Mixing
$C_p$ space topology
$begingroup$
If $X, Y$ are topological spaces then consider the set $C(X, Y) $ of continuous functions. Let $[x_1,...,x_n;O_1,...,O_n] ={fin C(X, Y) | f(x_i) in O_iin tau (Y), i=1,...,n}$. Then how to show that the collection $${Usubset C(X, Y) | forall fin U : exists nin mathbb{N}exists x_1,dots,x_n in X, O_1,ldots, O_n in tau (Y):\
fin [x_1,...,x_n;O_1,...,O_n] subset U }$$ is a topology on $C(X, Y)$? I can see that $emptyset $ is in there(vacuously) and $C(X, Y) $ is also in there since for any function $fin C(X, Y) $ we can take any $x_1,...,x_n$ and $O_i=Y, i=1,..., n$. But I'm stuck with the other two.
general-topology
edited Jan 20 at 7:38
Henno Brandsma
111k348119
asked Jan 19 at 19:43
Arpan DasArpan Das
1017
$endgroup$
add a comment |
$begingroup$
If $X, Y$ are topological spaces then consider the set $C(X, Y) $ of continuous functions. Let $[x_1,...,x_n;O_1,...,O_n] ={fin C(X, Y) | f(x_i) in O_iin tau (Y), i=1,...,n}$. Then how to show that the collection $${Usubset C(X, Y) | forall fin U : exists nin mathbb{N}exists x_1,dots,x_n in X, O_1,ldots, O_n in tau (Y):\
fin [x_1,...,x_n;O_1,...,O_n] subset U }$$ is a topology on $C(X, Y)$? I can see that $emptyset $ is in there(vacuously) and $C(X, Y) $ is also in there since for any function $fin C(X, Y) $ we can take any $x_1,...,x_n$ and $O_i=Y, i=1,..., n$. But I'm stuck with the other two.
general-topology
edited Jan 20 at 7:38
Henno Brandsma
111k348119
asked Jan 19 at 19:43
Arpan DasArpan Das
1017
$endgroup$
1
$begingroup$
This is just the product topology that $C(X,Y)$ inherits as a subspace of $Y^X$.
$endgroup$
– Henno Brandsma
Jan 19 at 21:45
$begingroup$
@ArpanDas Shouldn't the union be easy too? If ${U_i}$ is a collection of open sets in $C(X,Y)$, then any $fin bigcup_i U_i$ belongs to some $U_{i_0}$ and we are done.
$endgroup$
– Dog_69
Jan 19 at 21:50
$begingroup$
And for the intersection, I think that $[x_1,dots,x_n,O_1,dots,O_n]cap[y_1,dots,y_m,P_1,dots,P_m] = [x_1,dots,x_n,y_1,dots,y_m,O_1,dots,O_n,P_1,dots,P_m]$, so for $finbigcap_{i=1}^n U_i$ you can consider the intersection all sets $[x_1,dots,x_n,O_1,dots O_n]$ associated with each $U_i$.
$endgroup$
– Dog_69
Jan 19 at 22:01
$begingroup$
I thought the $x_i$ s and $O_i$s depend on U. So for union I got confused. Intersection works although.
$endgroup$
– Arpan Das
Jan 20 at 4:29
add a comment |
$begingroup$
If $X, Y$ are topological spaces then consider the set $C(X, Y) $ of continuous functions. Let $[x_1,...,x_n;O_1,...,O_n] ={fin C(X, Y) | f(x_i) in O_iin tau (Y), i=1,...,n}$. Then how to show that the collection $${Usubset C(X, Y) | forall fin U : exists nin mathbb{N}exists x_1,dots,x_n in X, O_1,ldots, O_n in tau (Y):\
fin [x_1,...,x_n;O_1,...,O_n] subset U }$$ is a topology on $C(X, Y)$? I can see that $emptyset $ is in there(vacuously) and $C(X, Y) $ is also in there since for any function $fin C(X, Y) $ we can take any $x_1,...,x_n$ and $O_i=Y, i=1,..., n$. But I'm stuck with the other two.
general-topology
edited Jan 20 at 7:38
Henno Brandsma
111k348119
asked Jan 19 at 19:43
Arpan DasArpan Das
1017
$endgroup$
If $X, Y$ are topological spaces then consider the set $C(X, Y) $ of continuous functions. Let $[x_1,...,x_n;O_1,...,O_n] ={fin C(X, Y) | f(x_i) in O_iin tau (Y), i=1,...,n}$. Then how to show that the collection $${Usubset C(X, Y) | forall fin U : exists nin mathbb{N}exists x_1,dots,x_n in X, O_1,ldots, O_n in tau (Y):\
fin [x_1,...,x_n;O_1,...,O_n] subset U }$$ is a topology on $C(X, Y)$? I can see that $emptyset $ is in there(vacuously) and $C(X, Y) $ is also in there since for any function $fin C(X, Y) $ we can take any $x_1,...,x_n$ and $O_i=Y, i=1,..., n$. But I'm stuck with the other two.
