{A}}: A rightarrow Y$ and $ f {restriction_{B}}: B rightarrow Y$ continuous.
Prove that $f$ is continuous.
Proposed proof:
It is sufficient to show that if $F subseteq Y$ is closed in $Y$, then $f^{-1}(F)$ is closed in $X$.
Let $F subseteq Y$ be closed in $Y$.
By continuity of $f{restriction_{A}}$ and $f{restriction_{B}}$, one has it that $f^{-1}{restriction_{A}}(F)$ and $f^{-1}{restriction_{B}}(F)$ are closed in their subspace topologies.
I.e. $f^{-1}{restriction_{A}}(F) = A backslash (bigcup_limits{{i in I}}U_i cap A) $ and similarly, $f^{-1}{restriction_{B}}(F) = B backslash (bigcup_limits{{j in J}}U_j cap B) $
Also being restrictions of $f$ we have that:
$$f^{-1}{restriction_{A}}(F) = f^{-1}(F) cap A$$
$$f^{-1}{restriction_{B}}(F) = f^{-1}(F) cap B$$
$Rightarrow f^{-1}{restriction_{A}}(F) cup f^{-1}{restriction_{B}}(F) = (f^{-1}(F) cap B) cup (f^{-1}(F) cap A) = f^{-1}(F) cap (A cup B) = f^{-1}(F)$
Since $A cup B = X$.
It now suffices to show that $f^{-1}{restriction_{A}}(F) = A backslash (bigcup_limits{{i in I}}U_i cap A) $ is closed in $X$ and similarly for $f^{-1}{restriction_{B}}(F)$, as the union of two closed sets is closed.
$$X backslash f^{-1}{restriction_{A}}(F) = X backslash (A backslash (bigcup_limits{{i in I}}U_i cap A)) $$
$$= X backslash ((A backslash bigcup_limits{{i in I}}U_i) cup (A backslash A) $$
$$= (X backslash A) cup bigcup_limits{{i in I}}U_i $$
Which is just the union of open sets in X (since A is closed by assumption).
Hence $X backslash f^{-1}{restriction_{A}}(F)$ is open $Rightarrow f^{-1}{restriction_{A}}(F)$ is closed.
One can show $f^{-1}{restriction_{B}}(F)$ is closed similarly, thereby completing the proof.
general-topology proof-verification
asked Nov 20 '18 at 12:52
Eugaurie
61
Let $A,B subseteq X$ be closed in $X$ with $A cup B = X$.
Suppose there exists a function, $f: X rightarrow Y$, with both
$f {restriction_{A}}: A rightarrow Y$ and $ f {restriction_{B}}: B rightarrow Y$ continuous.
Prove that $f$ is continuous.
Proposed proof:
It is sufficient to show that if $F subseteq Y$ is closed in $Y$, then $f^{-1}(F)$ is closed in $X$.
Let $F subseteq Y$ be closed in $Y$.
By continuity of $f{restriction_{A}}$ and $f{restriction_{B}}$, one has it that $f^{-1}{restriction_{A}}(F)$ and $f^{-1}{restriction_{B}}(F)$ are closed in their subspace topologies.
I.e. $f^{-1}{restriction_{A}}(F) = A backslash (bigcup_limits{{i in I}}U_i cap A) $ and similarly, $f^{-1}{restriction_{B}}(F) = B backslash (bigcup_limits{{j in J}}U_j cap B) $
Also being restrictions of $f$ we have that:
$$f^{-1}{restriction_{A}}(F) = f^{-1}(F) cap A$$
$$f^{-1}{restriction_{B}}(F) = f^{-1}(F) cap B$$
$Rightarrow f^{-1}{restriction_{A}}(F) cup f^{-1}{restriction_{B}}(F) = (f^{-1}(F) cap B) cup (f^{-1}(F) cap A) = f^{-1}(F) cap (A cup B) = f^{-1}(F)$
Since $A cup B = X$.
