Mixing
Continuous function (Topology)
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Theorem. Let $f:Xrightarrow{Y}$ be a continuous function. They are equivalent: $f(overline{A})subseteqoverline{f(A)}$ for all $Asubseteq X$ iff $f^{-1}(B^{circ})subseteq(f^{-1}(B))^{circ}$ for all $Bsubseteq Y$
Proof: Let $ Bsubseteq{Y} $ be. Let's see that $f^{-1}(B^{circ})subseteq{(f^{-1} (B))^{circ}} $. Note that $ B^{circ} = Ysetminus{(overline{Ysetminus {B}})} $. Then $$ f^{-1}(B^{circ})= f^{- 1}(Ysetminus {(overline {Ysetminus{B}})}) = X setminus{f^{-1} (overline {Ysetminus{B}})} $$ On the other hand, we have $$ (f^{- 1}(B))^{circ} = Xsetminus {overline { (X setminus {f^{- 1} (B)}})} = X setminus {overline {f ^ {- 1} (Y setminus {B})}} $$ Now, let's see that $ overline {f ^ {- 1} (Y setminus {B})} subseteq {f ^ {- 1} (overline {Y setminus {B}})} $. Let $ A = f ^ {- 1} (Y setminus {B}) subseteq {X} $ be. By hypothesis, we have $$ f (overline {f ^ {- 1} (Y setminus {B})}) subseteq {overline {f (f ^ {- 1} (Y setminus {B})) }} subseteq {overline {Y setminus {B}}} $$ For the above, we have $$ overline {f ^ {- 1} (Y setminus {B})} subseteq {f ^ {- 1} (f (overline {f ^ {- 1} (Y setminus {B})}))} subseteq {f ^ {- 1} (overline {Y setminus {B}})} $$ Therefore, $$ f ^ {- 1} (B ^ {circ}) subseteq {X setminus {f ^ {- 1} (overline {Y setminus {B}})}} subseteq { X setminus {overline {f ^ {- 1} (Y setminus {B})}}} = {(f ^ {- 1} (B)) ^ {circ}} $$ Reciprocally, Let $ A subseteq {X} $ be. Let's see that $ f (overline {A}) subseteq {overline {f (A)}} $. Note that $ overline {A} = X setminus {(X setminus {A}) ^ {circ}} $. Then $$ f (overline {A}) = f (X setminus {(X setminus {A}) ^ {circ}}) subseteq {Y setminus {f ((X setminus {A}) ^ {circ})}} $$ On the other hand, we have $$ overline {f (A)} = Y setminus {(Y setminus {f (A)}) ^ {circ}} $$ Now , let's see that $ (Y setminus {f (A)}) ^ {circ} subseteq {f ((X setminus {A}) ^ {circ})} $. Let $ B = Y setminus {f (A)} subseteq {Y} $ be. By hypothesis, we have $$ f ^ {- 1} ((Y setminus {f (A)}) ^ {circ}) subseteq {(f ^ {- 1} (Y setminus {f (A)} )) ^ {circ}} $$ Note that $ f ^ {- 1} (Y setminus {f (A)}) = X setminus {f ^ {- 1} (f (A))} subseteq {X setminus {A}} $. Then $$ f ^ {- 1} ((Y setminus {f (A)}) ^ {circ}) subseteq {(X setminus {A}) ^ {circ}} $$ Thus, $ f (f ^ {- 1} ((Y setminus {f (A)})^ {circ})) subseteq {f ((X setminus {A})^ {circ})} $
I don't know how to conclude...
general-topology functions continuity
asked Jan 28 at 4:52
DarkmasterDarkmaster
848
$endgroup$
add a comment |
$begingroup$
Theorem. Let $f:Xrightarrow{Y}$ be a continuous function. They are equivalent: $f(overline{A})subseteqoverline{f(A)}$ for all $Asubseteq X$ iff $f^{-1}(B^{circ})subseteq(f^{-1}(B))^{circ}$ for all $Bsubseteq Y$
Proof: Let $ Bsubseteq{Y} $ be. Let's see that $f^{-1}(B^{circ})subseteq{(f^{-1} (B))^{circ}} $. Note that $ B^{circ} = Ysetminus{(overline{Ysetminus {B}})} $. Then $$ f^{-1}(B^{circ})= f^{- 1}(Ysetminus {(overline {Ysetminus{B}})}) = X setminus{f^{-1} (overline {Ysetminus{B}})} $$ On the other hand, we have $$ (f^{- 1}(B))^{circ} = Xsetminus {overline { (X setminus {f^{- 1} (B)}})} = X setminus {overline {f ^ {- 1} (Y setminus {B})}} $$ Now, let's see that $ overline {f ^ {- 1} (Y setminus {B})} subseteq {f ^ {- 1} (overline {Y setminus {B}})} $. Let $ A = f ^ {- 1} (Y setminus {B}) subseteq {X} $ be. By hypothesis, we have $$ f (overline {f ^ {- 1} (Y setminus {B})}) subseteq {overline {f (f ^ {- 1} (Y setminus {B})) }} subseteq {overline {Y setminus {B}}} $$ For the above, we have $$ overline {f ^ {- 1} (Y setminus {B})} subseteq {f ^ {- 1} (f (overline {f ^ {- 1} (Y setminus {B})}))} subseteq {f ^ {- 1} (overline {Y setminus {B}})} $$ Therefore, $$ f ^ {- 1} (B ^ {circ}) subseteq {X setminus {f ^ {- 1} (overline {Y setminus {B}})}} subseteq { X setminus {overline {f ^ {- 1} (Y setminus {B})}}} = {(f ^ {- 1} (B)) ^ {circ}} $$ Reciprocally, Let $ A subseteq {X} $ be. Let's see that $ f (overline {A}) subseteq {overline {f (A)}} $. Note that $ overline {A} = X setminus {(X setminus {A}) ^ {circ}} $. Then $$ f (overline {A}) = f (X setminus {(X setminus {A}) ^ {circ}}) subseteq {Y setminus {f ((X setminus {A}) ^ {circ})}} $$ On the other hand, we have $$ overline {f (A)} = Y setminus {(Y setminus {f (A)}) ^ {circ}} $$ Now , let's see that $ (Y setminus {f (A)}) ^ {circ} subseteq {f ((X setminus {A}) ^ {circ})} $. Let $ B = Y setminus {f (A)} subseteq {Y} $ be. By hypothesis, we have $$ f ^ {- 1} ((Y setminus {f (A)}) ^ {circ}) subseteq {(f ^ {- 1} (Y setminus {f (A)} )) ^ {circ}} $$ Note that $ f ^ {- 1} (Y setminus {f (A)}) = X setminus {f ^ {- 1} (f (A))} subseteq {X setminus {A}} $. Then $$ f ^ {- 1} ((Y setminus {f (A)}) ^ {circ}) subseteq {(X setminus {A}) ^ {circ}} $$ Thus, $ f (f ^ {- 1} ((Y setminus {f (A)})^ {circ})) subseteq {f ((X setminus {A})^ {circ})} $
I don't know how to conclude...
