Mixing
Problem proving that topology product is a topology
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Given $(X,mathcal{F}),(Y,mathcal{F}')$ two topological spaces, we define for $X times Y$ the topology product as $$mathcal{F}_{prod} := {cup_{k in K}(U_k times V_k) : U_k in mathcal{F}, V_k in mathcal{F}'}.$$
How we can prove that verify the second condition of a topology (that is, the union of elements also belongs to $mathcal{F}_{prod}$)?
We get $cup_i (cup_{k_i} (U_{k_i} times V_{k_i})) = cup_{k_i} (cup_i (U_{k_i} times V_{k_i})) = ...$?
general-topology
asked 2 days ago
user540275
306
add a comment |
up vote
1
down vote
favorite
Given $(X,mathcal{F}),(Y,mathcal{F}')$ two topological spaces, we define for $X times Y$ the topology product as $$mathcal{F}_{prod} := {cup_{k in K}(U_k times V_k) : U_k in mathcal{F}, V_k in mathcal{F}'}.$$
How we can prove that verify the second condition of a topology (that is, the union of elements also belongs to $mathcal{F}_{prod}$)?
We get $cup_i (cup_{k_i} (U_{k_i} times V_{k_i})) = cup_{k_i} (cup_i (U_{k_i} times V_{k_i})) = ...$?
general-topology
asked 2 days ago
user540275
306
1
Really it is $(A times B) cup (A' times B') subset (A cup A') times (B cup B') = (A times B) cup (A' times B') cup (A times B') cup (A' times B).$ I think is not valid your argument.
– user540275
2 days ago
Yes. It is valid only for intersections
– Dog_69
2 days ago
So your argument is not valid, yeah? How it is for unions?
– user540275
2 days ago
1
If you let $bigsqcup_{iin I} K_i = { (i, k) mid i in I, k in K_i }$, then $bigcup_i (bigcup_{k_i} (U_{k_i} times V_{k_i})) = bigcup_{(i, k_i) in bigsqcup_{iin I} K_i} (U_{k_i} times V_{k_i})$.
– Daniel Schepler
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $(X,mathcal{F}),(Y,mathcal{F}')$ two topological spaces, we define for $X times Y$ the topology product as $$mathcal{F}_{prod} := {cup_{k in K}(U_k times V_k) : U_k in mathcal{F}, V_k in mathcal{F}'}.$$
How we can prove that verify the second condition of a topology (that is, the union of elements also belongs to $mathcal{F}_{prod}$)?
We get $cup_i (cup_{k_i} (U_{k_i} times V_{k_i})) = cup_{k_i} (cup_i (U_{k_i} times V_{k_i})) = ...$?
general-topology
asked 2 days ago
user540275
306
Given $(X,mathcal{F}),(Y,mathcal{F}')$ two topological spaces, we define for $X times Y$ the topology product as $$mathcal{F}_{prod} := {cup_{k in K}(U_k times V_k) : U_k in mathcal{F}, V_k in mathcal{F}'}.$$
How we can prove that verify the second condition of a topology (that is, the union of elements also belongs to $mathcal{F}_{prod}$)?
We get $cup_i (cup_{k_i} (U_{k_i} times V_{k_i})) = cup_{k_i} (cup_i (U_{k_i} times V_{k_i})) = ...$?
general-topology
general-topology
asked 2 days ago
user540275
306
asked 2 days ago
user540275
306
asked 2 days ago
user540275
306
asked 2 days ago
user540275
306
asked 2 days ago
user540275
306
306
1
Really it is $(A times B) cup (A' times B') subset (A cup A') times (B cup B') = (A times B) cup (A' times B') cup (A times B') cup (A' times B).$ I think is not valid your argument.
– user540275
2 days ago
Yes. It is valid only for intersections
– Dog_69
2 days ago
So your argument is not valid, yeah? How it is for unions?
– user540275
2 days ago
1
If you let $bigsqcup_{iin I} K_i = { (i, k) mid i in I, k in K_i }$, then $bigcup_i (bigcup_{k_i} (U_{k_i} times V_{k_i})) = bigcup_{(i, k_i) in bigsqcup_{iin I} K_i} (U_{k_i} times V_{k_i})$.
– Daniel Schepler
yesterday
add a comment |
1
Really it is $(A times B) cup (A' times B') subset (A cup A') times (B cup B') = (A times B) cup (A' times B') cup (A times B') cup (A' times B).$ I think is not valid your argument.
– user540275
2 days ago
Yes. It is valid only for intersections
– Dog_69
2 days ago
So your argument is not valid, yeah? How it is for unions?