It now suffices to show that $f^{-1}{restriction_{A}}(F) = A backslash (bigcup_limits{{i in I}}U_i cap A) $ is closed in $X$ and similarly for $f^{-1}{restriction_{B}}(F)$, as the union of two closed sets is closed.
$$X backslash f^{-1}{restriction_{A}}(F) = X backslash (A backslash (bigcup_limits{{i in I}}U_i cap A)) $$
$$= X backslash ((A backslash bigcup_limits{{i in I}}U_i) cup (A backslash A) $$
$$= (X backslash A) cup bigcup_limits{{i in I}}U_i $$
Which is just the union of open sets in X (since A is closed by assumption).
Hence $X backslash f^{-1}{restriction_{A}}(F)$ is open $Rightarrow f^{-1}{restriction_{A}}(F)$ is closed.
One can show $f^{-1}{restriction_{B}}(F)$ is closed similarly, thereby completing the proof.
general-topology proof-verification
general-topology proof-verification
asked Nov 20 '18 at 12:52
Eugaurie
61
asked Nov 20 '18 at 12:52
Eugaurie
61
asked Nov 20 '18 at 12:52
Eugaurie
61
asked Nov 20 '18 at 12:52
Eugaurie
61
asked Nov 20 '18 at 12:52
Eugaurie
61
61
add a comment |
add a comment |
2 Answers
2
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oldest
votes
What is the meaning of the family $(U_i)_{i in J}$ ???
And why so complicated ??
If $f^{-1}{restriction_{A}}(F)$ is closed in the subspace topology of $A$, then there is a closed subset $Z$ of $X$ such that $f^{-1}{restriction_{A}}(F)=A cap Z$ .
Since $A$ is closed (in $X$), we have that $f^{-1}{restriction_{A}}(F)=A cap Z$ is closed in $X$.
answered Nov 20 '18 at 13:14
Fred
44.2k1845
I mean to say that if a set is open in the subspace topology for A then it must be a union of open sets from the wider space intersect A. I don't know what these sets are per-se so I just call them a family. I certainly agree that it's not the most elegant proof. I think what you've written is basically the same as the proof, only you've avoided dealing with open sets and renamed the complement of the family of open sets in $X$, $Z$
– Eugaurie
Nov 20 '18 at 13:21
What I mean is: Since $A$ is closed in $X$, we have for a subset $C$ of $A$: $C$ is closed in $A iff C$ is closed in $X$.
– Fred
Nov 20 '18 at 13:27
Ah, I see. So does that basically just allow me to skip out the whole section of my proof wherein I prove that $C$ closed in $A$ implies $C$ closed in $X$?
– Eugaurie
Nov 20 '18 at 13:35
Yes ! Try a prof for $C$ is closed in $A iff C$ is closed in $X$. This is easy, since $A$ is closed (in $X$).
– Fred
Nov 20 '18 at 13:42
Thank you so much!
– Eugaurie
Nov 20 '18 at 13:42
add a comment |
If $R$ is a subset of (the closed set) $A$ that is closed in $A$ then also it will be closed in $X$.
This because $R$ is closed in $A$ iff it can be written as $Acap S$ where $S$ is a closed subset of $X$.
So $R$ appears to be the intersection of two sets that are both closed in $X$, hence $R$ is closed in $X$.
This can be applied here.
If $F$ is a closed subset of $Y$ then: $$f^{-1}(F)=(f{restriction_{A}})^{-1}(F)cup(f{restriction_{B}})^{-1}(F)$$where $(f{restriction_{A}})^{-1}(F)$ is closed in $A$ hence closed in $X$ and $(f{restriction_{B}})^{-1}(F)$ is closed in $B$ hence closed in $X$.
Then as a union of two sets both closed in $X$ the preimage $f^{-1}(F)$ is closed in $X$.
answered Nov 20 '18 at 13:30
drhab
97.9k544129
I think that's essentially what I seek to do, only I prove that: $f^{-1}{restriction_{A}}(F)$ closed in $A Rightarrow$ $f^{-1}{restriction_{A}}(F)$ closed in $X$
– Eugaurie
Nov 20 '18 at 13:32
Well, that is actually the first part of my answer. But more straightforward and avoiding open sets.