general-topology functions continuity
asked Jan 28 at 4:52
DarkmasterDarkmaster
848
$endgroup$
1
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The two conditions are both equivalent to the continuity of $f$.
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– Eclipse Sun
Jan 28 at 4:59
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@EclipseSun I know. I just want to know if there is a demonstration of that proposition as I wrote it.
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– Darkmaster
Jan 28 at 5:04
1
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I meant that you don't need to assume $f$ to be continuous at the very beginning.
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– Eclipse Sun
Jan 28 at 5:07
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@EclipseSun You're right. Leaving that aside, do you know how to do the proof of that theorem? Leaving aside the continuity ...
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– Darkmaster
Jan 28 at 5:10
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cf this question for one of these properties.
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– Henno Brandsma
Jan 28 at 5:45
add a comment |
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Theorem. Let $f:Xrightarrow{Y}$ be a continuous function. They are equivalent: $f(overline{A})subseteqoverline{f(A)}$ for all $Asubseteq X$ iff $f^{-1}(B^{circ})subseteq(f^{-1}(B))^{circ}$ for all $Bsubseteq Y$
Proof: Let $ Bsubseteq{Y} $ be. Let's see that $f^{-1}(B^{circ})subseteq{(f^{-1} (B))^{circ}} $. Note that $ B^{circ} = Ysetminus{(overline{Ysetminus {B}})} $. Then $$ f^{-1}(B^{circ})= f^{- 1}(Ysetminus {(overline {Ysetminus{B}})}) = X setminus{f^{-1} (overline {Ysetminus{B}})} $$ On the other hand, we have $$ (f^{- 1}(B))^{circ} = Xsetminus {overline { (X setminus {f^{- 1} (B)}})} = X setminus {overline {f ^ {- 1} (Y setminus {B})}} $$ Now, let's see that $ overline {f ^ {- 1} (Y setminus {B})} subseteq {f ^ {- 1} (overline {Y setminus {B}})} $. Let $ A = f ^ {- 1} (Y setminus {B}) subseteq {X} $ be. By hypothesis, we have $$ f (overline {f ^ {- 1} (Y setminus {B})}) subseteq {overline {f (f ^ {- 1} (Y setminus {B})) }} subseteq {overline {Y setminus {B}}} $$ For the above, we have $$ overline {f ^ {- 1} (Y setminus {B})} subseteq {f ^ {- 1} (f (overline {f ^ {- 1} (Y setminus {B})}))} subseteq {f ^ {- 1} (overline {Y setminus {B}})} $$ Therefore, $$ f ^ {- 1} (B ^ {circ}) subseteq {X setminus {f ^ {- 1} (overline {Y setminus {B}})}} subseteq { X setminus {overline {f ^ {- 1} (Y setminus {B})}}} = {(f ^ {- 1} (B)) ^ {circ}} $$ Reciprocally, Let $ A subseteq {X} $ be. Let's see that $ f (overline {A}) subseteq {overline {f (A)}} $. Note that $ overline {A} = X setminus {(X setminus {A}) ^ {circ}} $. Then $$ f (overline {A}) = f (X setminus {(X setminus {A}) ^ {circ}}) subseteq {Y setminus {f ((X setminus {A}) ^ {circ})}} $$ On the other hand, we have $$ overline {f (A)} = Y setminus {(Y setminus {f (A)}) ^ {circ}} $$ Now , let's see that $ (Y setminus {f (A)}) ^ {circ} subseteq {f ((X setminus {A}) ^ {circ})} $. Let $ B = Y setminus {f (A)} subseteq {Y} $ be. By hypothesis, we have $$ f ^ {- 1} ((Y setminus {f (A)}) ^ {circ}) subseteq {(f ^ {- 1} (Y setminus {f (A)} )) ^ {circ}} $$ Note that $ f ^ {- 1} (Y setminus {f (A)}) = X setminus {f ^ {- 1} (f (A))} subseteq {X setminus {A}} $. Then $$ f ^ {- 1} ((Y setminus {f (A)}) ^ {circ}) subseteq {(X setminus {A}) ^ {circ}} $$ Thus, $ f (f ^ {- 1} ((Y setminus {f (A)})^ {circ})) subseteq {f ((X setminus {A})^ {circ})} $
I don't know how to conclude...