– user540275
2 days ago
1
If you let $bigsqcup_{iin I} K_i = { (i, k) mid i in I, k in K_i }$, then $bigcup_i (bigcup_{k_i} (U_{k_i} times V_{k_i})) = bigcup_{(i, k_i) in bigsqcup_{iin I} K_i} (U_{k_i} times V_{k_i})$.
– Daniel Schepler
yesterday
1
1
Really it is $(A times B) cup (A' times B') subset (A cup A') times (B cup B') = (A times B) cup (A' times B') cup (A times B') cup (A' times B).$ I think is not valid your argument.
– user540275
2 days ago
Really it is $(A times B) cup (A' times B') subset (A cup A') times (B cup B') = (A times B) cup (A' times B') cup (A times B') cup (A' times B).$ I think is not valid your argument.
– user540275
2 days ago
Yes. It is valid only for intersections
– Dog_69
2 days ago
Yes. It is valid only for intersections
– Dog_69
2 days ago
So your argument is not valid, yeah? How it is for unions?
– user540275
2 days ago
So your argument is not valid, yeah? How it is for unions?
– user540275
2 days ago
1
1
If you let $bigsqcup_{iin I} K_i = { (i, k) mid i in I, k in K_i }$, then $bigcup_i (bigcup_{k_i} (U_{k_i} times V_{k_i})) = bigcup_{(i, k_i) in bigsqcup_{iin I} K_i} (U_{k_i} times V_{k_i})$.
– Daniel Schepler
yesterday
If you let $bigsqcup_{iin I} K_i = { (i, k) mid i in I, k in K_i }$, then $bigcup_i (bigcup_{k_i} (U_{k_i} times V_{k_i})) = bigcup_{(i, k_i) in bigsqcup_{iin I} K_i} (U_{k_i} times V_{k_i})$.
– Daniel Schepler
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
A union of unions of open boxes is again a union of open boxes.
That’s the whole essence as to unions. If you want formula:
Suppose we have $O_i = bigcup_{j in K_i} (U_j times V_j)$, where for each $i$ we have an index set $K_i$ that we take the union of open boxes over (we can always ensure that these index sets are pairwise disjoint for convenience, which we'll assume), then:
$$ bigcup_{i in I} O_i = bigcup_i left(bigcup_{j in K_i} (U_j times V_j)right) = bigcup_{j in bigcup_{iin I} K_i} (U_j times V_j)$$
For finite intersections observe that $$left(bigcup_{iin I} (U_i times V_i)right) cap left( bigcup_{j in J} (U_j times V_j) right)= bigcup_{(i,j) in I times J} (U_i cap U_j) times (V_i cap V_j) $$
which is again of the same form.
answered yesterday
Henno Brandsma
101k344107
add a comment |
up vote
0
down vote
There is something missing in your definition of the product topology: what is $K$? I think the definition you want is that, for every set $K$ and every function $U colon K ni k mapsto U_k in T_X$ and $V: K ni k mapsto V_k in T_Y$, the set $bigcup_{k in K} U_k times V_k$ is open in the product topology. (There is a little set theory to show this is a valid definition, but it seems to be what the notation is supposed to suggest.)
In this case, given one open set given by $K$ and one given by $K'$, we can assume wlog that $K$ and $K'$ are disjoint, let
$K'' = K cup K'$, let $U''_k = U_k$ for $k in K$ and $U''_k = U_{k}'$ for $ k in K'$, and similarly for $V''$, and then
the union is given by $bigcup_{k in k''} U''_k times V''_k$.
If we avoid the confused notation, what the problem is claiming is just that the family ${ U times V : u in T_X, V in T_Y}$ is a basis for the product topology.
answered yesterday
Carl Mummert
65.5k7130243
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
yesterday
@user540275: wlog we can assume $K$ and $K'$ are disjoint, so each $k in K times K'$ is only in one of them.
– Carl Mummert
yesterday
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
A union of unions of open boxes is again a union of open boxes.
That’s the whole essence as to unions. If you want formula:
Suppose we have $O_i = bigcup_{j in K_i} (U_j times V_j)$, where for each $i$ we have an index set $K_i$ that we take the union of open boxes over (we can always ensure that these index sets are pairwise disjoint for convenience, which we'll assume), then:
$$ bigcup_{i in I} O_i = bigcup_i left(bigcup_{j in K_i} (U_j times V_j)right) = bigcup_{j in bigcup_{iin I} K_i} (U_j times V_j)$$
For finite intersections observe that $$left(bigcup_{iin I} (U_i times V_i)right) cap left( bigcup_{j in J} (U_j times V_j) right)= bigcup_{(i,j) in I times J} (U_i cap U_j) times (V_i cap V_j) $$
which is again of the same form.
answered yesterday
Henno Brandsma
101k344107
add a comment |
up vote
1
down vote
A union of unions of open boxes is again a union of open boxes.