general-topology
general-topology
edited Jan 20 at 7:38
Henno Brandsma
111k348119
asked Jan 19 at 19:43
Arpan DasArpan Das
1017
edited Jan 20 at 7:38
Henno Brandsma
111k348119
asked Jan 19 at 19:43
Arpan DasArpan Das
1017
edited Jan 20 at 7:38
Henno Brandsma
111k348119
edited Jan 20 at 7:38
Henno Brandsma
111k348119
edited Jan 20 at 7:38
Henno Brandsma
111k348119
111k348119
asked Jan 19 at 19:43
Arpan DasArpan Das
1017
asked Jan 19 at 19:43
Arpan DasArpan Das
1017
asked Jan 19 at 19:43
Arpan DasArpan Das
1017
1017
1
$begingroup$
This is just the product topology that $C(X,Y)$ inherits as a subspace of $Y^X$.
$endgroup$
– Henno Brandsma
Jan 19 at 21:45
$begingroup$
@ArpanDas Shouldn't the union be easy too? If ${U_i}$ is a collection of open sets in $C(X,Y)$, then any $fin bigcup_i U_i$ belongs to some $U_{i_0}$ and we are done.
$endgroup$
– Dog_69
Jan 19 at 21:50
$begingroup$
And for the intersection, I think that $[x_1,dots,x_n,O_1,dots,O_n]cap[y_1,dots,y_m,P_1,dots,P_m] = [x_1,dots,x_n,y_1,dots,y_m,O_1,dots,O_n,P_1,dots,P_m]$, so for $finbigcap_{i=1}^n U_i$ you can consider the intersection all sets $[x_1,dots,x_n,O_1,dots O_n]$ associated with each $U_i$.
$endgroup$
– Dog_69
Jan 19 at 22:01
$begingroup$
I thought the $x_i$ s and $O_i$s depend on U. So for union I got confused. Intersection works although.
$endgroup$
– Arpan Das
Jan 20 at 4:29
add a comment |
1
$begingroup$
This is just the product topology that $C(X,Y)$ inherits as a subspace of $Y^X$.
$endgroup$
– Henno Brandsma
Jan 19 at 21:45
$begingroup$
@ArpanDas Shouldn't the union be easy too? If ${U_i}$ is a collection of open sets in $C(X,Y)$, then any $fin bigcup_i U_i$ belongs to some $U_{i_0}$ and we are done.
$endgroup$
– Dog_69
Jan 19 at 21:50
$begingroup$
And for the intersection, I think that $[x_1,dots,x_n,O_1,dots,O_n]cap[y_1,dots,y_m,P_1,dots,P_m] = [x_1,dots,x_n,y_1,dots,y_m,O_1,dots,O_n,P_1,dots,P_m]$, so for $finbigcap_{i=1}^n U_i$ you can consider the intersection all sets $[x_1,dots,x_n,O_1,dots O_n]$ associated with each $U_i$.
$endgroup$
– Dog_69
Jan 19 at 22:01
$begingroup$
I thought the $x_i$ s and $O_i$s depend on U. So for union I got confused. Intersection works although.
$endgroup$
– Arpan Das
Jan 20 at 4:29
1
1
$begingroup$
This is just the product topology that $C(X,Y)$ inherits as a subspace of $Y^X$.
$endgroup$
– Henno Brandsma
Jan 19 at 21:45
$begingroup$
This is just the product topology that $C(X,Y)$ inherits as a subspace of $Y^X$.
$endgroup$
– Henno Brandsma
Jan 19 at 21:45
$begingroup$
@ArpanDas Shouldn't the union be easy too? If ${U_i}$ is a collection of open sets in $C(X,Y)$, then any $fin bigcup_i U_i$ belongs to some $U_{i_0}$ and we are done.
$endgroup$
– Dog_69
Jan 19 at 21:50
$begingroup$
@ArpanDas Shouldn't the union be easy too? If ${U_i}$ is a collection of open sets in $C(X,Y)$, then any $fin bigcup_i U_i$ belongs to some $U_{i_0}$ and we are done.