general-topology functions continuity
asked Jan 28 at 4:52
DarkmasterDarkmaster
848
$endgroup$
Theorem. Let $f:Xrightarrow{Y}$ be a continuous function. They are equivalent: $f(overline{A})subseteqoverline{f(A)}$ for all $Asubseteq X$ iff $f^{-1}(B^{circ})subseteq(f^{-1}(B))^{circ}$ for all $Bsubseteq Y$
Proof: Let $ Bsubseteq{Y} $ be. Let's see that $f^{-1}(B^{circ})subseteq{(f^{-1} (B))^{circ}} $. Note that $ B^{circ} = Ysetminus{(overline{Ysetminus {B}})} $. Then $$ f^{-1}(B^{circ})= f^{- 1}(Ysetminus {(overline {Ysetminus{B}})}) = X setminus{f^{-1} (overline {Ysetminus{B}})} $$ On the other hand, we have $$ (f^{- 1}(B))^{circ} = Xsetminus {overline { (X setminus {f^{- 1} (B)}})} = X setminus {overline {f ^ {- 1} (Y setminus {B})}} $$ Now, let's see that $ overline {f ^ {- 1} (Y setminus {B})} subseteq {f ^ {- 1} (overline {Y setminus {B}})} $. Let $ A = f ^ {- 1} (Y setminus {B}) subseteq {X} $ be. By hypothesis, we have $$ f (overline {f ^ {- 1} (Y setminus {B})}) subseteq {overline {f (f ^ {- 1} (Y setminus {B})) }} subseteq {overline {Y setminus {B}}} $$ For the above, we have $$ overline {f ^ {- 1} (Y setminus {B})} subseteq {f ^ {- 1} (f (overline {f ^ {- 1} (Y setminus {B})}))} subseteq {f ^ {- 1} (overline {Y setminus {B}})} $$ Therefore, $$ f ^ {- 1} (B ^ {circ}) subseteq {X setminus {f ^ {- 1} (overline {Y setminus {B}})}} subseteq { X setminus {overline {f ^ {- 1} (Y setminus {B})}}} = {(f ^ {- 1} (B)) ^ {circ}} $$ Reciprocally, Let $ A subseteq {X} $ be. Let's see that $ f (overline {A}) subseteq {overline {f (A)}} $. Note that $ overline {A} = X setminus {(X setminus {A}) ^ {circ}} $. Then $$ f (overline {A}) = f (X setminus {(X setminus {A}) ^ {circ}}) subseteq {Y setminus {f ((X setminus {A}) ^ {circ})}} $$ On the other hand, we have $$ overline {f (A)} = Y setminus {(Y setminus {f (A)}) ^ {circ}} $$ Now , let's see that $ (Y setminus {f (A)}) ^ {circ} subseteq {f ((X setminus {A}) ^ {circ})} $. Let $ B = Y setminus {f (A)} subseteq {Y} $ be. By hypothesis, we have $$ f ^ {- 1} ((Y setminus {f (A)}) ^ {circ}) subseteq {(f ^ {- 1} (Y setminus {f (A)} )) ^ {circ}} $$ Note that $ f ^ {- 1} (Y setminus {f (A)}) = X setminus {f ^ {- 1} (f (A))} subseteq {X setminus {A}} $. Then $$ f ^ {- 1} ((Y setminus {f (A)}) ^ {circ}) subseteq {(X setminus {A}) ^ {circ}} $$ Thus, $ f (f ^ {- 1} ((Y setminus {f (A)})^ {circ})) subseteq {f ((X setminus {A})^ {circ})} $
I don't know how to conclude...
general-topology functions continuity
general-topology functions continuity
asked Jan 28 at 4:52
DarkmasterDarkmaster
848
asked Jan 28 at 4:52
DarkmasterDarkmaster
848
edited Jan 29 at 11:21
Darkmaster
asked Jan 28 at 4:52
DarkmasterDarkmaster
848
asked Jan 28 at 4:52
DarkmasterDarkmaster
848
asked Jan 28 at 4:52
DarkmasterDarkmaster
848
848
1
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The two conditions are both equivalent to the continuity of $f$.
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– Eclipse Sun
Jan 28 at 4:59
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@EclipseSun I know. I just want to know if there is a demonstration of that proposition as I wrote it.
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– Darkmaster
Jan 28 at 5:04
1
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I meant that you don't need to assume $f$ to be continuous at the very beginning.
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– Eclipse Sun
Jan 28 at 5:07
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@EclipseSun You're right. Leaving that aside, do you know how to do the proof of that theorem? Leaving aside the continuity ...
$endgroup$
– Darkmaster
Jan 28 at 5:10
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cf this question for one of these properties.
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– Henno Brandsma
Jan 28 at 5:45
add a comment |
1
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The two conditions are both equivalent to the continuity of $f$.
$endgroup$
– Eclipse Sun
Jan 28 at 4:59
$begingroup$
@EclipseSun I know. I just want to know if there is a demonstration of that proposition as I wrote it.
$endgroup$
– Darkmaster
Jan 28 at 5:04
1
$begingroup$
I meant that you don't need to assume $f$ to be continuous at the very beginning.
$endgroup$
– Eclipse Sun
Jan 28 at 5:07
$begingroup$
@EclipseSun You're right. Leaving that aside, do you know how to do the proof of that theorem? Leaving aside the continuity ...
$endgroup$
– Darkmaster
Jan 28 at 5:10
$begingroup$
cf this question for one of these properties.
$endgroup$
– Henno Brandsma
Jan 28 at 5:45
1
1
$begingroup$
The two conditions are both equivalent to the continuity of $f$.
$endgroup$
– Eclipse Sun
Jan 28 at 4:59
$begingroup$
The two conditions are both equivalent to the continuity of $f$.
