Show that $tau := {X setminus phi (A) mid A subset X }$ is a topology on $X$.
$begingroup$
Let $X$ be a set and $mathcal P (X)$ the powerset of $X$. Suppose $phi: mathcal P(X) to mathcal P(X)$ is a function such that
$phi(emptyset) = emptyset$,- For all $A subset X$: $A subset phi(A)$,
- For all $A subset X$: $phi(phi(A)) = phi(A)$ and
- For all $A, B in mathcal P(X)$: $phi(A cup B) = phi(A) cup phi(B)$.
Show that $tau := { X setminus phi(A) mid A subset X }$ is a topology on $X$.
Progress I've made:
$emptyset = X setminus phi(X) in tau$.
$X = X setminus phi(emptyset) in tau$.- Finite intersections: $(X setminus phi(A)) cap (X setminus phi(B)) = X setminus (phi(A) cup phi(B)) = X setminus phi(A cup B) in tau$.
- Infinite unions: This is what I'm struggling with. How do I complete the proof?
general-topology
asked Jan 3 at 19:26
Gilles CastelGilles Castel
1269
$endgroup$
add a comment |
$begingroup$
Let $X$ be a set and $mathcal P (X)$ the powerset of $X$. Suppose $phi: mathcal P(X) to mathcal P(X)$ is a function such that
$phi(emptyset) = emptyset$,- For all $A subset X$: $A subset phi(A)$,
- For all $A subset X$: $phi(phi(A)) = phi(A)$ and
- For all $A, B in mathcal P(X)$: $phi(A cup B) = phi(A) cup phi(B)$.
Show that $tau := { X setminus phi(A) mid A subset X }$ is a topology on $X$.
Progress I've made:
$emptyset = X setminus phi(X) in tau$.
$X = X setminus phi(emptyset) in tau$.- Finite intersections: $(X setminus phi(A)) cap (X setminus phi(B)) = X setminus (phi(A) cup phi(B)) = X setminus phi(A cup B) in tau$.
- Infinite unions: This is what I'm struggling with. How do I complete the proof?
general-topology
asked Jan 3 at 19:26
Gilles CastelGilles Castel
1269
$endgroup$
add a comment |
$begingroup$
Let $X$ be a set and $mathcal P (X)$ the powerset of $X$. Suppose $phi: mathcal P(X) to mathcal P(X)$ is a function such that
$phi(emptyset) = emptyset$,- For all $A subset X$: $A subset phi(A)$,
- For all $A subset X$: $phi(phi(A)) = phi(A)$ and
- For all $A, B in mathcal P(X)$: $phi(A cup B) = phi(A) cup phi(B)$.
Show that $tau := { X setminus phi(A) mid A subset X }$ is a topology on $X$.
Progress I've made:
$emptyset = X setminus phi(X) in tau$.
$X = X setminus phi(emptyset) in tau$.- Finite intersections: $(X setminus phi(A)) cap (X setminus phi(B)) = X setminus (phi(A) cup phi(B)) = X setminus phi(A cup B) in tau$.
- Infinite unions: This is what I'm struggling with. How do I complete the proof?
general-topology
asked Jan 3 at 19:26
Gilles CastelGilles Castel
1269
$endgroup$
Let $X$ be a set and $mathcal P (X)$ the powerset of $X$. Suppose $phi: mathcal P(X) to mathcal P(X)$ is a function such that
$phi(emptyset) = emptyset$,- For all $A subset X$: $A subset phi(A)$,
- For all $A subset X$: $phi(phi(A)) = phi(A)$ and
- For all $A, B in mathcal P(X)$: $phi(A cup B) = phi(A) cup phi(B)$.
Show that $tau := { X setminus phi(A) mid A subset X }$ is a topology on $X$.
Progress I've made:
$emptyset = X setminus phi(X) in tau$.
$X = X setminus phi(emptyset) in tau$.- Finite intersections: $(X setminus phi(A)) cap (X setminus phi(B)) = X setminus (phi(A) cup phi(B)) = X setminus phi(A cup B) in tau$.