$endgroup$
– Dog_69
Jan 19 at 21:50
$begingroup$
And for the intersection, I think that $[x_1,dots,x_n,O_1,dots,O_n]cap[y_1,dots,y_m,P_1,dots,P_m] = [x_1,dots,x_n,y_1,dots,y_m,O_1,dots,O_n,P_1,dots,P_m]$, so for $finbigcap_{i=1}^n U_i$ you can consider the intersection all sets $[x_1,dots,x_n,O_1,dots O_n]$ associated with each $U_i$.
$endgroup$
– Dog_69
Jan 19 at 22:01
$begingroup$
And for the intersection, I think that $[x_1,dots,x_n,O_1,dots,O_n]cap[y_1,dots,y_m,P_1,dots,P_m] = [x_1,dots,x_n,y_1,dots,y_m,O_1,dots,O_n,P_1,dots,P_m]$, so for $finbigcap_{i=1}^n U_i$ you can consider the intersection all sets $[x_1,dots,x_n,O_1,dots O_n]$ associated with each $U_i$.
$endgroup$
– Dog_69
Jan 19 at 22:01
$begingroup$
I thought the $x_i$ s and $O_i$s depend on U. So for union I got confused. Intersection works although.
$endgroup$
– Arpan Das
Jan 20 at 4:29
$begingroup$
I thought the $x_i$ s and $O_i$s depend on U. So for union I got confused. Intersection works although.
$endgroup$
– Arpan Das
Jan 20 at 4:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The union is trivial: if $U_i, i in I$ are all open and $f in U=bigcup_i U_i$, then for some $j in I$, $f in U_j$. By definition we then have $x_1,ldots x_n$ in $X$ and $O_1,ldots,O_n$ open in $Y$ such that $f in [x_1,ldots,x_n; O_1, O_2,ldots O_n] subseteq U_i subseteq U$.
For finite intersections we only need to check the intersection of two open sets $U_1$ and $U_2$ (finite then follows by induction, a standard argument), so let $f in U_1 cap U_2$. Then $f in U_1$ gives us $x_1,ldots x_n$ in $X$ and $O_1,ldots,O_n$ open in $Y$ such that $f in [x_1,ldots,x_n; O_1, O_2,ldots O_n] subseteq U_1$ and $f in U_2$ gives us $x'_1,ldots x'_m$ in $X$ and $O'_1,ldots,O'_m$ open in $Y$ such that $f in [x'_1,ldots,x'_m; O'_1, O'_2,ldots O'_m] subseteq U_2$. But then
$$f in [x_1,ldots,x_n, x'_1,ldots, x'_m; O_1,ldots,O_n, O'_1,ldots,O'_m] subseteq \
[x_1,ldots,x_n; O_1, O_2,ldots O_n] cap [x'_1,ldots,x'_m; O'_1, O'_2,ldots O'_m] subseteq U_1 cap U_2$$
as required.
You can also note that basic open neighbourhood of $f$ (the $[x_1,ldots,x_n; O_1, ldots, O_n]$) is of the form $B cap C(X,Y)$ where $B$ is a basic open subset of $Y^X$ in the product topology, given by a finite subset ${x_1, ldots,x_n}$ of "coordinates" from $X$ and finitely many $O_1, ldots, O_n$ in those coordinates. So e.g. $f_n to f$ in this topology iff $f_n(x) to f(x)$ for all $x in X$, as this holds for the product topology (and likewise for nets), hence the name "pointwise topology" for this space $C_p(X,Y)$. The "$p$" could also be seen as standing for "product", as it's just the product topology on the set, essentially.
answered Jan 20 at 7:57
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
add a comment |
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$begingroup$
The union is trivial: if $U_i, i in I$ are all open and $f in U=bigcup_i U_i$, then for some $j in I$, $f in U_j$. By definition we then have $x_1,ldots x_n$ in $X$ and $O_1,ldots,O_n$ open in $Y$ such that $f in [x_1,ldots,x_n; O_1, O_2,ldots O_n] subseteq U_i subseteq U$.
For finite intersections we only need to check the intersection of two open sets $U_1$ and $U_2$ (finite then follows by induction, a standard argument), so let $f in U_1 cap U_2$. Then $f in U_1$ gives us $x_1,ldots x_n$ in $X$ and $O_1,ldots,O_n$ open in $Y$ such that $f in [x_1,ldots,x_n; O_1, O_2,ldots O_n] subseteq U_1$ and $f in U_2$ gives us $x'_1,ldots x'_m$ in $X$ and $O'_1,ldots,O'_m$ open in $Y$ such that $f in [x'_1,ldots,x'_m; O'_1, O'_2,ldots O'_m] subseteq U_2$. But then
$$f in [x_1,ldots,x_n, x'_1,ldots, x'_m; O_1,ldots,O_n, O'_1,ldots,O'_m] subseteq \
[x_1,ldots,x_n; O_1, O_2,ldots O_n] cap [x'_1,ldots,x'_m; O'_1, O'_2,ldots O'_m] subseteq U_1 cap U_2$$
as required.