– drhab
Nov 20 '18 at 13:39
I see, that is much more straightforward. Thank you, I had not seen this before.
– Eugaurie
Nov 20 '18 at 13:42
add a comment |
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2 Answers
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2 Answers
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What is the meaning of the family $(U_i)_{i in J}$ ???
And why so complicated ??
If $f^{-1}{restriction_{A}}(F)$ is closed in the subspace topology of $A$, then there is a closed subset $Z$ of $X$ such that $f^{-1}{restriction_{A}}(F)=A cap Z$ .
Since $A$ is closed (in $X$), we have that $f^{-1}{restriction_{A}}(F)=A cap Z$ is closed in $X$.
answered Nov 20 '18 at 13:14
Fred
44.2k1845
I mean to say that if a set is open in the subspace topology for A then it must be a union of open sets from the wider space intersect A. I don't know what these sets are per-se so I just call them a family. I certainly agree that it's not the most elegant proof. I think what you've written is basically the same as the proof, only you've avoided dealing with open sets and renamed the complement of the family of open sets in $X$, $Z$
– Eugaurie
Nov 20 '18 at 13:21
What I mean is: Since $A$ is closed in $X$, we have for a subset $C$ of $A$: $C$ is closed in $A iff C$ is closed in $X$.
– Fred
Nov 20 '18 at 13:27
Ah, I see. So does that basically just allow me to skip out the whole section of my proof wherein I prove that $C$ closed in $A$ implies $C$ closed in $X$?
– Eugaurie
Nov 20 '18 at 13:35
Yes ! Try a prof for $C$ is closed in $A iff C$ is closed in $X$. This is easy, since $A$ is closed (in $X$).
– Fred
Nov 20 '18 at 13:42
Thank you so much!
– Eugaurie
Nov 20 '18 at 13:42
add a comment |
What is the meaning of the family $(U_i)_{i in J}$ ???
And why so complicated ??
If $f^{-1}{restriction_{A}}(F)$ is closed in the subspace topology of $A$, then there is a closed subset $Z$ of $X$ such that $f^{-1}{restriction_{A}}(F)=A cap Z$ .
Since $A$ is closed (in $X$), we have that $f^{-1}{restriction_{A}}(F)=A cap Z$ is closed in $X$.
answered Nov 20 '18 at 13:14
Fred
44.2k1845
I mean to say that if a set is open in the subspace topology for A then it must be a union of open sets from the wider space intersect A. I don't know what these sets are per-se so I just call them a family. I certainly agree that it's not the most elegant proof. I think what you've written is basically the same as the proof, only you've avoided dealing with open sets and renamed the complement of the family of open sets in $X$, $Z$
– Eugaurie
Nov 20 '18 at 13:21
What I mean is: Since $A$ is closed in $X$, we have for a subset $C$ of $A$: $C$ is closed in $A iff C$ is closed in $X$.
– Fred
Nov 20 '18 at 13:27
Ah, I see. So does that basically just allow me to skip out the whole section of my proof wherein I prove that $C$ closed in $A$ implies $C$ closed in $X$?
– Eugaurie
Nov 20 '18 at 13:35
Yes ! Try a prof for $C$ is closed in $A iff C$ is closed in $X$. This is easy, since $A$ is closed (in $X$).
– Fred
Nov 20 '18 at 13:42
Thank you so much!
– Eugaurie
Nov 20 '18 at 13:42
add a comment |
What is the meaning of the family $(U_i)_{i in J}$ ???
And why so complicated ??
If $f^{-1}{restriction_{A}}(F)$ is closed in the subspace topology of $A$, then there is a closed subset $Z$ of $X$ such that $f^{-1}{restriction_{A}}(F)=A cap Z$ .
Since $A$ is closed (in $X$), we have that $f^{-1}{restriction_{A}}(F)=A cap Z$ is closed in $X$.
answered Nov 20 '18 at 13:14
Fred
44.2k1845
What is the meaning of the family $(U_i)_{i in J}$ ???