That’s the whole essence as to unions. If you want formula:
Suppose we have $O_i = bigcup_{j in K_i} (U_j times V_j)$, where for each $i$ we have an index set $K_i$ that we take the union of open boxes over (we can always ensure that these index sets are pairwise disjoint for convenience, which we'll assume), then:
$$ bigcup_{i in I} O_i = bigcup_i left(bigcup_{j in K_i} (U_j times V_j)right) = bigcup_{j in bigcup_{iin I} K_i} (U_j times V_j)$$
For finite intersections observe that $$left(bigcup_{iin I} (U_i times V_i)right) cap left( bigcup_{j in J} (U_j times V_j) right)= bigcup_{(i,j) in I times J} (U_i cap U_j) times (V_i cap V_j) $$
which is again of the same form.
answered yesterday
Henno Brandsma
101k344107
add a comment |
up vote
1
down vote
up vote
1
down vote
A union of unions of open boxes is again a union of open boxes.
That’s the whole essence as to unions. If you want formula:
Suppose we have $O_i = bigcup_{j in K_i} (U_j times V_j)$, where for each $i$ we have an index set $K_i$ that we take the union of open boxes over (we can always ensure that these index sets are pairwise disjoint for convenience, which we'll assume), then:
$$ bigcup_{i in I} O_i = bigcup_i left(bigcup_{j in K_i} (U_j times V_j)right) = bigcup_{j in bigcup_{iin I} K_i} (U_j times V_j)$$
For finite intersections observe that $$left(bigcup_{iin I} (U_i times V_i)right) cap left( bigcup_{j in J} (U_j times V_j) right)= bigcup_{(i,j) in I times J} (U_i cap U_j) times (V_i cap V_j) $$
which is again of the same form.
answered yesterday
Henno Brandsma
101k344107
A union of unions of open boxes is again a union of open boxes.
That’s the whole essence as to unions. If you want formula:
Suppose we have $O_i = bigcup_{j in K_i} (U_j times V_j)$, where for each $i$ we have an index set $K_i$ that we take the union of open boxes over (we can always ensure that these index sets are pairwise disjoint for convenience, which we'll assume), then:
$$ bigcup_{i in I} O_i = bigcup_i left(bigcup_{j in K_i} (U_j times V_j)right) = bigcup_{j in bigcup_{iin I} K_i} (U_j times V_j)$$
For finite intersections observe that $$left(bigcup_{iin I} (U_i times V_i)right) cap left( bigcup_{j in J} (U_j times V_j) right)= bigcup_{(i,j) in I times J} (U_i cap U_j) times (V_i cap V_j) $$
which is again of the same form.
answered yesterday
Henno Brandsma
101k344107
edited yesterday
answered yesterday
Henno Brandsma
101k344107
answered yesterday
Henno Brandsma
101k344107
answered yesterday
Henno Brandsma
101k344107
101k344107
add a comment |
add a comment |
up vote
0
down vote
There is something missing in your definition of the product topology: what is $K$? I think the definition you want is that, for every set $K$ and every function $U colon K ni k mapsto U_k in T_X$ and $V: K ni k mapsto V_k in T_Y$, the set $bigcup_{k in K} U_k times V_k$ is open in the product topology. (There is a little set theory to show this is a valid definition, but it seems to be what the notation is supposed to suggest.)
In this case, given one open set given by $K$ and one given by $K'$, we can assume wlog that $K$ and $K'$ are disjoint, let
$K'' = K cup K'$, let $U''_k = U_k$ for $k in K$ and $U''_k = U_{k}'$ for $ k in K'$, and similarly for $V''$, and then
the union is given by $bigcup_{k in k''} U''_k times V''_k$.
If we avoid the confused notation, what the problem is claiming is just that the family ${ U times V : u in T_X, V in T_Y}$ is a basis for the product topology.
answered yesterday
Carl Mummert
65.5k7130243
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
yesterday
@user540275: wlog we can assume $K$ and $K'$ are disjoint, so each $k in K times K'$ is only in one of them.
– Carl Mummert
yesterday
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
yesterday
add a comment |
up vote
0
down vote
There is something missing in your definition of the product topology: what is $K$? I think the definition you want is that, for every set $K$ and every function $U colon K ni k mapsto U_k in T_X$ and $V: K ni k mapsto V_k in T_Y$, the set $bigcup_{k in K} U_k times V_k$ is open in the product topology. (There is a little set theory to show this is a valid definition, but it seems to be what the notation is supposed to suggest.)