You can also note that basic open neighbourhood of $f$ (the $[x_1,ldots,x_n; O_1, ldots, O_n]$) is of the form $B cap C(X,Y)$ where $B$ is a basic open subset of $Y^X$ in the product topology, given by a finite subset ${x_1, ldots,x_n}$ of "coordinates" from $X$ and finitely many $O_1, ldots, O_n$ in those coordinates. So e.g. $f_n to f$ in this topology iff $f_n(x) to f(x)$ for all $x in X$, as this holds for the product topology (and likewise for nets), hence the name "pointwise topology" for this space $C_p(X,Y)$. The "$p$" could also be seen as standing for "product", as it's just the product topology on the set, essentially.
answered Jan 20 at 7:57
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
add a comment |
$begingroup$
The union is trivial: if $U_i, i in I$ are all open and $f in U=bigcup_i U_i$, then for some $j in I$, $f in U_j$. By definition we then have $x_1,ldots x_n$ in $X$ and $O_1,ldots,O_n$ open in $Y$ such that $f in [x_1,ldots,x_n; O_1, O_2,ldots O_n] subseteq U_i subseteq U$.
For finite intersections we only need to check the intersection of two open sets $U_1$ and $U_2$ (finite then follows by induction, a standard argument), so let $f in U_1 cap U_2$. Then $f in U_1$ gives us $x_1,ldots x_n$ in $X$ and $O_1,ldots,O_n$ open in $Y$ such that $f in [x_1,ldots,x_n; O_1, O_2,ldots O_n] subseteq U_1$ and $f in U_2$ gives us $x'_1,ldots x'_m$ in $X$ and $O'_1,ldots,O'_m$ open in $Y$ such that $f in [x'_1,ldots,x'_m; O'_1, O'_2,ldots O'_m] subseteq U_2$. But then
$$f in [x_1,ldots,x_n, x'_1,ldots, x'_m; O_1,ldots,O_n, O'_1,ldots,O'_m] subseteq \
[x_1,ldots,x_n; O_1, O_2,ldots O_n] cap [x'_1,ldots,x'_m; O'_1, O'_2,ldots O'_m] subseteq U_1 cap U_2$$
as required.
You can also note that basic open neighbourhood of $f$ (the $[x_1,ldots,x_n; O_1, ldots, O_n]$) is of the form $B cap C(X,Y)$ where $B$ is a basic open subset of $Y^X$ in the product topology, given by a finite subset ${x_1, ldots,x_n}$ of "coordinates" from $X$ and finitely many $O_1, ldots, O_n$ in those coordinates. So e.g. $f_n to f$ in this topology iff $f_n(x) to f(x)$ for all $x in X$, as this holds for the product topology (and likewise for nets), hence the name "pointwise topology" for this space $C_p(X,Y)$. The "$p$" could also be seen as standing for "product", as it's just the product topology on the set, essentially.
answered Jan 20 at 7:57
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
add a comment |
$begingroup$
The union is trivial: if $U_i, i in I$ are all open and $f in U=bigcup_i U_i$, then for some $j in I$, $f in U_j$. By definition we then have $x_1,ldots x_n$ in $X$ and $O_1,ldots,O_n$ open in $Y$ such that $f in [x_1,ldots,x_n; O_1, O_2,ldots O_n] subseteq U_i subseteq U$.
For finite intersections we only need to check the intersection of two open sets $U_1$ and $U_2$ (finite then follows by induction, a standard argument), so let $f in U_1 cap U_2$. Then $f in U_1$ gives us $x_1,ldots x_n$ in $X$ and $O_1,ldots,O_n$ open in $Y$ such that $f in [x_1,ldots,x_n; O_1, O_2,ldots O_n] subseteq U_1$ and $f in U_2$ gives us $x'_1,ldots x'_m$ in $X$ and $O'_1,ldots,O'_m$ open in $Y$ such that $f in [x'_1,ldots,x'_m; O'_1, O'_2,ldots O'_m] subseteq U_2$. But then
$$f in [x_1,ldots,x_n, x'_1,ldots, x'_m; O_1,ldots,O_n, O'_1,ldots,O'_m] subseteq \
[x_1,ldots,x_n; O_1, O_2,ldots O_n] cap [x'_1,ldots,x'_m; O'_1, O'_2,ldots O'_m] subseteq U_1 cap U_2$$
as required.