- Infinite unions: This is what I'm struggling with. How do I complete the proof?
general-topology
general-topology
asked Jan 3 at 19:26
Gilles CastelGilles Castel
1269
asked Jan 3 at 19:26
Gilles CastelGilles Castel
1269
asked Jan 3 at 19:26
Gilles CastelGilles Castel
1269
asked Jan 3 at 19:26
Gilles CastelGilles Castel
1269
asked Jan 3 at 19:26
Gilles CastelGilles Castel
1269
1269
add a comment |
add a comment |
1 Answer
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$begingroup$
Easy fact from the axioms: $C subseteq D$ implies $phi(C) subseteq phi(D)$:
$C subseteq D$ iff $C cup D=D$ which implies $phi(C) cup phi(D) = phi(C cup D) = phi(D)$ which holds iff $phi(C) subseteq phi(D)$. QED
Suppose the subsets $B_i, i in I$ of $X$ obey $phi(B_i) = B_i$ for all $i$.
Then define $B=bigcap_{i in I} B_i$ and note that by the second axiom $B subseteq phi(B)$. On the other hand, for all $i$ $B subseteq B_i$ so that by the first fact we have $forall i: phi(B) subseteq phi(B_i) = B_i$ so $phi(B) subseteq bigcap_{i in I} B_i = B$ and so $phi(B) = B$.
Now if $Xsetminus phi(A_i), i in I$ are in $tau$ then:
$$bigcup_i (Xsetminus phi(A_i)) = Xsetminus bigcap_{i in I} phi(A_i)$$ and as $phi(phi(A_i)) = phi(A_i)$ we apply the last paragraph's fact (with $B_i = phi(A_i)$ on intersections to see that $bigcap_{i in I} phi(A_i) = phileft(bigcap_{i in I} phi(A_i)right)$ and so $$bigcup_i (Xsetminus phi(A_i)) = X setminus phileft(bigcap_{i in I} phi(A_i)right) in tau$$
as required. The proof idea is clear: sets $phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections.
answered Jan 3 at 22:52
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
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$begingroup$
Easy fact from the axioms: $C subseteq D$ implies $phi(C) subseteq phi(D)$:
$C subseteq D$ iff $C cup D=D$ which implies $phi(C) cup phi(D) = phi(C cup D) = phi(D)$ which holds iff $phi(C) subseteq phi(D)$. QED
Suppose the subsets $B_i, i in I$ of $X$ obey $phi(B_i) = B_i$ for all $i$.
Then define $B=bigcap_{i in I} B_i$ and note that by the second axiom $B subseteq phi(B)$. On the other hand, for all $i$ $B subseteq B_i$ so that by the first fact we have $forall i: phi(B) subseteq phi(B_i) = B_i$ so $phi(B) subseteq bigcap_{i in I} B_i = B$ and so $phi(B) = B$.
Now if $Xsetminus phi(A_i), i in I$ are in $tau$ then:
$$bigcup_i (Xsetminus phi(A_i)) = Xsetminus bigcap_{i in I} phi(A_i)$$ and as $phi(phi(A_i)) = phi(A_i)$ we apply the last paragraph's fact (with $B_i = phi(A_i)$ on intersections to see that $bigcap_{i in I} phi(A_i) = phileft(bigcap_{i in I} phi(A_i)right)$ and so $$bigcup_i (Xsetminus phi(A_i)) = X setminus phileft(bigcap_{i in I} phi(A_i)right) in tau$$
as required. The proof idea is clear: sets $phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections.
answered Jan 3 at 22:52
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
$begingroup$
Easy fact from the axioms: $C subseteq D$ implies $phi(C) subseteq phi(D)$:
$C subseteq D$ iff $C cup D=D$ which implies $phi(C) cup phi(D) = phi(C cup D) = phi(D)$ which holds iff $phi(C) subseteq phi(D)$. QED
Suppose the subsets $B_i, i in I$ of $X$ obey $phi(B_i) = B_i$ for all $i$.
Then define $B=bigcap_{i in I} B_i$ and note that by the second axiom $B subseteq phi(B)$. On the other hand, for all $i$ $B subseteq B_i$ so that by the first fact we have $forall i: phi(B) subseteq phi(B_i) = B_i$ so $phi(B) subseteq bigcap_{i in I} B_i = B$ and so $phi(B) = B$.
