Mixing
Intuition/clarification for product topology in Munkres' Topology
$begingroup$
In Topology, the second edition by Munkres, in section 19, on page 113 he says the following (when talking about how to impose topologies on set):
"Another way to proceed is to generalize the subbasis formulation of the definition, given in §15. In this case, we take as a subbasis all the sets of the form $pi^{-1}_i(U_i)$, where $i$ is any index and $U_i$ is an open set of $X_i$. We shall call this topology the product topology. How do these topologies differ? (Munkres is referring to box vs product) Consider the typical basis element $B$ for the second topology. It is a finite intersection of subbasis elements $pi^{-1}_i(U_i)$, say for $i=i_1, dots i_k$. Then a point $boldsymbol{x}$ belongs to $B$ if and only if $pi_i(boldsymbol{x})$ belongs to $U_i$ for $i=i_1, dots i_k$; there is no restriction on $pi_i(boldsymbol{x})$ for other values of $i$."
Questions I have:
1. $pi^{-1}_i(U_i)$ is the set $U_i times Y ?$ Because he defines $pi_1:X times Y rightarrow X$
2. He says: "In this case, we take as a subbasis all the sets of the form $pi^{-1}_i(U_i)$". Why is this a subbasis? And what do the basis elements look like?
3. If it's possible to explain what he means after "How do these topologies differ?". Somehow from that explanation, the box topology and the product topology agree for the finite cartesian product and differ for the infinite product. I find this hard to understand.
real-analysis general-topology intuition
asked Feb 1 at 23:45
urxwrturxwrt
144
$endgroup$
add a comment |
$begingroup$
In Topology, the second edition by Munkres, in section 19, on page 113 he says the following (when talking about how to impose topologies on set):
"Another way to proceed is to generalize the subbasis formulation of the definition, given in §15. In this case, we take as a subbasis all the sets of the form $pi^{-1}_i(U_i)$, where $i$ is any index and $U_i$ is an open set of $X_i$. We shall call this topology the product topology. How do these topologies differ? (Munkres is referring to box vs product) Consider the typical basis element $B$ for the second topology. It is a finite intersection of subbasis elements $pi^{-1}_i(U_i)$, say for $i=i_1, dots i_k$. Then a point $boldsymbol{x}$ belongs to $B$ if and only if $pi_i(boldsymbol{x})$ belongs to $U_i$ for $i=i_1, dots i_k$; there is no restriction on $pi_i(boldsymbol{x})$ for other values of $i$."
Questions I have:
1. $pi^{-1}_i(U_i)$ is the set $U_i times Y ?$ Because he defines $pi_1:X times Y rightarrow X$
2. He says: "In this case, we take as a subbasis all the sets of the form $pi^{-1}_i(U_i)$". Why is this a subbasis? And what do the basis elements look like?
3. If it's possible to explain what he means after "How do these topologies differ?". Somehow from that explanation, the box topology and the product topology agree for the finite cartesian product and differ for the infinite product. I find this hard to understand.
real-analysis general-topology intuition
asked Feb 1 at 23:45
urxwrturxwrt
144
$endgroup$
add a comment |
$begingroup$
In Topology, the second edition by Munkres, in section 19, on page 113 he says the following (when talking about how to impose topologies on set):
"Another way to proceed is to generalize the subbasis formulation of the definition, given in §15. In this case, we take as a subbasis all the sets of the form $pi^{-1}_i(U_i)$, where $i$ is any index and $U_i$ is an open set of $X_i$. We shall call this topology the product topology. How do these topologies differ? (Munkres is referring to box vs product) Consider the typical basis element $B$ for the second topology. It is a finite intersection of subbasis elements $pi^{-1}_i(U_i)$, say for $i=i_1, dots i_k$. Then a point $boldsymbol{x}$ belongs to $B$ if and only if $pi_i(boldsymbol{x})$ belongs to $U_i$ for $i=i_1, dots i_k$; there is no restriction on $pi_i(boldsymbol{x})$ for other values of $i$."
Questions I have:
1. $pi^{-1}_i(U_i)$ is the set $U_i times Y ?$ Because he defines $pi_1:X times Y rightarrow X$
2. He says: "In this case, we take as a subbasis all the sets of the form $pi^{-1}_i(U_i)$". Why is this a subbasis? And what do the basis elements look like?