In this case, given one open set given by $K$ and one given by $K'$, we can assume wlog that $K$ and $K'$ are disjoint, let
$K'' = K cup K'$, let $U''_k = U_k$ for $k in K$ and $U''_k = U_{k}'$ for $ k in K'$, and similarly for $V''$, and then
the union is given by $bigcup_{k in k''} U''_k times V''_k$.
If we avoid the confused notation, what the problem is claiming is just that the family ${ U times V : u in T_X, V in T_Y}$ is a basis for the product topology.
answered yesterday
Carl Mummert
65.5k7130243
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
yesterday
@user540275: wlog we can assume $K$ and $K'$ are disjoint, so each $k in K times K'$ is only in one of them.
– Carl Mummert
yesterday
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
There is something missing in your definition of the product topology: what is $K$? I think the definition you want is that, for every set $K$ and every function $U colon K ni k mapsto U_k in T_X$ and $V: K ni k mapsto V_k in T_Y$, the set $bigcup_{k in K} U_k times V_k$ is open in the product topology. (There is a little set theory to show this is a valid definition, but it seems to be what the notation is supposed to suggest.)
In this case, given one open set given by $K$ and one given by $K'$, we can assume wlog that $K$ and $K'$ are disjoint, let
$K'' = K cup K'$, let $U''_k = U_k$ for $k in K$ and $U''_k = U_{k}'$ for $ k in K'$, and similarly for $V''$, and then
the union is given by $bigcup_{k in k''} U''_k times V''_k$.
If we avoid the confused notation, what the problem is claiming is just that the family ${ U times V : u in T_X, V in T_Y}$ is a basis for the product topology.
answered yesterday
Carl Mummert
65.5k7130243
There is something missing in your definition of the product topology: what is $K$? I think the definition you want is that, for every set $K$ and every function $U colon K ni k mapsto U_k in T_X$ and $V: K ni k mapsto V_k in T_Y$, the set $bigcup_{k in K} U_k times V_k$ is open in the product topology. (There is a little set theory to show this is a valid definition, but it seems to be what the notation is supposed to suggest.)
In this case, given one open set given by $K$ and one given by $K'$, we can assume wlog that $K$ and $K'$ are disjoint, let
$K'' = K cup K'$, let $U''_k = U_k$ for $k in K$ and $U''_k = U_{k}'$ for $ k in K'$, and similarly for $V''$, and then
the union is given by $bigcup_{k in k''} U''_k times V''_k$.
If we avoid the confused notation, what the problem is claiming is just that the family ${ U times V : u in T_X, V in T_Y}$ is a basis for the product topology.
answered yesterday
Carl Mummert
65.5k7130243
edited yesterday
answered yesterday
Carl Mummert
65.5k7130243
answered yesterday
Carl Mummert
65.5k7130243
answered yesterday
Carl Mummert
65.5k7130243
65.5k7130243
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
yesterday
@user540275: wlog we can assume $K$ and $K'$ are disjoint, so each $k in K times K'$ is only in one of them.
– Carl Mummert
yesterday
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
yesterday
add a comment |
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
yesterday
@user540275: wlog we can assume $K$ and $K'$ are disjoint, so each $k in K times K'$ is only in one of them.
– Carl Mummert
yesterday
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
yesterday
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
yesterday
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
yesterday
@user540275: wlog we can assume $K$ and $K'$ are disjoint, so each $k in K times K'$ is only in one of them.
– Carl Mummert
yesterday
@user540275: wlog we can assume $K$ and $K'$ are disjoint, so each $k in K times K'$ is only in one of them.
– Carl Mummert
yesterday
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
yesterday
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
yesterday
add a comment |
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1
Really it is $(A times B) cup (A' times B') subset (A cup A') times (B cup B') = (A times B) cup (A' times B') cup (A times B') cup (A' times B).$ I think is not valid your argument.
– user540275
2 days ago
Yes. It is valid only for intersections
– Dog_69
2 days ago
So your argument is not valid, yeah? How it is for unions?
– user540275
2 days ago
1
If you let $bigsqcup_{iin I} K_i = { (i, k) mid i in I, k in K_i }$, then $bigcup_i (bigcup_{k_i} (U_{k_i} times V_{k_i})) = bigcup_{(i, k_i) in bigsqcup_{iin I} K_i} (U_{k_i} times V_{k_i})$.
– Daniel Schepler
yesterday