You can also note that basic open neighbourhood of $f$ (the $[x_1,ldots,x_n; O_1, ldots, O_n]$) is of the form $B cap C(X,Y)$ where $B$ is a basic open subset of $Y^X$ in the product topology, given by a finite subset ${x_1, ldots,x_n}$ of "coordinates" from $X$ and finitely many $O_1, ldots, O_n$ in those coordinates. So e.g. $f_n to f$ in this topology iff $f_n(x) to f(x)$ for all $x in X$, as this holds for the product topology (and likewise for nets), hence the name "pointwise topology" for this space $C_p(X,Y)$. The "$p$" could also be seen as standing for "product", as it's just the product topology on the set, essentially.
answered Jan 20 at 7:57
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
The union is trivial: if $U_i, i in I$ are all open and $f in U=bigcup_i U_i$, then for some $j in I$, $f in U_j$. By definition we then have $x_1,ldots x_n$ in $X$ and $O_1,ldots,O_n$ open in $Y$ such that $f in [x_1,ldots,x_n; O_1, O_2,ldots O_n] subseteq U_i subseteq U$.
For finite intersections we only need to check the intersection of two open sets $U_1$ and $U_2$ (finite then follows by induction, a standard argument), so let $f in U_1 cap U_2$. Then $f in U_1$ gives us $x_1,ldots x_n$ in $X$ and $O_1,ldots,O_n$ open in $Y$ such that $f in [x_1,ldots,x_n; O_1, O_2,ldots O_n] subseteq U_1$ and $f in U_2$ gives us $x'_1,ldots x'_m$ in $X$ and $O'_1,ldots,O'_m$ open in $Y$ such that $f in [x'_1,ldots,x'_m; O'_1, O'_2,ldots O'_m] subseteq U_2$. But then
$$f in [x_1,ldots,x_n, x'_1,ldots, x'_m; O_1,ldots,O_n, O'_1,ldots,O'_m] subseteq \
[x_1,ldots,x_n; O_1, O_2,ldots O_n] cap [x'_1,ldots,x'_m; O'_1, O'_2,ldots O'_m] subseteq U_1 cap U_2$$
as required.
You can also note that basic open neighbourhood of $f$ (the $[x_1,ldots,x_n; O_1, ldots, O_n]$) is of the form $B cap C(X,Y)$ where $B$ is a basic open subset of $Y^X$ in the product topology, given by a finite subset ${x_1, ldots,x_n}$ of "coordinates" from $X$ and finitely many $O_1, ldots, O_n$ in those coordinates. So e.g. $f_n to f$ in this topology iff $f_n(x) to f(x)$ for all $x in X$, as this holds for the product topology (and likewise for nets), hence the name "pointwise topology" for this space $C_p(X,Y)$. The "$p$" could also be seen as standing for "product", as it's just the product topology on the set, essentially.
answered Jan 20 at 7:57
Henno BrandsmaHenno Brandsma
111k348119
answered Jan 20 at 7:57
Henno BrandsmaHenno Brandsma
111k348119
answered Jan 20 at 7:57
Henno BrandsmaHenno Brandsma
111k348119
answered Jan 20 at 7:57
Henno BrandsmaHenno Brandsma
111k348119
111k348119
add a comment |
add a comment |
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This is just the product topology that $C(X,Y)$ inherits as a subspace of $Y^X$.
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– Henno Brandsma
Jan 19 at 21:45
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@ArpanDas Shouldn't the union be easy too? If ${U_i}$ is a collection of open sets in $C(X,Y)$, then any $fin bigcup_i U_i$ belongs to some $U_{i_0}$ and we are done.
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– Dog_69
Jan 19 at 21:50
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And for the intersection, I think that $[x_1,dots,x_n,O_1,dots,O_n]cap[y_1,dots,y_m,P_1,dots,P_m] = [x_1,dots,x_n,y_1,dots,y_m,O_1,dots,O_n,P_1,dots,P_m]$, so for $finbigcap_{i=1}^n U_i$ you can consider the intersection all sets $[x_1,dots,x_n,O_1,dots O_n]$ associated with each $U_i$.
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– Dog_69
Jan 19 at 22:01
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I thought the $x_i$ s and $O_i$s depend on U. So for union I got confused. Intersection works although.
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– Arpan Das
Jan 20 at 4:29