3. If it's possible to explain what he means after "How do these topologies differ?". Somehow from that explanation, the box topology and the product topology agree for the finite cartesian product and differ for the infinite product. I find this hard to understand.
real-analysis general-topology intuition
asked Feb 1 at 23:45
urxwrturxwrt
144
$endgroup$
In Topology, the second edition by Munkres, in section 19, on page 113 he says the following (when talking about how to impose topologies on set):
"Another way to proceed is to generalize the subbasis formulation of the definition, given in §15. In this case, we take as a subbasis all the sets of the form $pi^{-1}_i(U_i)$, where $i$ is any index and $U_i$ is an open set of $X_i$. We shall call this topology the product topology. How do these topologies differ? (Munkres is referring to box vs product) Consider the typical basis element $B$ for the second topology. It is a finite intersection of subbasis elements $pi^{-1}_i(U_i)$, say for $i=i_1, dots i_k$. Then a point $boldsymbol{x}$ belongs to $B$ if and only if $pi_i(boldsymbol{x})$ belongs to $U_i$ for $i=i_1, dots i_k$; there is no restriction on $pi_i(boldsymbol{x})$ for other values of $i$."
Questions I have:
1. $pi^{-1}_i(U_i)$ is the set $U_i times Y ?$ Because he defines $pi_1:X times Y rightarrow X$
2. He says: "In this case, we take as a subbasis all the sets of the form $pi^{-1}_i(U_i)$". Why is this a subbasis? And what do the basis elements look like?
3. If it's possible to explain what he means after "How do these topologies differ?". Somehow from that explanation, the box topology and the product topology agree for the finite cartesian product and differ for the infinite product. I find this hard to understand.
real-analysis general-topology intuition
real-analysis general-topology intuition
asked Feb 1 at 23:45
urxwrturxwrt
144
asked Feb 1 at 23:45
urxwrturxwrt
144
edited Feb 3 at 3:06
urxwrt
asked Feb 1 at 23:45
urxwrturxwrt
144
asked Feb 1 at 23:45
urxwrturxwrt
144
asked Feb 1 at 23:45
urxwrturxwrt
144
144
add a comment |
add a comment |
1 Answer
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Question 1
I'm not sure what you mean by $Y$ here, but suppose you have a family of topological space $(X_alpha)_{alpha in A}$, and we define the product $Y = prod_{alpha in A} X_alpha$ (our index set $A$ need not be countable). The elements of $Y$ are essentially functions $f$, mapping from $A$ to the union of all the $X_alpha$s, where $f(alpha) in X_alpha$. For each $alpha in A$, we can naturally define a projection map $pi_alpha$, which takes $f in Y$, and maps it to $f(alpha)$, giving us a map from $Y$ to $X_alpha$.
(In the case of a finite product, we can make our index set $A = {1, 2, ldots, n}$ for some $n$. We can also think of our elements of $Y$ as $n$-tuples. Then, given an $n$-tuple $y in Y$, $pi_i(y)$ is the $i$th coordinate of $y$. Outside of finite/countable settings, it makes more sense to think of the elements as functions rather than tuples.)
If $U_alpha$ is open in $X_alpha$, the set $pi_{alpha}^{-1}(U_alpha)$ is the set of functions in $Y$ satisfying $f(alpha) in U_alpha$. For other indices $beta neq alpha$, $f(beta)$ is unrestricted, other than the usual restriction of $f(beta) in X_beta$. So we have
$$pi^{-1}_alpha(U_alpha) = prod_{beta in A} X'_beta,$$
where
$$X'_beta = begin{cases} X_beta & text{if } beta neq alpha \ U_alpha & text{if } beta = alpha end{cases}.$$
Question 2
This is a subbasis because it is not necessarily the case that the intersection of two of these sets will be a union of basis elements. Even if we restrict ourselves to $mathbb{R} times mathbb{R}$, these sets will not be a basis. For example, we can express an open square in the form:
$$(0, 1) times (0, 1) = ((0, 1) times mathbb{R}) cap (mathbb{R} times (0, 1)) = pi_1^{-1}(0, 1) cap pi_2^{-1}(0, 1),$$
making it the intersection of two (presumably) open sets in our (supposed) topology, which should make it open. But, this is a bounded set, and $pi_i^{-1}(U)$ is unbounded for any open $U subseteq mathbb{R}$ and $i in {0, 1}$. So, there's no way to express the square as a union of our supposedly basic sets, which makes the sets not basic!
To form a basis from a subbasis, all you do is take all finite intersections of your subbasic sets. So, our basis will be sets of the of the form
$$pi_{alpha_1}^{-1}(U_{alpha_1}) cap ldots cap pi_{alpha_n}^{-1}(U_{alpha_n}),$$
where $alpha_1, ldots, alpha_n in A$ and $U_{alpha_i} in X_{alpha_i}$ for all $i$. Equivalently, they are sets of the form
$$prod_{alpha in A} X'_alpha,$$
where $X'_alpha = X_alpha$ for all $alpha in A$ except for a number of $alpha$s. For such $alpha$s, $X'_alpha$ is an open subset of $X_alpha$.