And why so complicated ??
If $f^{-1}{restriction_{A}}(F)$ is closed in the subspace topology of $A$, then there is a closed subset $Z$ of $X$ such that $f^{-1}{restriction_{A}}(F)=A cap Z$ .
Since $A$ is closed (in $X$), we have that $f^{-1}{restriction_{A}}(F)=A cap Z$ is closed in $X$.
answered Nov 20 '18 at 13:14
Fred
44.2k1845
answered Nov 20 '18 at 13:14
Fred
44.2k1845
answered Nov 20 '18 at 13:14
Fred
44.2k1845
answered Nov 20 '18 at 13:14
Fred
44.2k1845
44.2k1845
I mean to say that if a set is open in the subspace topology for A then it must be a union of open sets from the wider space intersect A. I don't know what these sets are per-se so I just call them a family. I certainly agree that it's not the most elegant proof. I think what you've written is basically the same as the proof, only you've avoided dealing with open sets and renamed the complement of the family of open sets in $X$, $Z$
– Eugaurie
Nov 20 '18 at 13:21
What I mean is: Since $A$ is closed in $X$, we have for a subset $C$ of $A$: $C$ is closed in $A iff C$ is closed in $X$.
– Fred
Nov 20 '18 at 13:27
Ah, I see. So does that basically just allow me to skip out the whole section of my proof wherein I prove that $C$ closed in $A$ implies $C$ closed in $X$?
– Eugaurie
Nov 20 '18 at 13:35
Yes ! Try a prof for $C$ is closed in $A iff C$ is closed in $X$. This is easy, since $A$ is closed (in $X$).
– Fred
Nov 20 '18 at 13:42
Thank you so much!
– Eugaurie
Nov 20 '18 at 13:42
add a comment |
I mean to say that if a set is open in the subspace topology for A then it must be a union of open sets from the wider space intersect A. I don't know what these sets are per-se so I just call them a family. I certainly agree that it's not the most elegant proof. I think what you've written is basically the same as the proof, only you've avoided dealing with open sets and renamed the complement of the family of open sets in $X$, $Z$
– Eugaurie
Nov 20 '18 at 13:21
What I mean is: Since $A$ is closed in $X$, we have for a subset $C$ of $A$: $C$ is closed in $A iff C$ is closed in $X$.
– Fred
Nov 20 '18 at 13:27
Ah, I see. So does that basically just allow me to skip out the whole section of my proof wherein I prove that $C$ closed in $A$ implies $C$ closed in $X$?
– Eugaurie
Nov 20 '18 at 13:35
Yes ! Try a prof for $C$ is closed in $A iff C$ is closed in $X$. This is easy, since $A$ is closed (in $X$).
– Fred
Nov 20 '18 at 13:42
Thank you so much!
– Eugaurie
Nov 20 '18 at 13:42
I mean to say that if a set is open in the subspace topology for A then it must be a union of open sets from the wider space intersect A. I don't know what these sets are per-se so I just call them a family. I certainly agree that it's not the most elegant proof. I think what you've written is basically the same as the proof, only you've avoided dealing with open sets and renamed the complement of the family of open sets in $X$, $Z$
– Eugaurie
Nov 20 '18 at 13:21
I mean to say that if a set is open in the subspace topology for A then it must be a union of open sets from the wider space intersect A. I don't know what these sets are per-se so I just call them a family. I certainly agree that it's not the most elegant proof. I think what you've written is basically the same as the proof, only you've avoided dealing with open sets and renamed the complement of the family of open sets in $X$, $Z$
– Eugaurie
Nov 20 '18 at 13:21
What I mean is: Since $A$ is closed in $X$, we have for a subset $C$ of $A$: $C$ is closed in $A iff C$ is closed in $X$.
– Fred
Nov 20 '18 at 13:27
What I mean is: Since $A$ is closed in $X$, we have for a subset $C$ of $A$: $C$ is closed in $A iff C$ is closed in $X$.