$endgroup$
– Eclipse Sun
Jan 28 at 4:59
$begingroup$
@EclipseSun I know. I just want to know if there is a demonstration of that proposition as I wrote it.
$endgroup$
– Darkmaster
Jan 28 at 5:04
$begingroup$
@EclipseSun I know. I just want to know if there is a demonstration of that proposition as I wrote it.
$endgroup$
– Darkmaster
Jan 28 at 5:04
1
1
$begingroup$
I meant that you don't need to assume $f$ to be continuous at the very beginning.
$endgroup$
– Eclipse Sun
Jan 28 at 5:07
$begingroup$
I meant that you don't need to assume $f$ to be continuous at the very beginning.
$endgroup$
– Eclipse Sun
Jan 28 at 5:07
$begingroup$
@EclipseSun You're right. Leaving that aside, do you know how to do the proof of that theorem? Leaving aside the continuity ...
$endgroup$
– Darkmaster
Jan 28 at 5:10
$begingroup$
@EclipseSun You're right. Leaving that aside, do you know how to do the proof of that theorem? Leaving aside the continuity ...
$endgroup$
– Darkmaster
Jan 28 at 5:10
$begingroup$
cf this question for one of these properties.
$endgroup$
– Henno Brandsma
Jan 28 at 5:45
$begingroup$
cf this question for one of these properties.
$endgroup$
– Henno Brandsma
Jan 28 at 5:45
add a comment |
2 Answers
2
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Referring to the link: Relation between interior and exterior.
Given that $f^{-1}$ exists, so $f$ is bijective and $f$ is given continuous so $f^{-1}$ is open.
Let, $Asubset X$ and $f(overline A)subset overline{f(A)}$ holds.
Now, let $Bsubset Y$ then $$B^circ=Y-overline{Y-B}\ implies f^{-1}(B^circ)=f^{-1}(Y-overline{Y-B})\=X-f^{-1}(overline{Y-B})\ subset X-overline {f^{-1}(Y-B)}\=X-overline{X-f^{-1}(B)}\=f^{-1}(B)^circ$$
Now repeat the same process by setting $overline A=X-(X-A)^circ$.
answered Jan 28 at 5:08
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
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It is not given that $f^{-1}$ exists, this is just the notation for the inverse image...
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– Henno Brandsma
Jan 28 at 5:28
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@SujitBhattacharyya $X-f^{-1}(overline{Y-B})subset X-overline {f^{-1}(Y-B)}$? Why?
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– Darkmaster
Jan 28 at 5:32
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@HennoBrandsma $f^{-1}$ is the inverse image of which map? $f$ of course. That's why I have written that $f^{-1}$ exists.
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– Sujit Bhattacharyya
Jan 28 at 5:44
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For any function $f:X to Y$, $f^{-1}(A)={x in X: f(x) in A}$ if $A subseteq Y$. No need for an inverse function.
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– Henno Brandsma
Jan 28 at 5:46
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@Darkmaster Set, $f(A)=Bimplies A=f^{-1}(B)$, given $f(overline A)subset overline {f(A)}implies overline Asubset f^{-1}(overline {f(A)})implies overline{f^{-1}(B)}subset f^{-1}(overline B)$. For similarity replace $B$ by $Y-B$ and observe its complement.
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– Sujit Bhattacharyya
Jan 28 at 5:48
|
show 4 more comments
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The following two statements about a function $f:X to Y$ between topological spaces are equivalent:
$forall A subseteq X: f[overline{A}] subseteq overline{f[A]}$.- For all closed $C subseteq Y$: $f^{-1}[C]$ is closed in $X$.
For 1 implies 2: let $C$ be closed in $Y$. Define $D = f^{-1}[C]$ and apply fact 1 to it: $$f[overline{D}] subseteq overline{f[D]} = overline{f[f^{-1}[C]]}subseteq overline{C} = C$$ as $C$ is closed. But this implies that $overline{D} subseteq f^{-1}[C] =D$ (all points of $overline{D}$ map into $C$) and so $D = f^{-1}[C]$ is closed, as required.
For 2 implies 1: Let $A subseteq X$ and note we have $$A subseteq f^{-1}[f[A]] subseteq f^{-1}[overline{f[A]}]$$ and by 2. the latter set is closed. Hence
$overline{A} subseteq f^{-1}[overline{f[A]}]$ as well and this means that $f[overline{A}]subseteq overline{f[A]}$ by the definitions of inverse images.
The following two statements about a function $f:X to Y$ between topological spaces are equivalent:
$forall B subseteq Y: f^{-1}[B^circ] subseteq f^{-1}[B]^circ$.- For all $O subseteq Y$ open, $f^{-1}[O]$ is open in $X$.
For 3 implies 4: let $O$ be open in $Y$, then $O^circ = O$ and by 3:
$$f^{-1}[O]= f^{-1}[O^circ] subseteq f^{-1}[O]^circ subseteq f^{-1}[O]$$
so $f^{-1}[O]^circ = f^{-1}[O]$ and $f^{-1}[O]$ is open.
For 4 implies 3: if $B subseteq Y$ is any subset $B^circ subseteq B$ so
also $f^{-1}[B^circ] subseteq f^{-1}[B]$. The former is an open subset (by 4) contained in $f^{-1}[B]$ and the interior of a set is its largest open subset so it follows that $f^{-1}[B^circ] subseteq f^{-1}[B]^circ$, as required.
Now note that 2 and 4 are clearly equivalent as $f^{-1}[Ysetminus A]= Xsetminus f^{-1}[A]$ for all subsets $B$ of $Y$, and open and closed sets are related via complements.