Question 3
The box topology produces sets of the form
$$prod_{alpha in A} X'_alpha,$$
where $X'_alpha$ is an open subset of $X_alpha$ for all $alpha in A$. That is, we don't require $X'_alpha = X_alpha$ for all but a finite number of $alpha$s. Every coordinate can be restricted at the same time.
For example, take the product $Y = prod_{i=1}^infty mathbb{R}$, and the subset $U = prod_{i=1}^infty (-i, i)$. Since $(-i, i) subseteq mathbb{R}$ is open for all $i$, $U$ is open in $Y$ under the box topology. But, $(-i, i) neq mathbb{R}$ for infinitely many $i$ (in fact, for all $i$), so $U$ is not a basic element in the product topology.
This doesn't necessarily mean it's not open in the product topology, but this is true as well. If it were open in the product topology, it would have to contain a non-empty basic set. Such a set would have to take the form
$$U_1 times U_2 times ldots times U_n times mathbb{R} times mathbb{R} times ldots,$$
where $U_1, ldots, U_n$ are open subsets of $mathbb{R}$ (possibly equal to $mathbb{R}$). But, this means that the $n + 1$th coordinate is unrestricted, but yet is still confined to $(-n - 1, n + 1)$, which is a contradiction. So, no basic set can be contained in $U$, and hence $U$ is not open with respect to the product topology.
answered Feb 2 at 0:47
Theo BenditTheo Bendit
20.9k12355
$endgroup$
$begingroup$
Hello, sorry for the late reply. For question 1, I am confused how The elements of Y are essentially functions $f$ I clarified what I mean by $Y$ in the edit.
$endgroup$
– urxwrt
Feb 3 at 3:16
$begingroup$
It's another way of thinking about "tuples" that doesn't assume your index set is finite, countable, or ordered in any way. If you have a product of, say, $3$ spaces, you can think of the points in the product as ordered triples, where the first entry belongs to the first space, the second entry to the second, and the third to the third. But, when you have infinitely many spaces, say, one for each element of $mathbb{R}$, it makes more sense to think of functions instead. An element of, say ${1,0}$ multiplied to itself $|mathbb{R}|$ times is a function from $mathbb{R}$ to ${1,0}.$
$endgroup$
– Theo Bendit
Feb 3 at 5:32
$begingroup$
You couldn't think of this as "tuples" in any sensible way. There's no "first" entry, and certainly no "next" entry (unless you wish to well-order the reals using the axiom of choice). You might have an element of this product space that assigns, to the "$x$th" entry (where $x in mathbb{R}$), $1$ when $x$ is rational, and $0$ when $x$ is irrational. That would be one element of the product. It's not possible to write it as an $infty$-tuple, but it makes a lot of sense as a function.
$endgroup$
– Theo Bendit
Feb 3 at 5:34
add a comment |
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$begingroup$
Question 1
I'm not sure what you mean by $Y$ here, but suppose you have a family of topological space $(X_alpha)_{alpha in A}$, and we define the product $Y = prod_{alpha in A} X_alpha$ (our index set $A$ need not be countable). The elements of $Y$ are essentially functions $f$, mapping from $A$ to the union of all the $X_alpha$s, where $f(alpha) in X_alpha$. For each $alpha in A$, we can naturally define a projection map $pi_alpha$, which takes $f in Y$, and maps it to $f(alpha)$, giving us a map from $Y$ to $X_alpha$.
(In the case of a finite product, we can make our index set $A = {1, 2, ldots, n}$ for some $n$. We can also think of our elements of $Y$ as $n$-tuples. Then, given an $n$-tuple $y in Y$, $pi_i(y)$ is the $i$th coordinate of $y$. Outside of finite/countable settings, it makes more sense to think of the elements as functions rather than tuples.)
If $U_alpha$ is open in $X_alpha$, the set $pi_{alpha}^{-1}(U_alpha)$ is the set of functions in $Y$ satisfying $f(alpha) in U_alpha$. For other indices $beta neq alpha$, $f(beta)$ is unrestricted, other than the usual restriction of $f(beta) in X_beta$. So we have
$$pi^{-1}_alpha(U_alpha) = prod_{beta in A} X'_beta,$$
where
$$X'_beta = begin{cases} X_beta & text{if } beta neq alpha \ U_alpha & text{if } beta = alpha end{cases}.$$
Question 2
This is a subbasis because it is not necessarily the case that the intersection of two of these sets will be a union of basis elements. Even if we restrict ourselves to $mathbb{R} times mathbb{R}$, these sets will not be a basis. For example, we can express an open square in the form:
$$(0, 1) times (0, 1) = ((0, 1) times mathbb{R}) cap (mathbb{R} times (0, 1)) = pi_1^{-1}(0, 1) cap pi_2^{-1}(0, 1),$$
making it the intersection of two (presumably) open sets in our (supposed) topology, which should make it open. But, this is a bounded set, and $pi_i^{-1}(U)$ is unbounded for any open $U subseteq mathbb{R}$ and $i in {0, 1}$. So, there's no way to express the square as a union of our supposedly basic sets, which makes the sets not basic!