– Fred
Nov 20 '18 at 13:27
Ah, I see. So does that basically just allow me to skip out the whole section of my proof wherein I prove that $C$ closed in $A$ implies $C$ closed in $X$?
– Eugaurie
Nov 20 '18 at 13:35
Ah, I see. So does that basically just allow me to skip out the whole section of my proof wherein I prove that $C$ closed in $A$ implies $C$ closed in $X$?
– Eugaurie
Nov 20 '18 at 13:35
Yes ! Try a prof for $C$ is closed in $A iff C$ is closed in $X$. This is easy, since $A$ is closed (in $X$).
– Fred
Nov 20 '18 at 13:42
Yes ! Try a prof for $C$ is closed in $A iff C$ is closed in $X$. This is easy, since $A$ is closed (in $X$).
– Fred
Nov 20 '18 at 13:42
Thank you so much!
– Eugaurie
Nov 20 '18 at 13:42
Thank you so much!
– Eugaurie
Nov 20 '18 at 13:42
add a comment |
If $R$ is a subset of (the closed set) $A$ that is closed in $A$ then also it will be closed in $X$.
This because $R$ is closed in $A$ iff it can be written as $Acap S$ where $S$ is a closed subset of $X$.
So $R$ appears to be the intersection of two sets that are both closed in $X$, hence $R$ is closed in $X$.
This can be applied here.
If $F$ is a closed subset of $Y$ then: $$f^{-1}(F)=(f{restriction_{A}})^{-1}(F)cup(f{restriction_{B}})^{-1}(F)$$where $(f{restriction_{A}})^{-1}(F)$ is closed in $A$ hence closed in $X$ and $(f{restriction_{B}})^{-1}(F)$ is closed in $B$ hence closed in $X$.
Then as a union of two sets both closed in $X$ the preimage $f^{-1}(F)$ is closed in $X$.
answered Nov 20 '18 at 13:30
drhab
97.9k544129
I think that's essentially what I seek to do, only I prove that: $f^{-1}{restriction_{A}}(F)$ closed in $A Rightarrow$ $f^{-1}{restriction_{A}}(F)$ closed in $X$
– Eugaurie
Nov 20 '18 at 13:32
Well, that is actually the first part of my answer. But more straightforward and avoiding open sets.
– drhab
Nov 20 '18 at 13:39
I see, that is much more straightforward. Thank you, I had not seen this before.
– Eugaurie
Nov 20 '18 at 13:42
add a comment |
If $R$ is a subset of (the closed set) $A$ that is closed in $A$ then also it will be closed in $X$.
This because $R$ is closed in $A$ iff it can be written as $Acap S$ where $S$ is a closed subset of $X$.
So $R$ appears to be the intersection of two sets that are both closed in $X$, hence $R$ is closed in $X$.
This can be applied here.
If $F$ is a closed subset of $Y$ then: $$f^{-1}(F)=(f{restriction_{A}})^{-1}(F)cup(f{restriction_{B}})^{-1}(F)$$where $(f{restriction_{A}})^{-1}(F)$ is closed in $A$ hence closed in $X$ and $(f{restriction_{B}})^{-1}(F)$ is closed in $B$ hence closed in $X$.
Then as a union of two sets both closed in $X$ the preimage $f^{-1}(F)$ is closed in $X$.
answered Nov 20 '18 at 13:30
drhab
97.9k544129
I think that's essentially what I seek to do, only I prove that: $f^{-1}{restriction_{A}}(F)$ closed in $A Rightarrow$ $f^{-1}{restriction_{A}}(F)$ closed in $X$
– Eugaurie
Nov 20 '18 at 13:32
Well, that is actually the first part of my answer. But more straightforward and avoiding open sets.
– drhab
Nov 20 '18 at 13:39
I see, that is much more straightforward. Thank you, I had not seen this before.
– Eugaurie
Nov 20 '18 at 13:42
add a comment |
If $R$ is a subset of (the closed set) $A$ that is closed in $A$ then also it will be closed in $X$.
This because $R$ is closed in $A$ iff it can be written as $Acap S$ where $S$ is a closed subset of $X$.
