Understanding definition of subbasis of product topology
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According to Munkres, the product topology has a subbasis which the union of all $S_i$ such that:
$$S_i ={pi_i^{-1}(U): U text{ is open in }X_i }$$
I am quite certain that I am misunderstanding this definition. To me this looks like $S_i$ contains all Cartesian products such that the i'th element is open in $X_i$. However, that would make the union of all $S_i$ the power set of the product space. It looks to me like Munkres, and other resources as well, are treating this definition as one where $S_i$ comprises all sets of the form
$$dots X_{i-3}times X_{i-2} times X_{i-1} times U times X_{i+1} times X_{i+2} times dots$$
Where U is an open set in $X_i$. All other sets in this product are the entire space $X_j$. Why is this the case? Why, for instance, doesn't $S_i$ contain this set:
$$dots U_{i-3}times U_{i-2} times U_{i-1} times U times U_{i+1} times U_{i+2} times dots$$
Where $U_j$ may or may not be open in $X_j$ for $j neq i$?
general-topology
asked Nov 19 at 22:01
Avatrin
6581420
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According to Munkres, the product topology has a subbasis which the union of all $S_i$ such that:
$$S_i ={pi_i^{-1}(U): U text{ is open in }X_i }$$
I am quite certain that I am misunderstanding this definition. To me this looks like $S_i$ contains all Cartesian products such that the i'th element is open in $X_i$. However, that would make the union of all $S_i$ the power set of the product space. It looks to me like Munkres, and other resources as well, are treating this definition as one where $S_i$ comprises all sets of the form
$$dots X_{i-3}times X_{i-2} times X_{i-1} times U times X_{i+1} times X_{i+2} times dots$$
Where U is an open set in $X_i$. All other sets in this product are the entire space $X_j$. Why is this the case? Why, for instance, doesn't $S_i$ contain this set:
$$dots U_{i-3}times U_{i-2} times U_{i-1} times U times U_{i+1} times U_{i+2} times dots$$
Where $U_j$ may or may not be open in $X_j$ for $j neq i$?
general-topology
asked Nov 19 at 22:01
Avatrin
6581420
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
According to Munkres, the product topology has a subbasis which the union of all $S_i$ such that:
$$S_i ={pi_i^{-1}(U): U text{ is open in }X_i }$$
I am quite certain that I am misunderstanding this definition. To me this looks like $S_i$ contains all Cartesian products such that the i'th element is open in $X_i$. However, that would make the union of all $S_i$ the power set of the product space. It looks to me like Munkres, and other resources as well, are treating this definition as one where $S_i$ comprises all sets of the form
$$dots X_{i-3}times X_{i-2} times X_{i-1} times U times X_{i+1} times X_{i+2} times dots$$
Where U is an open set in $X_i$. All other sets in this product are the entire space $X_j$. Why is this the case? Why, for instance, doesn't $S_i$ contain this set:
$$dots U_{i-3}times U_{i-2} times U_{i-1} times U times U_{i+1} times U_{i+2} times dots$$
Where $U_j$ may or may not be open in $X_j$ for $j neq i$?
general-topology
asked Nov 19 at 22:01
Avatrin
6581420
According to Munkres, the product topology has a subbasis which the union of all $S_i$ such that:
$$S_i ={pi_i^{-1}(U): U text{ is open in }X_i }$$
I am quite certain that I am misunderstanding this definition. To me this looks like $S_i$ contains all Cartesian products such that the i'th element is open in $X_i$. However, that would make the union of all $S_i$ the power set of the product space. It looks to me like Munkres, and other resources as well, are treating this definition as one where $S_i$ comprises all sets of the form
$$dots X_{i-3}times X_{i-2} times X_{i-1} times U times X_{i+1} times X_{i+2} times dots$$
Where U is an open set in $X_i$. All other sets in this product are the entire space $X_j$. Why is this the case? Why, for instance, doesn't $S_i$ contain this set:
$$dots U_{i-3}times U_{i-2} times U_{i-1} times U times U_{i+1} times U_{i+2} times dots$$
Where $U_j$ may or may not be open in $X_j$ for $j neq i$?
general-topology
general-topology
asked Nov 19 at 22:01
Avatrin
6581420
asked Nov 19 at 22:01
Avatrin
6581420
edited Nov 20 at 12:09
asked Nov 19 at 22:01
Avatrin
6581420
asked Nov 19 at 22:01
Avatrin
6581420
asked Nov 19 at 22:01
Avatrin
6581420
6581420
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add a comment |
2 Answers
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You are misunderstanding $pi_i^{-1}(U)$. This is the set which has $X_j$ in the $j^{th}$ slot (for $jneq i$) and $U$ in the $i^{th}$ slot.
For example, in $mathbb R^3$, one has $pi_1^{-1}(A) = Atimes{mathbb R}times{mathbb R}$.
In other words, $pi_i^{-1}(U)$ is the single set of all points in the product which project into $U$ under the action of $pi_i$.
answered Nov 19 at 22:06
MPW
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But your first formula is exactly what $pi_i^{-1}[U]$ means: the only condition for a point to be in that set, is that the $i$-th coordinate of that point is in $U$; all other coordinates are completely free.