To form a basis from a subbasis, all you do is take all finite intersections of your subbasic sets. So, our basis will be sets of the of the form
$$pi_{alpha_1}^{-1}(U_{alpha_1}) cap ldots cap pi_{alpha_n}^{-1}(U_{alpha_n}),$$
where $alpha_1, ldots, alpha_n in A$ and $U_{alpha_i} in X_{alpha_i}$ for all $i$. Equivalently, they are sets of the form
$$prod_{alpha in A} X'_alpha,$$
where $X'_alpha = X_alpha$ for all $alpha in A$ except for a number of $alpha$s. For such $alpha$s, $X'_alpha$ is an open subset of $X_alpha$.
Question 3
The box topology produces sets of the form
$$prod_{alpha in A} X'_alpha,$$
where $X'_alpha$ is an open subset of $X_alpha$ for all $alpha in A$. That is, we don't require $X'_alpha = X_alpha$ for all but a finite number of $alpha$s. Every coordinate can be restricted at the same time.
For example, take the product $Y = prod_{i=1}^infty mathbb{R}$, and the subset $U = prod_{i=1}^infty (-i, i)$. Since $(-i, i) subseteq mathbb{R}$ is open for all $i$, $U$ is open in $Y$ under the box topology. But, $(-i, i) neq mathbb{R}$ for infinitely many $i$ (in fact, for all $i$), so $U$ is not a basic element in the product topology.
This doesn't necessarily mean it's not open in the product topology, but this is true as well. If it were open in the product topology, it would have to contain a non-empty basic set. Such a set would have to take the form
$$U_1 times U_2 times ldots times U_n times mathbb{R} times mathbb{R} times ldots,$$
where $U_1, ldots, U_n$ are open subsets of $mathbb{R}$ (possibly equal to $mathbb{R}$). But, this means that the $n + 1$th coordinate is unrestricted, but yet is still confined to $(-n - 1, n + 1)$, which is a contradiction. So, no basic set can be contained in $U$, and hence $U$ is not open with respect to the product topology.
answered Feb 2 at 0:47
Theo BenditTheo Bendit
20.9k12355
$endgroup$
$begingroup$
Hello, sorry for the late reply. For question 1, I am confused how The elements of Y are essentially functions $f$ I clarified what I mean by $Y$ in the edit.
$endgroup$
– urxwrt
Feb 3 at 3:16
$begingroup$
It's another way of thinking about "tuples" that doesn't assume your index set is finite, countable, or ordered in any way. If you have a product of, say, $3$ spaces, you can think of the points in the product as ordered triples, where the first entry belongs to the first space, the second entry to the second, and the third to the third. But, when you have infinitely many spaces, say, one for each element of $mathbb{R}$, it makes more sense to think of functions instead. An element of, say ${1,0}$ multiplied to itself $|mathbb{R}|$ times is a function from $mathbb{R}$ to ${1,0}.$
$endgroup$
– Theo Bendit
Feb 3 at 5:32
$begingroup$
You couldn't think of this as "tuples" in any sensible way. There's no "first" entry, and certainly no "next" entry (unless you wish to well-order the reals using the axiom of choice). You might have an element of this product space that assigns, to the "$x$th" entry (where $x in mathbb{R}$), $1$ when $x$ is rational, and $0$ when $x$ is irrational. That would be one element of the product. It's not possible to write it as an $infty$-tuple, but it makes a lot of sense as a function.
$endgroup$
– Theo Bendit
Feb 3 at 5:34
add a comment |
$begingroup$
Question 1
I'm not sure what you mean by $Y$ here, but suppose you have a family of topological space $(X_alpha)_{alpha in A}$, and we define the product $Y = prod_{alpha in A} X_alpha$ (our index set $A$ need not be countable). The elements of $Y$ are essentially functions $f$, mapping from $A$ to the union of all the $X_alpha$s, where $f(alpha) in X_alpha$. For each $alpha in A$, we can naturally define a projection map $pi_alpha$, which takes $f in Y$, and maps it to $f(alpha)$, giving us a map from $Y$ to $X_alpha$.
(In the case of a finite product, we can make our index set $A = {1, 2, ldots, n}$ for some $n$. We can also think of our elements of $Y$ as $n$-tuples. Then, given an $n$-tuple $y in Y$, $pi_i(y)$ is the $i$th coordinate of $y$. Outside of finite/countable settings, it makes more sense to think of the elements as functions rather than tuples.)
