$mathbb{R} times mathbb{R} times … times mathbb{R}$ ($n$ times) is homeomorphic to $mathbb{R}^{n}$
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Here is a short sentence from the book "General Topology" of Stephen Willard :
"[...] In particular, $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology is homeomorphic to $mathbb{R}^{n}$."
Recently, I asked a question about the fact that we don't have exactly $mathbb{R}^{n} = mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times). Thus, I know now that these two sets are in bijection.
Now, my problem is the following : in the book in question, the author sometimes says things like "the usual topology on $mathbb{R}^{2}$" and I was wondering :
- If we consider $mathbb{R}^{n}$, is the "usual topology" not the Tychonoff topology (product topology) ? Which is the same as the ones induced by any norm (considering $mathbb{R}^{n}$ as a vector space) or as the box topology ?
- If we consider $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times), then :
- Considering that what I said previously is correct (which is maybe not the case), why does the author say that we consider "$mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology" ?
- Again, considering that what I said is correct, if we assume that $mathbb{R}^{n}$ is endowed with its product topology, what is in fact the topology on the set $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) that makes the author says that these two topological spaces are homeomorphic ?
I hope it is clear enough. Thank you for your future answers.
general-topology
asked Nov 22 at 14:32
deeppinkwater
618
add a comment |
up vote
0
down vote
favorite
Here is a short sentence from the book "General Topology" of Stephen Willard :
"[...] In particular, $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology is homeomorphic to $mathbb{R}^{n}$."
Recently, I asked a question about the fact that we don't have exactly $mathbb{R}^{n} = mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times). Thus, I know now that these two sets are in bijection.
Now, my problem is the following : in the book in question, the author sometimes says things like "the usual topology on $mathbb{R}^{2}$" and I was wondering :
- If we consider $mathbb{R}^{n}$, is the "usual topology" not the Tychonoff topology (product topology) ? Which is the same as the ones induced by any norm (considering $mathbb{R}^{n}$ as a vector space) or as the box topology ?
- If we consider $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times), then :
- Considering that what I said previously is correct (which is maybe not the case), why does the author say that we consider "$mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology" ?
- Again, considering that what I said is correct, if we assume that $mathbb{R}^{n}$ is endowed with its product topology, what is in fact the topology on the set $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) that makes the author says that these two topological spaces are homeomorphic ?
I hope it is clear enough. Thank you for your future answers.
general-topology
asked Nov 22 at 14:32
deeppinkwater
618
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Here is a short sentence from the book "General Topology" of Stephen Willard :
"[...] In particular, $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology is homeomorphic to $mathbb{R}^{n}$."
Recently, I asked a question about the fact that we don't have exactly $mathbb{R}^{n} = mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times). Thus, I know now that these two sets are in bijection.
Now, my problem is the following : in the book in question, the author sometimes says things like "the usual topology on $mathbb{R}^{2}$" and I was wondering :
- If we consider $mathbb{R}^{n}$, is the "usual topology" not the Tychonoff topology (product topology) ? Which is the same as the ones induced by any norm (considering $mathbb{R}^{n}$ as a vector space) or as the box topology ?
- If we consider $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times), then :
- Considering that what I said previously is correct (which is maybe not the case), why does the author say that we consider "$mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology" ?
- Again, considering that what I said is correct, if we assume that $mathbb{R}^{n}$ is endowed with its product topology, what is in fact the topology on the set $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) that makes the author says that these two topological spaces are homeomorphic ?
I hope it is clear enough. Thank you for your future answers.
general-topology
asked Nov 22 at 14:32
deeppinkwater
618
Here is a short sentence from the book "General Topology" of Stephen Willard :
"[...] In particular, $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology is homeomorphic to $mathbb{R}^{n}$."
Recently, I asked a question about the fact that we don't have exactly $mathbb{R}^{n} = mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times). Thus, I know now that these two sets are in bijection.
Now, my problem is the following : in the book in question, the author sometimes says things like "the usual topology on $mathbb{R}^{2}$" and I was wondering :
- If we consider $mathbb{R}^{n}$, is the "usual topology" not the Tychonoff topology (product topology) ? Which is the same as the ones induced by any norm (considering $mathbb{R}^{n}$ as a vector space) or as the box topology ?
- If we consider $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times), then :
- Considering that what I said previously is correct (which is maybe not the case), why does the author say that we consider "$mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology" ?
- Again, considering that what I said is correct, if we assume that $mathbb{R}^{n}$ is endowed with its product topology, what is in fact the topology on the set $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) that makes the author says that these two topological spaces are homeomorphic ?
I hope it is clear enough. Thank you for your future answers.
general-topology
general-topology
asked Nov 22 at 14:32
deeppinkwater
618
asked Nov 22 at 14:32
deeppinkwater
618
asked Nov 22 at 14:32
deeppinkwater
618
asked Nov 22 at 14:32
deeppinkwater
618
asked Nov 22 at 14:32
deeppinkwater
618
618
add a comment |
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The author is simply making a distinction between the topological product and the set product. By $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ he actually means the ( topological) product $(mathbb{R},tau_0)times(mathbb{R},tau_0)timesdotsbtimes(mathbb{R},tau_0)$ with $tau_0$ being the order topology on $mathbb{R}$. On the other hand by $mathbb{R}^n$ he means the $n$-dimensional topological space i.e. the topology $(mathbb{R}^n,tau_1)$ where $mathbb{R^n}$ is the set product $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ and $tau_1$ is the topology generated by the open balls.