So all conditions are equivalent to $f$ being continuous.
answered Jan 28 at 9:45
Henno BrandsmaHenno Brandsma
114k348123
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@SujitBhattacharyya Thanks! You're great!
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– Darkmaster
Jan 28 at 16:21
1
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@Darkmaster did you mean as a comment to the other answer?
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– Henno Brandsma
Jan 28 at 16:51
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Referring to the link: Relation between interior and exterior.
Given that $f^{-1}$ exists, so $f$ is bijective and $f$ is given continuous so $f^{-1}$ is open.
Let, $Asubset X$ and $f(overline A)subset overline{f(A)}$ holds.
Now, let $Bsubset Y$ then $$B^circ=Y-overline{Y-B}\ implies f^{-1}(B^circ)=f^{-1}(Y-overline{Y-B})\=X-f^{-1}(overline{Y-B})\ subset X-overline {f^{-1}(Y-B)}\=X-overline{X-f^{-1}(B)}\=f^{-1}(B)^circ$$
Now repeat the same process by setting $overline A=X-(X-A)^circ$.
answered Jan 28 at 5:08
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
$begingroup$
It is not given that $f^{-1}$ exists, this is just the notation for the inverse image...
$endgroup$
– Henno Brandsma
Jan 28 at 5:28
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@SujitBhattacharyya $X-f^{-1}(overline{Y-B})subset X-overline {f^{-1}(Y-B)}$? Why?
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– Darkmaster
Jan 28 at 5:32
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@HennoBrandsma $f^{-1}$ is the inverse image of which map? $f$ of course. That's why I have written that $f^{-1}$ exists.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:44
$begingroup$
For any function $f:X to Y$, $f^{-1}(A)={x in X: f(x) in A}$ if $A subseteq Y$. No need for an inverse function.
$endgroup$
– Henno Brandsma
Jan 28 at 5:46
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@Darkmaster Set, $f(A)=Bimplies A=f^{-1}(B)$, given $f(overline A)subset overline {f(A)}implies overline Asubset f^{-1}(overline {f(A)})implies overline{f^{-1}(B)}subset f^{-1}(overline B)$. For similarity replace $B$ by $Y-B$ and observe its complement.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:48
|
show 4 more comments
$begingroup$
Referring to the link: Relation between interior and exterior.
Given that $f^{-1}$ exists, so $f$ is bijective and $f$ is given continuous so $f^{-1}$ is open.
Let, $Asubset X$ and $f(overline A)subset overline{f(A)}$ holds.
Now, let $Bsubset Y$ then $$B^circ=Y-overline{Y-B}\ implies f^{-1}(B^circ)=f^{-1}(Y-overline{Y-B})\=X-f^{-1}(overline{Y-B})\ subset X-overline {f^{-1}(Y-B)}\=X-overline{X-f^{-1}(B)}\=f^{-1}(B)^circ$$
Now repeat the same process by setting $overline A=X-(X-A)^circ$.
answered Jan 28 at 5:08
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
$begingroup$
It is not given that $f^{-1}$ exists, this is just the notation for the inverse image...
$endgroup$
– Henno Brandsma
Jan 28 at 5:28
$begingroup$
@SujitBhattacharyya $X-f^{-1}(overline{Y-B})subset X-overline {f^{-1}(Y-B)}$? Why?
$endgroup$
– Darkmaster
Jan 28 at 5:32
$begingroup$
@HennoBrandsma $f^{-1}$ is the inverse image of which map? $f$ of course. That's why I have written that $f^{-1}$ exists.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:44
$begingroup$
For any function $f:X to Y$, $f^{-1}(A)={x in X: f(x) in A}$ if $A subseteq Y$. No need for an inverse function.
$endgroup$
– Henno Brandsma
Jan 28 at 5:46
$begingroup$
@Darkmaster Set, $f(A)=Bimplies A=f^{-1}(B)$, given $f(overline A)subset overline {f(A)}implies overline Asubset f^{-1}(overline {f(A)})implies overline{f^{-1}(B)}subset f^{-1}(overline B)$. For similarity replace $B$ by $Y-B$ and observe its complement.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:48
|
show 4 more comments
$begingroup$
Referring to the link: Relation between interior and exterior.
Given that $f^{-1}$ exists, so $f$ is bijective and $f$ is given continuous so $f^{-1}$ is open.
Let, $Asubset X$ and $f(overline A)subset overline{f(A)}$ holds.
Now, let $Bsubset Y$ then $$B^circ=Y-overline{Y-B}\ implies f^{-1}(B^circ)=f^{-1}(Y-overline{Y-B})\=X-f^{-1}(overline{Y-B})\ subset X-overline {f^{-1}(Y-B)}\=X-overline{X-f^{-1}(B)}\=f^{-1}(B)^circ$$
Now repeat the same process by setting $overline A=X-(X-A)^circ$.
answered Jan 28 at 5:08
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
Referring to the link: Relation between interior and exterior.
Given that $f^{-1}$ exists, so $f$ is bijective and $f$ is given continuous so $f^{-1}$ is open.
Let, $Asubset X$ and $f(overline A)subset overline{f(A)}$ holds.
Now, let $Bsubset Y$ then $$B^circ=Y-overline{Y-B}\ implies f^{-1}(B^circ)=f^{-1}(Y-overline{Y-B})\=X-f^{-1}(overline{Y-B})\ subset X-overline {f^{-1}(Y-B)}\=X-overline{X-f^{-1}(B)}\=f^{-1}(B)^circ$$
Now repeat the same process by setting $overline A=X-(X-A)^circ$.
answered Jan 28 at 5:08
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
answered Jan 28 at 5:08
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
answered Jan 28 at 5:08
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
answered Jan 28 at 5:08
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
1,580519
$begingroup$
It is not given that $f^{-1}$ exists, this is just the notation for the inverse image...