If $U_alpha$ is open in $X_alpha$, the set $pi_{alpha}^{-1}(U_alpha)$ is the set of functions in $Y$ satisfying $f(alpha) in U_alpha$. For other indices $beta neq alpha$, $f(beta)$ is unrestricted, other than the usual restriction of $f(beta) in X_beta$. So we have
$$pi^{-1}_alpha(U_alpha) = prod_{beta in A} X'_beta,$$
where
$$X'_beta = begin{cases} X_beta & text{if } beta neq alpha \ U_alpha & text{if } beta = alpha end{cases}.$$
Question 2
This is a subbasis because it is not necessarily the case that the intersection of two of these sets will be a union of basis elements. Even if we restrict ourselves to $mathbb{R} times mathbb{R}$, these sets will not be a basis. For example, we can express an open square in the form:
$$(0, 1) times (0, 1) = ((0, 1) times mathbb{R}) cap (mathbb{R} times (0, 1)) = pi_1^{-1}(0, 1) cap pi_2^{-1}(0, 1),$$
making it the intersection of two (presumably) open sets in our (supposed) topology, which should make it open. But, this is a bounded set, and $pi_i^{-1}(U)$ is unbounded for any open $U subseteq mathbb{R}$ and $i in {0, 1}$. So, there's no way to express the square as a union of our supposedly basic sets, which makes the sets not basic!
To form a basis from a subbasis, all you do is take all finite intersections of your subbasic sets. So, our basis will be sets of the of the form
$$pi_{alpha_1}^{-1}(U_{alpha_1}) cap ldots cap pi_{alpha_n}^{-1}(U_{alpha_n}),$$
where $alpha_1, ldots, alpha_n in A$ and $U_{alpha_i} in X_{alpha_i}$ for all $i$. Equivalently, they are sets of the form
$$prod_{alpha in A} X'_alpha,$$
where $X'_alpha = X_alpha$ for all $alpha in A$ except for a number of $alpha$s. For such $alpha$s, $X'_alpha$ is an open subset of $X_alpha$.
Question 3
The box topology produces sets of the form
$$prod_{alpha in A} X'_alpha,$$
where $X'_alpha$ is an open subset of $X_alpha$ for all $alpha in A$. That is, we don't require $X'_alpha = X_alpha$ for all but a finite number of $alpha$s. Every coordinate can be restricted at the same time.
For example, take the product $Y = prod_{i=1}^infty mathbb{R}$, and the subset $U = prod_{i=1}^infty (-i, i)$. Since $(-i, i) subseteq mathbb{R}$ is open for all $i$, $U$ is open in $Y$ under the box topology. But, $(-i, i) neq mathbb{R}$ for infinitely many $i$ (in fact, for all $i$), so $U$ is not a basic element in the product topology.
This doesn't necessarily mean it's not open in the product topology, but this is true as well. If it were open in the product topology, it would have to contain a non-empty basic set. Such a set would have to take the form
$$U_1 times U_2 times ldots times U_n times mathbb{R} times mathbb{R} times ldots,$$
where $U_1, ldots, U_n$ are open subsets of $mathbb{R}$ (possibly equal to $mathbb{R}$). But, this means that the $n + 1$th coordinate is unrestricted, but yet is still confined to $(-n - 1, n + 1)$, which is a contradiction. So, no basic set can be contained in $U$, and hence $U$ is not open with respect to the product topology.
answered Feb 2 at 0:47
Theo BenditTheo Bendit
20.9k12355
$endgroup$
$begingroup$
Hello, sorry for the late reply. For question 1, I am confused how The elements of Y are essentially functions $f$ I clarified what I mean by $Y$ in the edit.
$endgroup$
– urxwrt
Feb 3 at 3:16
$begingroup$
It's another way of thinking about "tuples" that doesn't assume your index set is finite, countable, or ordered in any way. If you have a product of, say, $3$ spaces, you can think of the points in the product as ordered triples, where the first entry belongs to the first space, the second entry to the second, and the third to the third. But, when you have infinitely many spaces, say, one for each element of $mathbb{R}$, it makes more sense to think of functions instead. An element of, say ${1,0}$ multiplied to itself $|mathbb{R}|$ times is a function from $mathbb{R}$ to ${1,0}.$
$endgroup$
– Theo Bendit
Feb 3 at 5:32
$begingroup$
You couldn't think of this as "tuples" in any sensible way. There's no "first" entry, and certainly no "next" entry (unless you wish to well-order the reals using the axiom of choice). You might have an element of this product space that assigns, to the "$x$th" entry (where $x in mathbb{R}$), $1$ when $x$ is rational, and $0$ when $x$ is irrational. That would be one element of the product. It's not possible to write it as an $infty$-tuple, but it makes a lot of sense as a function.