So $R$ appears to be the intersection of two sets that are both closed in $X$, hence $R$ is closed in $X$.
This can be applied here.
If $F$ is a closed subset of $Y$ then: $$f^{-1}(F)=(f{restriction_{A}})^{-1}(F)cup(f{restriction_{B}})^{-1}(F)$$where $(f{restriction_{A}})^{-1}(F)$ is closed in $A$ hence closed in $X$ and $(f{restriction_{B}})^{-1}(F)$ is closed in $B$ hence closed in $X$.
Then as a union of two sets both closed in $X$ the preimage $f^{-1}(F)$ is closed in $X$.
answered Nov 20 '18 at 13:30
drhab
97.9k544129
If $R$ is a subset of (the closed set) $A$ that is closed in $A$ then also it will be closed in $X$.
This because $R$ is closed in $A$ iff it can be written as $Acap S$ where $S$ is a closed subset of $X$.
So $R$ appears to be the intersection of two sets that are both closed in $X$, hence $R$ is closed in $X$.
This can be applied here.
If $F$ is a closed subset of $Y$ then: $$f^{-1}(F)=(f{restriction_{A}})^{-1}(F)cup(f{restriction_{B}})^{-1}(F)$$where $(f{restriction_{A}})^{-1}(F)$ is closed in $A$ hence closed in $X$ and $(f{restriction_{B}})^{-1}(F)$ is closed in $B$ hence closed in $X$.
Then as a union of two sets both closed in $X$ the preimage $f^{-1}(F)$ is closed in $X$.
answered Nov 20 '18 at 13:30
drhab
97.9k544129
edited Nov 20 '18 at 14:25
answered Nov 20 '18 at 13:30
drhab
97.9k544129
answered Nov 20 '18 at 13:30
drhab
97.9k544129
answered Nov 20 '18 at 13:30
drhab
97.9k544129
97.9k544129
I think that's essentially what I seek to do, only I prove that: $f^{-1}{restriction_{A}}(F)$ closed in $A Rightarrow$ $f^{-1}{restriction_{A}}(F)$ closed in $X$
– Eugaurie
Nov 20 '18 at 13:32
Well, that is actually the first part of my answer. But more straightforward and avoiding open sets.
– drhab
Nov 20 '18 at 13:39
I see, that is much more straightforward. Thank you, I had not seen this before.
– Eugaurie
Nov 20 '18 at 13:42
add a comment |
I think that's essentially what I seek to do, only I prove that: $f^{-1}{restriction_{A}}(F)$ closed in $A Rightarrow$ $f^{-1}{restriction_{A}}(F)$ closed in $X$
– Eugaurie
Nov 20 '18 at 13:32
Well, that is actually the first part of my answer. But more straightforward and avoiding open sets.
– drhab
Nov 20 '18 at 13:39
I see, that is much more straightforward. Thank you, I had not seen this before.
– Eugaurie
Nov 20 '18 at 13:42
I think that's essentially what I seek to do, only I prove that: $f^{-1}{restriction_{A}}(F)$ closed in $A Rightarrow$ $f^{-1}{restriction_{A}}(F)$ closed in $X$
– Eugaurie
Nov 20 '18 at 13:32
I think that's essentially what I seek to do, only I prove that: $f^{-1}{restriction_{A}}(F)$ closed in $A Rightarrow$ $f^{-1}{restriction_{A}}(F)$ closed in $X$
– Eugaurie
Nov 20 '18 at 13:32
Well, that is actually the first part of my answer. But more straightforward and avoiding open sets.
– drhab
Nov 20 '18 at 13:39
Well, that is actually the first part of my answer. But more straightforward and avoiding open sets.
– drhab
Nov 20 '18 at 13:39
I see, that is much more straightforward. Thank you, I had not seen this before.
– Eugaurie
Nov 20 '18 at 13:42
I see, that is much more straightforward. Thank you, I had not seen this before.
– Eugaurie
Nov 20 '18 at 13:42
add a comment |
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