$endgroup$
– Henno Brandsma
Jan 28 at 5:28
$begingroup$
@SujitBhattacharyya $X-f^{-1}(overline{Y-B})subset X-overline {f^{-1}(Y-B)}$? Why?
$endgroup$
– Darkmaster
Jan 28 at 5:32
$begingroup$
@HennoBrandsma $f^{-1}$ is the inverse image of which map? $f$ of course. That's why I have written that $f^{-1}$ exists.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:44
$begingroup$
For any function $f:X to Y$, $f^{-1}(A)={x in X: f(x) in A}$ if $A subseteq Y$. No need for an inverse function.
$endgroup$
– Henno Brandsma
Jan 28 at 5:46
$begingroup$
@Darkmaster Set, $f(A)=Bimplies A=f^{-1}(B)$, given $f(overline A)subset overline {f(A)}implies overline Asubset f^{-1}(overline {f(A)})implies overline{f^{-1}(B)}subset f^{-1}(overline B)$. For similarity replace $B$ by $Y-B$ and observe its complement.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:48
|
show 4 more comments
$begingroup$
It is not given that $f^{-1}$ exists, this is just the notation for the inverse image...
$endgroup$
– Henno Brandsma
Jan 28 at 5:28
$begingroup$
@SujitBhattacharyya $X-f^{-1}(overline{Y-B})subset X-overline {f^{-1}(Y-B)}$? Why?
$endgroup$
– Darkmaster
Jan 28 at 5:32
$begingroup$
@HennoBrandsma $f^{-1}$ is the inverse image of which map? $f$ of course. That's why I have written that $f^{-1}$ exists.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:44
$begingroup$
For any function $f:X to Y$, $f^{-1}(A)={x in X: f(x) in A}$ if $A subseteq Y$. No need for an inverse function.
$endgroup$
– Henno Brandsma
Jan 28 at 5:46
$begingroup$
@Darkmaster Set, $f(A)=Bimplies A=f^{-1}(B)$, given $f(overline A)subset overline {f(A)}implies overline Asubset f^{-1}(overline {f(A)})implies overline{f^{-1}(B)}subset f^{-1}(overline B)$. For similarity replace $B$ by $Y-B$ and observe its complement.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:48
$begingroup$
It is not given that $f^{-1}$ exists, this is just the notation for the inverse image...
$endgroup$
– Henno Brandsma
Jan 28 at 5:28
$begingroup$
It is not given that $f^{-1}$ exists, this is just the notation for the inverse image...
$endgroup$
– Henno Brandsma
Jan 28 at 5:28
$begingroup$
@SujitBhattacharyya $X-f^{-1}(overline{Y-B})subset X-overline {f^{-1}(Y-B)}$? Why?
$endgroup$
– Darkmaster
Jan 28 at 5:32
$begingroup$
@SujitBhattacharyya $X-f^{-1}(overline{Y-B})subset X-overline {f^{-1}(Y-B)}$? Why?
$endgroup$
– Darkmaster
Jan 28 at 5:32
$begingroup$
@HennoBrandsma $f^{-1}$ is the inverse image of which map? $f$ of course. That's why I have written that $f^{-1}$ exists.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:44
$begingroup$
@HennoBrandsma $f^{-1}$ is the inverse image of which map? $f$ of course. That's why I have written that $f^{-1}$ exists.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:44
$begingroup$
For any function $f:X to Y$, $f^{-1}(A)={x in X: f(x) in A}$ if $A subseteq Y$. No need for an inverse function.
$endgroup$
– Henno Brandsma
Jan 28 at 5:46
$begingroup$
For any function $f:X to Y$, $f^{-1}(A)={x in X: f(x) in A}$ if $A subseteq Y$. No need for an inverse function.
$endgroup$
– Henno Brandsma
Jan 28 at 5:46
$begingroup$
@Darkmaster Set, $f(A)=Bimplies A=f^{-1}(B)$, given $f(overline A)subset overline {f(A)}implies overline Asubset f^{-1}(overline {f(A)})implies overline{f^{-1}(B)}subset f^{-1}(overline B)$. For similarity replace $B$ by $Y-B$ and observe its complement.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:48
$begingroup$
@Darkmaster Set, $f(A)=Bimplies A=f^{-1}(B)$, given $f(overline A)subset overline {f(A)}implies overline Asubset f^{-1}(overline {f(A)})implies overline{f^{-1}(B)}subset f^{-1}(overline B)$. For similarity replace $B$ by $Y-B$ and observe its complement.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:48
|
show 4 more comments
$begingroup$
The following two statements about a function $f:X to Y$ between topological spaces are equivalent:
$forall A subseteq X: f[overline{A}] subseteq overline{f[A]}$.- For all closed $C subseteq Y$: $f^{-1}[C]$ is closed in $X$.
For 1 implies 2: let $C$ be closed in $Y$. Define $D = f^{-1}[C]$ and apply fact 1 to it: $$f[overline{D}] subseteq overline{f[D]} = overline{f[f^{-1}[C]]}subseteq overline{C} = C$$ as $C$ is closed. But this implies that $overline{D} subseteq f^{-1}[C] =D$ (all points of $overline{D}$ map into $C$) and so $D = f^{-1}[C]$ is closed, as required.
For 2 implies 1: Let $A subseteq X$ and note we have $$A subseteq f^{-1}[f[A]] subseteq f^{-1}[overline{f[A]}]$$ and by 2. the latter set is closed. Hence
$overline{A} subseteq f^{-1}[overline{f[A]}]$ as well and this means that $f[overline{A}]subseteq overline{f[A]}$ by the definitions of inverse images.