$endgroup$
– Theo Bendit
Feb 3 at 5:34
add a comment |
$begingroup$
Question 1
I'm not sure what you mean by $Y$ here, but suppose you have a family of topological space $(X_alpha)_{alpha in A}$, and we define the product $Y = prod_{alpha in A} X_alpha$ (our index set $A$ need not be countable). The elements of $Y$ are essentially functions $f$, mapping from $A$ to the union of all the $X_alpha$s, where $f(alpha) in X_alpha$. For each $alpha in A$, we can naturally define a projection map $pi_alpha$, which takes $f in Y$, and maps it to $f(alpha)$, giving us a map from $Y$ to $X_alpha$.
(In the case of a finite product, we can make our index set $A = {1, 2, ldots, n}$ for some $n$. We can also think of our elements of $Y$ as $n$-tuples. Then, given an $n$-tuple $y in Y$, $pi_i(y)$ is the $i$th coordinate of $y$. Outside of finite/countable settings, it makes more sense to think of the elements as functions rather than tuples.)
If $U_alpha$ is open in $X_alpha$, the set $pi_{alpha}^{-1}(U_alpha)$ is the set of functions in $Y$ satisfying $f(alpha) in U_alpha$. For other indices $beta neq alpha$, $f(beta)$ is unrestricted, other than the usual restriction of $f(beta) in X_beta$. So we have
$$pi^{-1}_alpha(U_alpha) = prod_{beta in A} X'_beta,$$
where
$$X'_beta = begin{cases} X_beta & text{if } beta neq alpha \ U_alpha & text{if } beta = alpha end{cases}.$$
Question 2
This is a subbasis because it is not necessarily the case that the intersection of two of these sets will be a union of basis elements. Even if we restrict ourselves to $mathbb{R} times mathbb{R}$, these sets will not be a basis. For example, we can express an open square in the form:
$$(0, 1) times (0, 1) = ((0, 1) times mathbb{R}) cap (mathbb{R} times (0, 1)) = pi_1^{-1}(0, 1) cap pi_2^{-1}(0, 1),$$
making it the intersection of two (presumably) open sets in our (supposed) topology, which should make it open. But, this is a bounded set, and $pi_i^{-1}(U)$ is unbounded for any open $U subseteq mathbb{R}$ and $i in {0, 1}$. So, there's no way to express the square as a union of our supposedly basic sets, which makes the sets not basic!
To form a basis from a subbasis, all you do is take all finite intersections of your subbasic sets. So, our basis will be sets of the of the form
$$pi_{alpha_1}^{-1}(U_{alpha_1}) cap ldots cap pi_{alpha_n}^{-1}(U_{alpha_n}),$$
where $alpha_1, ldots, alpha_n in A$ and $U_{alpha_i} in X_{alpha_i}$ for all $i$. Equivalently, they are sets of the form
$$prod_{alpha in A} X'_alpha,$$
where $X'_alpha = X_alpha$ for all $alpha in A$ except for a number of $alpha$s. For such $alpha$s, $X'_alpha$ is an open subset of $X_alpha$.
Question 3
The box topology produces sets of the form
$$prod_{alpha in A} X'_alpha,$$
where $X'_alpha$ is an open subset of $X_alpha$ for all $alpha in A$. That is, we don't require $X'_alpha = X_alpha$ for all but a finite number of $alpha$s. Every coordinate can be restricted at the same time.
For example, take the product $Y = prod_{i=1}^infty mathbb{R}$, and the subset $U = prod_{i=1}^infty (-i, i)$. Since $(-i, i) subseteq mathbb{R}$ is open for all $i$, $U$ is open in $Y$ under the box topology. But, $(-i, i) neq mathbb{R}$ for infinitely many $i$ (in fact, for all $i$), so $U$ is not a basic element in the product topology.
This doesn't necessarily mean it's not open in the product topology, but this is true as well. If it were open in the product topology, it would have to contain a non-empty basic set. Such a set would have to take the form
$$U_1 times U_2 times ldots times U_n times mathbb{R} times mathbb{R} times ldots,$$
where $U_1, ldots, U_n$ are open subsets of $mathbb{R}$ (possibly equal to $mathbb{R}$). But, this means that the $n + 1$th coordinate is unrestricted, but yet is still confined to $(-n - 1, n + 1)$, which is a contradiction. So, no basic set can be contained in $U$, and hence $U$ is not open with respect to the product topology.
answered Feb 2 at 0:47
Theo BenditTheo Bendit
20.9k12355
$endgroup$
Question 1
I'm not sure what you mean by $Y$ here, but suppose you have a family of topological space $(X_alpha)_{alpha in A}$, and we define the product $Y = prod_{alpha in A} X_alpha$ (our index set $A$ need not be countable). The elements of $Y$ are essentially functions $f$, mapping from $A$ to the union of all the $X_alpha$s, where $f(alpha) in X_alpha$. For each $alpha in A$, we can naturally define a projection map $pi_alpha$, which takes $f in Y$, and maps it to $f(alpha)$, giving us a map from $Y$ to $X_alpha$.