Now if $Xsetminus phi(A_i), i in I$ are in $tau$ then:
$$bigcup_i (Xsetminus phi(A_i)) = Xsetminus bigcap_{i in I} phi(A_i)$$ and as $phi(phi(A_i)) = phi(A_i)$ we apply the last paragraph's fact (with $B_i = phi(A_i)$ on intersections to see that $bigcap_{i in I} phi(A_i) = phileft(bigcap_{i in I} phi(A_i)right)$ and so $$bigcup_i (Xsetminus phi(A_i)) = X setminus phileft(bigcap_{i in I} phi(A_i)right) in tau$$
as required. The proof idea is clear: sets $phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections.
answered Jan 3 at 22:52
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
$begingroup$
Easy fact from the axioms: $C subseteq D$ implies $phi(C) subseteq phi(D)$:
$C subseteq D$ iff $C cup D=D$ which implies $phi(C) cup phi(D) = phi(C cup D) = phi(D)$ which holds iff $phi(C) subseteq phi(D)$. QED
Suppose the subsets $B_i, i in I$ of $X$ obey $phi(B_i) = B_i$ for all $i$.
Then define $B=bigcap_{i in I} B_i$ and note that by the second axiom $B subseteq phi(B)$. On the other hand, for all $i$ $B subseteq B_i$ so that by the first fact we have $forall i: phi(B) subseteq phi(B_i) = B_i$ so $phi(B) subseteq bigcap_{i in I} B_i = B$ and so $phi(B) = B$.
Now if $Xsetminus phi(A_i), i in I$ are in $tau$ then:
$$bigcup_i (Xsetminus phi(A_i)) = Xsetminus bigcap_{i in I} phi(A_i)$$ and as $phi(phi(A_i)) = phi(A_i)$ we apply the last paragraph's fact (with $B_i = phi(A_i)$ on intersections to see that $bigcap_{i in I} phi(A_i) = phileft(bigcap_{i in I} phi(A_i)right)$ and so $$bigcup_i (Xsetminus phi(A_i)) = X setminus phileft(bigcap_{i in I} phi(A_i)right) in tau$$
as required. The proof idea is clear: sets $phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections.
answered Jan 3 at 22:52
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
Easy fact from the axioms: $C subseteq D$ implies $phi(C) subseteq phi(D)$:
$C subseteq D$ iff $C cup D=D$ which implies $phi(C) cup phi(D) = phi(C cup D) = phi(D)$ which holds iff $phi(C) subseteq phi(D)$. QED
Suppose the subsets $B_i, i in I$ of $X$ obey $phi(B_i) = B_i$ for all $i$.
Then define $B=bigcap_{i in I} B_i$ and note that by the second axiom $B subseteq phi(B)$. On the other hand, for all $i$ $B subseteq B_i$ so that by the first fact we have $forall i: phi(B) subseteq phi(B_i) = B_i$ so $phi(B) subseteq bigcap_{i in I} B_i = B$ and so $phi(B) = B$.
Now if $Xsetminus phi(A_i), i in I$ are in $tau$ then:
$$bigcup_i (Xsetminus phi(A_i)) = Xsetminus bigcap_{i in I} phi(A_i)$$ and as $phi(phi(A_i)) = phi(A_i)$ we apply the last paragraph's fact (with $B_i = phi(A_i)$ on intersections to see that $bigcap_{i in I} phi(A_i) = phileft(bigcap_{i in I} phi(A_i)right)$ and so $$bigcup_i (Xsetminus phi(A_i)) = X setminus phileft(bigcap_{i in I} phi(A_i)right) in tau$$
as required. The proof idea is clear: sets $phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections.
answered Jan 3 at 22:52
Henno BrandsmaHenno Brandsma
115k349125
answered Jan 3 at 22:52
Henno BrandsmaHenno Brandsma
115k349125
answered Jan 3 at 22:52
Henno BrandsmaHenno Brandsma
115k349125
answered Jan 3 at 22:52
Henno BrandsmaHenno Brandsma
115k349125
115k349125
add a comment |
add a comment |
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