The following two statements about a function $f:X to Y$ between topological spaces are equivalent:
$forall B subseteq Y: f^{-1}[B^circ] subseteq f^{-1}[B]^circ$.- For all $O subseteq Y$ open, $f^{-1}[O]$ is open in $X$.
For 3 implies 4: let $O$ be open in $Y$, then $O^circ = O$ and by 3:
$$f^{-1}[O]= f^{-1}[O^circ] subseteq f^{-1}[O]^circ subseteq f^{-1}[O]$$
so $f^{-1}[O]^circ = f^{-1}[O]$ and $f^{-1}[O]$ is open.
For 4 implies 3: if $B subseteq Y$ is any subset $B^circ subseteq B$ so
also $f^{-1}[B^circ] subseteq f^{-1}[B]$. The former is an open subset (by 4) contained in $f^{-1}[B]$ and the interior of a set is its largest open subset so it follows that $f^{-1}[B^circ] subseteq f^{-1}[B]^circ$, as required.
Now note that 2 and 4 are clearly equivalent as $f^{-1}[Ysetminus A]= Xsetminus f^{-1}[A]$ for all subsets $B$ of $Y$, and open and closed sets are related via complements.
So all conditions are equivalent to $f$ being continuous.
answered Jan 28 at 9:45
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
$begingroup$
@SujitBhattacharyya Thanks! You're great!
$endgroup$
– Darkmaster
Jan 28 at 16:21
1
$begingroup$
@Darkmaster did you mean as a comment to the other answer?
$endgroup$
– Henno Brandsma
Jan 28 at 16:51
add a comment |
$begingroup$
The following two statements about a function $f:X to Y$ between topological spaces are equivalent:
$forall A subseteq X: f[overline{A}] subseteq overline{f[A]}$.- For all closed $C subseteq Y$: $f^{-1}[C]$ is closed in $X$.
For 1 implies 2: let $C$ be closed in $Y$. Define $D = f^{-1}[C]$ and apply fact 1 to it: $$f[overline{D}] subseteq overline{f[D]} = overline{f[f^{-1}[C]]}subseteq overline{C} = C$$ as $C$ is closed. But this implies that $overline{D} subseteq f^{-1}[C] =D$ (all points of $overline{D}$ map into $C$) and so $D = f^{-1}[C]$ is closed, as required.
For 2 implies 1: Let $A subseteq X$ and note we have $$A subseteq f^{-1}[f[A]] subseteq f^{-1}[overline{f[A]}]$$ and by 2. the latter set is closed. Hence
$overline{A} subseteq f^{-1}[overline{f[A]}]$ as well and this means that $f[overline{A}]subseteq overline{f[A]}$ by the definitions of inverse images.
The following two statements about a function $f:X to Y$ between topological spaces are equivalent:
$forall B subseteq Y: f^{-1}[B^circ] subseteq f^{-1}[B]^circ$.- For all $O subseteq Y$ open, $f^{-1}[O]$ is open in $X$.
For 3 implies 4: let $O$ be open in $Y$, then $O^circ = O$ and by 3:
$$f^{-1}[O]= f^{-1}[O^circ] subseteq f^{-1}[O]^circ subseteq f^{-1}[O]$$
so $f^{-1}[O]^circ = f^{-1}[O]$ and $f^{-1}[O]$ is open.
For 4 implies 3: if $B subseteq Y$ is any subset $B^circ subseteq B$ so
also $f^{-1}[B^circ] subseteq f^{-1}[B]$. The former is an open subset (by 4) contained in $f^{-1}[B]$ and the interior of a set is its largest open subset so it follows that $f^{-1}[B^circ] subseteq f^{-1}[B]^circ$, as required.
Now note that 2 and 4 are clearly equivalent as $f^{-1}[Ysetminus A]= Xsetminus f^{-1}[A]$ for all subsets $B$ of $Y$, and open and closed sets are related via complements.
So all conditions are equivalent to $f$ being continuous.
answered Jan 28 at 9:45
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
$begingroup$
@SujitBhattacharyya Thanks! You're great!
$endgroup$
– Darkmaster
Jan 28 at 16:21
1
$begingroup$
@Darkmaster did you mean as a comment to the other answer?
$endgroup$
– Henno Brandsma
Jan 28 at 16:51
add a comment |
$begingroup$
The following two statements about a function $f:X to Y$ between topological spaces are equivalent:
$forall A subseteq X: f[overline{A}] subseteq overline{f[A]}$.- For all closed $C subseteq Y$: $f^{-1}[C]$ is closed in $X$.
For 1 implies 2: let $C$ be closed in $Y$. Define $D = f^{-1}[C]$ and apply fact 1 to it: $$f[overline{D}] subseteq overline{f[D]} = overline{f[f^{-1}[C]]}subseteq overline{C} = C$$ as $C$ is closed. But this implies that $overline{D} subseteq f^{-1}[C] =D$ (all points of $overline{D}$ map into $C$) and so $D = f^{-1}[C]$ is closed, as required.
For 2 implies 1: Let $A subseteq X$ and note we have $$A subseteq f^{-1}[f[A]] subseteq f^{-1}[overline{f[A]}]$$ and by 2. the latter set is closed. Hence
$overline{A} subseteq f^{-1}[overline{f[A]}]$ as well and this means that $f[overline{A}]subseteq overline{f[A]}$ by the definitions of inverse images.
The following two statements about a function $f:X to Y$ between topological spaces are equivalent:
$forall B subseteq Y: f^{-1}[B^circ] subseteq f^{-1}[B]^circ$.- For all $O subseteq Y$ open, $f^{-1}[O]$ is open in $X$.