(In the case of a finite product, we can make our index set $A = {1, 2, ldots, n}$ for some $n$. We can also think of our elements of $Y$ as $n$-tuples. Then, given an $n$-tuple $y in Y$, $pi_i(y)$ is the $i$th coordinate of $y$. Outside of finite/countable settings, it makes more sense to think of the elements as functions rather than tuples.)
If $U_alpha$ is open in $X_alpha$, the set $pi_{alpha}^{-1}(U_alpha)$ is the set of functions in $Y$ satisfying $f(alpha) in U_alpha$. For other indices $beta neq alpha$, $f(beta)$ is unrestricted, other than the usual restriction of $f(beta) in X_beta$. So we have
$$pi^{-1}_alpha(U_alpha) = prod_{beta in A} X'_beta,$$
where
$$X'_beta = begin{cases} X_beta & text{if } beta neq alpha \ U_alpha & text{if } beta = alpha end{cases}.$$
Question 2
This is a subbasis because it is not necessarily the case that the intersection of two of these sets will be a union of basis elements. Even if we restrict ourselves to $mathbb{R} times mathbb{R}$, these sets will not be a basis. For example, we can express an open square in the form:
$$(0, 1) times (0, 1) = ((0, 1) times mathbb{R}) cap (mathbb{R} times (0, 1)) = pi_1^{-1}(0, 1) cap pi_2^{-1}(0, 1),$$
making it the intersection of two (presumably) open sets in our (supposed) topology, which should make it open. But, this is a bounded set, and $pi_i^{-1}(U)$ is unbounded for any open $U subseteq mathbb{R}$ and $i in {0, 1}$. So, there's no way to express the square as a union of our supposedly basic sets, which makes the sets not basic!
To form a basis from a subbasis, all you do is take all finite intersections of your subbasic sets. So, our basis will be sets of the of the form
$$pi_{alpha_1}^{-1}(U_{alpha_1}) cap ldots cap pi_{alpha_n}^{-1}(U_{alpha_n}),$$
where $alpha_1, ldots, alpha_n in A$ and $U_{alpha_i} in X_{alpha_i}$ for all $i$. Equivalently, they are sets of the form
$$prod_{alpha in A} X'_alpha,$$
where $X'_alpha = X_alpha$ for all $alpha in A$ except for a number of $alpha$s. For such $alpha$s, $X'_alpha$ is an open subset of $X_alpha$.
Question 3
The box topology produces sets of the form
$$prod_{alpha in A} X'_alpha,$$
where $X'_alpha$ is an open subset of $X_alpha$ for all $alpha in A$. That is, we don't require $X'_alpha = X_alpha$ for all but a finite number of $alpha$s. Every coordinate can be restricted at the same time.
For example, take the product $Y = prod_{i=1}^infty mathbb{R}$, and the subset $U = prod_{i=1}^infty (-i, i)$. Since $(-i, i) subseteq mathbb{R}$ is open for all $i$, $U$ is open in $Y$ under the box topology. But, $(-i, i) neq mathbb{R}$ for infinitely many $i$ (in fact, for all $i$), so $U$ is not a basic element in the product topology.
This doesn't necessarily mean it's not open in the product topology, but this is true as well. If it were open in the product topology, it would have to contain a non-empty basic set. Such a set would have to take the form
$$U_1 times U_2 times ldots times U_n times mathbb{R} times mathbb{R} times ldots,$$
where $U_1, ldots, U_n$ are open subsets of $mathbb{R}$ (possibly equal to $mathbb{R}$). But, this means that the $n + 1$th coordinate is unrestricted, but yet is still confined to $(-n - 1, n + 1)$, which is a contradiction. So, no basic set can be contained in $U$, and hence $U$ is not open with respect to the product topology.
answered Feb 2 at 0:47
Theo BenditTheo Bendit
20.9k12355
answered Feb 2 at 0:47
Theo BenditTheo Bendit
20.9k12355
answered Feb 2 at 0:47
Theo BenditTheo Bendit
20.9k12355
answered Feb 2 at 0:47
Theo BenditTheo Bendit
20.9k12355
20.9k12355
$begingroup$
Hello, sorry for the late reply. For question 1, I am confused how The elements of Y are essentially functions $f$ I clarified what I mean by $Y$ in the edit.