Author's like using these sorts of abbreviations but this can sometimes cause confusion the first time you see it. To avoid this, in fact, I've seen some authors (cannot immediately think of an example) that use $E_n$ to refer to the topological space $(mathbb{R}^n,tau_1)$ above.
answered Nov 22 at 14:52
Jean-Pierre de Villiers
415
Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
– deeppinkwater
Nov 22 at 15:11
I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
– Jean-Pierre de Villiers
Nov 22 at 15:39
add a comment |
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The author is simply making a distinction between the topological product and the set product. By $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ he actually means the ( topological) product $(mathbb{R},tau_0)times(mathbb{R},tau_0)timesdotsbtimes(mathbb{R},tau_0)$ with $tau_0$ being the order topology on $mathbb{R}$. On the other hand by $mathbb{R}^n$ he means the $n$-dimensional topological space i.e. the topology $(mathbb{R}^n,tau_1)$ where $mathbb{R^n}$ is the set product $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ and $tau_1$ is the topology generated by the open balls.
Author's like using these sorts of abbreviations but this can sometimes cause confusion the first time you see it. To avoid this, in fact, I've seen some authors (cannot immediately think of an example) that use $E_n$ to refer to the topological space $(mathbb{R}^n,tau_1)$ above.
answered Nov 22 at 14:52
Jean-Pierre de Villiers
415
Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
– deeppinkwater
Nov 22 at 15:11
I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
– Jean-Pierre de Villiers
Nov 22 at 15:39
add a comment |
up vote
1
down vote
The author is simply making a distinction between the topological product and the set product. By $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ he actually means the ( topological) product $(mathbb{R},tau_0)times(mathbb{R},tau_0)timesdotsbtimes(mathbb{R},tau_0)$ with $tau_0$ being the order topology on $mathbb{R}$. On the other hand by $mathbb{R}^n$ he means the $n$-dimensional topological space i.e. the topology $(mathbb{R}^n,tau_1)$ where $mathbb{R^n}$ is the set product $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ and $tau_1$ is the topology generated by the open balls.
Author's like using these sorts of abbreviations but this can sometimes cause confusion the first time you see it. To avoid this, in fact, I've seen some authors (cannot immediately think of an example) that use $E_n$ to refer to the topological space $(mathbb{R}^n,tau_1)$ above.
answered Nov 22 at 14:52
Jean-Pierre de Villiers
415
Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
– deeppinkwater
Nov 22 at 15:11
I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
– Jean-Pierre de Villiers
Nov 22 at 15:39
add a comment |
up vote
1
down vote
up vote
1
down vote
The author is simply making a distinction between the topological product and the set product. By $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ he actually means the ( topological) product $(mathbb{R},tau_0)times(mathbb{R},tau_0)timesdotsbtimes(mathbb{R},tau_0)$ with $tau_0$ being the order topology on $mathbb{R}$. On the other hand by $mathbb{R}^n$ he means the $n$-dimensional topological space i.e. the topology $(mathbb{R}^n,tau_1)$ where $mathbb{R^n}$ is the set product $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ and $tau_1$ is the topology generated by the open balls.
Author's like using these sorts of abbreviations but this can sometimes cause confusion the first time you see it. To avoid this, in fact, I've seen some authors (cannot immediately think of an example) that use $E_n$ to refer to the topological space $(mathbb{R}^n,tau_1)$ above.
answered Nov 22 at 14:52
Jean-Pierre de Villiers
415
The author is simply making a distinction between the topological product and the set product. By $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ he actually means the ( topological) product $(mathbb{R},tau_0)times(mathbb{R},tau_0)timesdotsbtimes(mathbb{R},tau_0)$ with $tau_0$ being the order topology on $mathbb{R}$. On the other hand by $mathbb{R}^n$ he means the $n$-dimensional topological space i.e. the topology $(mathbb{R}^n,tau_1)$ where $mathbb{R^n}$ is the set product $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ and $tau_1$ is the topology generated by the open balls.
Author's like using these sorts of abbreviations but this can sometimes cause confusion the first time you see it. To avoid this, in fact, I've seen some authors (cannot immediately think of an example) that use $E_n$ to refer to the topological space $(mathbb{R}^n,tau_1)$ above.
answered Nov 22 at 14:52
Jean-Pierre de Villiers
415
answered Nov 22 at 14:52
Jean-Pierre de Villiers
415
answered Nov 22 at 14:52
Jean-Pierre de Villiers
415
answered Nov 22 at 14:52
Jean-Pierre de Villiers
415
415
Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
– deeppinkwater
Nov 22 at 15:11
I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
– Jean-Pierre de Villiers
Nov 22 at 15:39
add a comment |
Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
– deeppinkwater
Nov 22 at 15:11
I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
– Jean-Pierre de Villiers
Nov 22 at 15:39
Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
– deeppinkwater
Nov 22 at 15:11
Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
– deeppinkwater
Nov 22 at 15:11
I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
– Jean-Pierre de Villiers
Nov 22 at 15:39
I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
– Jean-Pierre de Villiers
Nov 22 at 15:39
add a comment |
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