For 3 implies 4: let $O$ be open in $Y$, then $O^circ = O$ and by 3:
$$f^{-1}[O]= f^{-1}[O^circ] subseteq f^{-1}[O]^circ subseteq f^{-1}[O]$$
so $f^{-1}[O]^circ = f^{-1}[O]$ and $f^{-1}[O]$ is open.
For 4 implies 3: if $B subseteq Y$ is any subset $B^circ subseteq B$ so
also $f^{-1}[B^circ] subseteq f^{-1}[B]$. The former is an open subset (by 4) contained in $f^{-1}[B]$ and the interior of a set is its largest open subset so it follows that $f^{-1}[B^circ] subseteq f^{-1}[B]^circ$, as required.
Now note that 2 and 4 are clearly equivalent as $f^{-1}[Ysetminus A]= Xsetminus f^{-1}[A]$ for all subsets $B$ of $Y$, and open and closed sets are related via complements.
So all conditions are equivalent to $f$ being continuous.
answered Jan 28 at 9:45
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
The following two statements about a function $f:X to Y$ between topological spaces are equivalent:
$forall A subseteq X: f[overline{A}] subseteq overline{f[A]}$.- For all closed $C subseteq Y$: $f^{-1}[C]$ is closed in $X$.
For 1 implies 2: let $C$ be closed in $Y$. Define $D = f^{-1}[C]$ and apply fact 1 to it: $$f[overline{D}] subseteq overline{f[D]} = overline{f[f^{-1}[C]]}subseteq overline{C} = C$$ as $C$ is closed. But this implies that $overline{D} subseteq f^{-1}[C] =D$ (all points of $overline{D}$ map into $C$) and so $D = f^{-1}[C]$ is closed, as required.
For 2 implies 1: Let $A subseteq X$ and note we have $$A subseteq f^{-1}[f[A]] subseteq f^{-1}[overline{f[A]}]$$ and by 2. the latter set is closed. Hence
$overline{A} subseteq f^{-1}[overline{f[A]}]$ as well and this means that $f[overline{A}]subseteq overline{f[A]}$ by the definitions of inverse images.
The following two statements about a function $f:X to Y$ between topological spaces are equivalent:
$forall B subseteq Y: f^{-1}[B^circ] subseteq f^{-1}[B]^circ$.- For all $O subseteq Y$ open, $f^{-1}[O]$ is open in $X$.
For 3 implies 4: let $O$ be open in $Y$, then $O^circ = O$ and by 3:
$$f^{-1}[O]= f^{-1}[O^circ] subseteq f^{-1}[O]^circ subseteq f^{-1}[O]$$
so $f^{-1}[O]^circ = f^{-1}[O]$ and $f^{-1}[O]$ is open.
For 4 implies 3: if $B subseteq Y$ is any subset $B^circ subseteq B$ so
also $f^{-1}[B^circ] subseteq f^{-1}[B]$. The former is an open subset (by 4) contained in $f^{-1}[B]$ and the interior of a set is its largest open subset so it follows that $f^{-1}[B^circ] subseteq f^{-1}[B]^circ$, as required.
Now note that 2 and 4 are clearly equivalent as $f^{-1}[Ysetminus A]= Xsetminus f^{-1}[A]$ for all subsets $B$ of $Y$, and open and closed sets are related via complements.
So all conditions are equivalent to $f$ being continuous.
answered Jan 28 at 9:45
Henno BrandsmaHenno Brandsma
114k348123
answered Jan 28 at 9:45
Henno BrandsmaHenno Brandsma
114k348123
answered Jan 28 at 9:45
Henno BrandsmaHenno Brandsma
114k348123
answered Jan 28 at 9:45
Henno BrandsmaHenno Brandsma
114k348123
114k348123
$begingroup$
@SujitBhattacharyya Thanks! You're great!
$endgroup$
– Darkmaster
Jan 28 at 16:21
1
$begingroup$
@Darkmaster did you mean as a comment to the other answer?
$endgroup$
– Henno Brandsma
Jan 28 at 16:51
add a comment |
$begingroup$
@SujitBhattacharyya Thanks! You're great!
$endgroup$
– Darkmaster
Jan 28 at 16:21
1
$begingroup$
@Darkmaster did you mean as a comment to the other answer?
$endgroup$
– Henno Brandsma
Jan 28 at 16:51
$begingroup$
@SujitBhattacharyya Thanks! You're great!
$endgroup$
– Darkmaster
Jan 28 at 16:21
$begingroup$
@SujitBhattacharyya Thanks! You're great!
$endgroup$
– Darkmaster
Jan 28 at 16:21
1
1
$begingroup$
@Darkmaster did you mean as a comment to the other answer?
$endgroup$
– Henno Brandsma
Jan 28 at 16:51
$begingroup$
@Darkmaster did you mean as a comment to the other answer?
$endgroup$
– Henno Brandsma
Jan 28 at 16:51
add a comment |
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$begingroup$
The two conditions are both equivalent to the continuity of $f$.
$endgroup$
– Eclipse Sun
Jan 28 at 4:59
$begingroup$
@EclipseSun I know. I just want to know if there is a demonstration of that proposition as I wrote it.
$endgroup$
– Darkmaster
Jan 28 at 5:04
1
$begingroup$
I meant that you don't need to assume $f$ to be continuous at the very beginning.
$endgroup$
– Eclipse Sun
Jan 28 at 5:07
$begingroup$
@EclipseSun You're right. Leaving that aside, do you know how to do the proof of that theorem? Leaving aside the continuity ...
$endgroup$
– Darkmaster
Jan 28 at 5:10
$begingroup$
cf this question for one of these properties.
$endgroup$
– Henno Brandsma
Jan 28 at 5:45