$endgroup$
– urxwrt
Feb 3 at 3:16
$begingroup$
It's another way of thinking about "tuples" that doesn't assume your index set is finite, countable, or ordered in any way. If you have a product of, say, $3$ spaces, you can think of the points in the product as ordered triples, where the first entry belongs to the first space, the second entry to the second, and the third to the third. But, when you have infinitely many spaces, say, one for each element of $mathbb{R}$, it makes more sense to think of functions instead. An element of, say ${1,0}$ multiplied to itself $|mathbb{R}|$ times is a function from $mathbb{R}$ to ${1,0}.$
$endgroup$
– Theo Bendit
Feb 3 at 5:32
$begingroup$
You couldn't think of this as "tuples" in any sensible way. There's no "first" entry, and certainly no "next" entry (unless you wish to well-order the reals using the axiom of choice). You might have an element of this product space that assigns, to the "$x$th" entry (where $x in mathbb{R}$), $1$ when $x$ is rational, and $0$ when $x$ is irrational. That would be one element of the product. It's not possible to write it as an $infty$-tuple, but it makes a lot of sense as a function.
$endgroup$
– Theo Bendit
Feb 3 at 5:34
add a comment |
$begingroup$
Hello, sorry for the late reply. For question 1, I am confused how The elements of Y are essentially functions $f$ I clarified what I mean by $Y$ in the edit.
$endgroup$
– urxwrt
Feb 3 at 3:16
$begingroup$
It's another way of thinking about "tuples" that doesn't assume your index set is finite, countable, or ordered in any way. If you have a product of, say, $3$ spaces, you can think of the points in the product as ordered triples, where the first entry belongs to the first space, the second entry to the second, and the third to the third. But, when you have infinitely many spaces, say, one for each element of $mathbb{R}$, it makes more sense to think of functions instead. An element of, say ${1,0}$ multiplied to itself $|mathbb{R}|$ times is a function from $mathbb{R}$ to ${1,0}.$
$endgroup$
– Theo Bendit
Feb 3 at 5:32
$begingroup$
You couldn't think of this as "tuples" in any sensible way. There's no "first" entry, and certainly no "next" entry (unless you wish to well-order the reals using the axiom of choice). You might have an element of this product space that assigns, to the "$x$th" entry (where $x in mathbb{R}$), $1$ when $x$ is rational, and $0$ when $x$ is irrational. That would be one element of the product. It's not possible to write it as an $infty$-tuple, but it makes a lot of sense as a function.
$endgroup$
– Theo Bendit
Feb 3 at 5:34
$begingroup$
Hello, sorry for the late reply. For question 1, I am confused how The elements of Y are essentially functions $f$ I clarified what I mean by $Y$ in the edit.
$endgroup$
– urxwrt
Feb 3 at 3:16
$begingroup$
Hello, sorry for the late reply. For question 1, I am confused how The elements of Y are essentially functions $f$ I clarified what I mean by $Y$ in the edit.
$endgroup$
– urxwrt
Feb 3 at 3:16
$begingroup$
It's another way of thinking about "tuples" that doesn't assume your index set is finite, countable, or ordered in any way. If you have a product of, say, $3$ spaces, you can think of the points in the product as ordered triples, where the first entry belongs to the first space, the second entry to the second, and the third to the third. But, when you have infinitely many spaces, say, one for each element of $mathbb{R}$, it makes more sense to think of functions instead. An element of, say ${1,0}$ multiplied to itself $|mathbb{R}|$ times is a function from $mathbb{R}$ to ${1,0}.$
$endgroup$
– Theo Bendit
Feb 3 at 5:32
$begingroup$
It's another way of thinking about "tuples" that doesn't assume your index set is finite, countable, or ordered in any way. If you have a product of, say, $3$ spaces, you can think of the points in the product as ordered triples, where the first entry belongs to the first space, the second entry to the second, and the third to the third. But, when you have infinitely many spaces, say, one for each element of $mathbb{R}$, it makes more sense to think of functions instead. An element of, say ${1,0}$ multiplied to itself $|mathbb{R}|$ times is a function from $mathbb{R}$ to ${1,0}.$
$endgroup$
– Theo Bendit
Feb 3 at 5:32
$begingroup$
You couldn't think of this as "tuples" in any sensible way. There's no "first" entry, and certainly no "next" entry (unless you wish to well-order the reals using the axiom of choice). You might have an element of this product space that assigns, to the "$x$th" entry (where $x in mathbb{R}$), $1$ when $x$ is rational, and $0$ when $x$ is irrational. That would be one element of the product. It's not possible to write it as an $infty$-tuple, but it makes a lot of sense as a function.
$endgroup$
– Theo Bendit
Feb 3 at 5:34
$begingroup$
You couldn't think of this as "tuples" in any sensible way. There's no "first" entry, and certainly no "next" entry (unless you wish to well-order the reals using the axiom of choice). You might have an element of this product space that assigns, to the "$x$th" entry (where $x in mathbb{R}$), $1$ when $x$ is rational, and $0$ when $x$ is irrational. That would be one element of the product. It's not possible to write it as an $infty$-tuple, but it makes a lot of sense as a function.
$endgroup$
– Theo Bendit
Feb 3 at 5:34
add a comment |
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