It contains the other set as a subset, but topologies aren't closed under subsets, so that doesn't mean anything. It can be written as $bigcap_{i in I} pi_i^{-1}[U_i]$, but that will be an infinite intersection, and topologies are only guaranteed to be closed under finite intersections.
answered Nov 19 at 22:05
Henno Brandsma
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2 Answers
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2 Answers
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You are misunderstanding $pi_i^{-1}(U)$. This is the set which has $X_j$ in the $j^{th}$ slot (for $jneq i$) and $U$ in the $i^{th}$ slot.
For example, in $mathbb R^3$, one has $pi_1^{-1}(A) = Atimes{mathbb R}times{mathbb R}$.
In other words, $pi_i^{-1}(U)$ is the single set of all points in the product which project into $U$ under the action of $pi_i$.
answered Nov 19 at 22:06
MPW
29.8k11956
add a comment |
up vote
1
down vote
You are misunderstanding $pi_i^{-1}(U)$. This is the set which has $X_j$ in the $j^{th}$ slot (for $jneq i$) and $U$ in the $i^{th}$ slot.
For example, in $mathbb R^3$, one has $pi_1^{-1}(A) = Atimes{mathbb R}times{mathbb R}$.
In other words, $pi_i^{-1}(U)$ is the single set of all points in the product which project into $U$ under the action of $pi_i$.
answered Nov 19 at 22:06
MPW
29.8k11956
add a comment |
up vote
1
down vote
up vote
1
down vote
You are misunderstanding $pi_i^{-1}(U)$. This is the set which has $X_j$ in the $j^{th}$ slot (for $jneq i$) and $U$ in the $i^{th}$ slot.
For example, in $mathbb R^3$, one has $pi_1^{-1}(A) = Atimes{mathbb R}times{mathbb R}$.
In other words, $pi_i^{-1}(U)$ is the single set of all points in the product which project into $U$ under the action of $pi_i$.
answered Nov 19 at 22:06
MPW
29.8k11956
You are misunderstanding $pi_i^{-1}(U)$. This is the set which has $X_j$ in the $j^{th}$ slot (for $jneq i$) and $U$ in the $i^{th}$ slot.
For example, in $mathbb R^3$, one has $pi_1^{-1}(A) = Atimes{mathbb R}times{mathbb R}$.
In other words, $pi_i^{-1}(U)$ is the single set of all points in the product which project into $U$ under the action of $pi_i$.
answered Nov 19 at 22:06
MPW
29.8k11956
edited Nov 19 at 22:13
answered Nov 19 at 22:06
MPW
29.8k11956
answered Nov 19 at 22:06
MPW
29.8k11956
answered Nov 19 at 22:06
MPW
29.8k11956
29.8k11956
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up vote
1
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But your first formula is exactly what $pi_i^{-1}[U]$ means: the only condition for a point to be in that set, is that the $i$-th coordinate of that point is in $U$; all other coordinates are completely free.
It contains the other set as a subset, but topologies aren't closed under subsets, so that doesn't mean anything. It can be written as $bigcap_{i in I} pi_i^{-1}[U_i]$, but that will be an infinite intersection, and topologies are only guaranteed to be closed under finite intersections.
answered Nov 19 at 22:05
Henno Brandsma
103k345112
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up vote
1
down vote
But your first formula is exactly what $pi_i^{-1}[U]$ means: the only condition for a point to be in that set, is that the $i$-th coordinate of that point is in $U$; all other coordinates are completely free.
It contains the other set as a subset, but topologies aren't closed under subsets, so that doesn't mean anything. It can be written as $bigcap_{i in I} pi_i^{-1}[U_i]$, but that will be an infinite intersection, and topologies are only guaranteed to be closed under finite intersections.
answered Nov 19 at 22:05
Henno Brandsma
103k345112
add a comment |
up vote
1
down vote
up vote
1
down vote
But your first formula is exactly what $pi_i^{-1}[U]$ means: the only condition for a point to be in that set, is that the $i$-th coordinate of that point is in $U$; all other coordinates are completely free.
It contains the other set as a subset, but topologies aren't closed under subsets, so that doesn't mean anything. It can be written as $bigcap_{i in I} pi_i^{-1}[U_i]$, but that will be an infinite intersection, and topologies are only guaranteed to be closed under finite intersections.
answered Nov 19 at 22:05
Henno Brandsma
103k345112
But your first formula is exactly what $pi_i^{-1}[U]$ means: the only condition for a point to be in that set, is that the $i$-th coordinate of that point is in $U$; all other coordinates are completely free.
It contains the other set as a subset, but topologies aren't closed under subsets, so that doesn't mean anything. It can be written as $bigcap_{i in I} pi_i^{-1}[U_i]$, but that will be an infinite intersection, and topologies are only guaranteed to be closed under finite intersections.
answered Nov 19 at 22:05
Henno Brandsma
103k345112
edited Nov 19 at 22:23
answered Nov 19 at 22:05
Henno Brandsma
103k345112
answered Nov 19 at 22:05
Henno Brandsma
103k345112
answered Nov 19 at 22:05
Henno Brandsma
103k345112
103k345112
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