Problem proving that topology product is a topology
up vote
1
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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 Nov 19 at 18:22
user540275
597
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 Nov 19 at 18:22
user540275
597
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
Nov 19 at 19:53
Yes. It is valid only for intersections
– Dog_69
Nov 19 at 20:37
So your argument is not valid, yeah? How it is for unions?
– user540275
Nov 19 at 20:52
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
Nov 19 at 21:54
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 Nov 19 at 18:22
user540275
597
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 Nov 19 at 18:22
user540275
597
asked Nov 19 at 18:22
user540275
597
asked Nov 19 at 18:22
user540275
597
asked Nov 19 at 18:22
user540275
597
asked Nov 19 at 18:22
user540275
597
597
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
Nov 19 at 19:53
Yes. It is valid only for intersections
– Dog_69
Nov 19 at 20:37
So your argument is not valid, yeah? How it is for unions?
– user540275
Nov 19 at 20:52
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
Nov 19 at 21:54
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
Nov 19 at 19:53
Yes. It is valid only for intersections
– Dog_69
Nov 19 at 20:37
So your argument is not valid, yeah? How it is for unions?
– user540275
Nov 19 at 20:52
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
Nov 19 at 21:54
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
Nov 19 at 19:53
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
Nov 19 at 19:53
Yes. It is valid only for intersections
– Dog_69
Nov 19 at 20:37
Yes. It is valid only for intersections
– Dog_69
Nov 19 at 20:37
So your argument is not valid, yeah? How it is for unions?
– user540275
Nov 19 at 20:52
So your argument is not valid, yeah? How it is for unions?
– user540275
Nov 19 at 20:52
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
Nov 19 at 21:54
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
Nov 19 at 21:54
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 Nov 19 at 21:18
Henno Brandsma
103k345112
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 Nov 19 at 21:43
Carl Mummert
65.8k7131246
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
Nov 19 at 21:51
@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
Nov 19 at 21:52
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
Nov 19 at 21:55
add a comment |
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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 Nov 19 at 21:18
Henno Brandsma
103k345112
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 Nov 19 at 21:18
Henno Brandsma
103k345112
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 Nov 19 at 21:18
Henno Brandsma
103k345112
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 Nov 19 at 21:18
Henno Brandsma
103k345112
edited Nov 19 at 22:00
answered Nov 19 at 21:18
Henno Brandsma
103k345112
answered Nov 19 at 21:18
Henno Brandsma
103k345112
answered Nov 19 at 21:18
Henno Brandsma
103k345112
103k345112
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 Nov 19 at 21:43
Carl Mummert
65.8k7131246
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
Nov 19 at 21:51
@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
Nov 19 at 21:52
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
Nov 19 at 21:55
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 Nov 19 at 21:43
Carl Mummert
65.8k7131246
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
Nov 19 at 21:51
@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
Nov 19 at 21:52
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
Nov 19 at 21:55
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 Nov 19 at 21:43
Carl Mummert
65.8k7131246
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 Nov 19 at 21:43
Carl Mummert
65.8k7131246
edited Nov 19 at 21:52
answered Nov 19 at 21:43
Carl Mummert
65.8k7131246
answered Nov 19 at 21:43
Carl Mummert
65.8k7131246
answered Nov 19 at 21:43
Carl Mummert
65.8k7131246
65.8k7131246
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
Nov 19 at 21:51
@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
Nov 19 at 21:52
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
Nov 19 at 21:55
add a comment |
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
Nov 19 at 21:51
@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
Nov 19 at 21:52
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
Nov 19 at 21:55
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
Nov 19 at 21:51
How you are defined $U_k'' = U_k$ and $U_k''=U_{k'}$ simultaneosly?
– user540275
Nov 19 at 21:51
@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
Nov 19 at 21:52
@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
Nov 19 at 21:52
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
Nov 19 at 21:55
Maybe it makes more sense for me to write it as $U'_k$, the function corresponding to $K'$ evaluated at $k$.
– Carl Mummert
Nov 19 at 21:55
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
Nov 19 at 19:53
Yes. It is valid only for intersections
– Dog_69
Nov 19 at 20:37
So your argument is not valid, yeah? How it is for unions?
– user540275
Nov 19 at 20:52
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
Nov 19 at 21:54