Mixing
What is {empty set} x something with the product topology?
$begingroup$
The product topology is defined as the topology induced by the basis of the product of open sets from each of the original topologies.
From what I understand, then ${emptyset}x]a,b[$ is an open set of $IR^2$ with the product topology, right? But what is this geometrically? I can't see what this is.
general-topology product-space
edited Jan 5 at 19:54
Lehs
7,00831662
asked Jan 5 at 19:48
J. DionisioJ. Dionisio
9511
$endgroup$
add a comment |
$begingroup$
The product topology is defined as the topology induced by the basis of the product of open sets from each of the original topologies.
From what I understand, then ${emptyset}x]a,b[$ is an open set of $IR^2$ with the product topology, right? But what is this geometrically? I can't see what this is.
general-topology product-space
edited Jan 5 at 19:54
Lehs
7,00831662
asked Jan 5 at 19:48
J. DionisioJ. Dionisio
9511
$endgroup$
3
$begingroup$
Empty cross anything is empty.
$endgroup$
– Randall
Jan 5 at 19:48
$begingroup$
The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
$endgroup$
– leibnewtz
Jan 5 at 19:59
$begingroup$
Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
$endgroup$
– leibnewtz
Jan 5 at 20:21
add a comment |
$begingroup$
The product topology is defined as the topology induced by the basis of the product of open sets from each of the original topologies.
From what I understand, then ${emptyset}x]a,b[$ is an open set of $IR^2$ with the product topology, right? But what is this geometrically? I can't see what this is.
general-topology product-space
edited Jan 5 at 19:54
Lehs
7,00831662
asked Jan 5 at 19:48
J. DionisioJ. Dionisio
9511
$endgroup$
The product topology is defined as the topology induced by the basis of the product of open sets from each of the original topologies.
From what I understand, then ${emptyset}x]a,b[$ is an open set of $IR^2$ with the product topology, right? But what is this geometrically? I can't see what this is.
general-topology product-space
general-topology product-space
edited Jan 5 at 19:54
Lehs
7,00831662
asked Jan 5 at 19:48
J. DionisioJ. Dionisio
9511
edited Jan 5 at 19:54
Lehs
7,00831662
asked Jan 5 at 19:48
J. DionisioJ. Dionisio
9511
edited Jan 5 at 19:54
Lehs
7,00831662
edited Jan 5 at 19:54
Lehs
7,00831662
edited Jan 5 at 19:54
Lehs
7,00831662
7,00831662
asked Jan 5 at 19:48
J. DionisioJ. Dionisio
9511
asked Jan 5 at 19:48
J. DionisioJ. Dionisio
9511
asked Jan 5 at 19:48
J. DionisioJ. Dionisio
9511
9511
3
$begingroup$
Empty cross anything is empty.
$endgroup$
– Randall
Jan 5 at 19:48
$begingroup$
The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
$endgroup$
– leibnewtz
Jan 5 at 19:59
$begingroup$
Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
$endgroup$
– leibnewtz
Jan 5 at 20:21
add a comment |
3
$begingroup$
Empty cross anything is empty.
$endgroup$
– Randall
Jan 5 at 19:48
$begingroup$
The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
$endgroup$
– leibnewtz
Jan 5 at 19:59
$begingroup$
Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
$endgroup$
– leibnewtz
Jan 5 at 20:21
3
3
$begingroup$
Empty cross anything is empty.
$endgroup$
– Randall
Jan 5 at 19:48
$begingroup$
Empty cross anything is empty.
$endgroup$
– Randall
Jan 5 at 19:48
$begingroup$
The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
$endgroup$
– leibnewtz
Jan 5 at 19:59
$begingroup$
The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
$endgroup$
– leibnewtz
Jan 5 at 19:59
$begingroup$
Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
$endgroup$
– leibnewtz
Jan 5 at 20:21
$begingroup$
Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
$endgroup$
– leibnewtz
Jan 5 at 20:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By definition, the elements of $Atimes B$ (for sets $A$ and $B$) are the ordered pairs $langle a,brangle$ such that $ain A$ and $bin B.$ Since $ainemptyset$ is impossible, then $emptysettimes B$ has no elements, meaning $emptysettimes B=emptyset,$ regardless of the set $B.$
The open sets of $Bbb R^2$ in the product topology can be obtained from the basis ${Atimes B:A,Btext{ are open subsets of }Bbb R}.$ This will include elements of the form $emptysettimes B=emptyset,$ as well as $Atimesemptyset=emptyset.$ However, it will not have any elements of the form ${emptyset}times B,$ since $emptyset$ isn't an element of $Bbb R.$
answered Jan 5 at 19:52
Cameron BuieCameron Buie
85.1k771155
$endgroup$
$begingroup$
The question is asking about ${emptyset } times A$, not $emptyset times A$
$endgroup$
– leibnewtz
Jan 5 at 19:58
$begingroup$
Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
$endgroup$
– J. Dionisio
Jan 5 at 20:00
$begingroup$
Oh, dear! I misunderstood. I'll edit accordingly.
$endgroup$
– Cameron Buie
Jan 5 at 20:03
$begingroup$
Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
$endgroup$
– J. Dionisio
Jan 5 at 20:23
add a comment |
$begingroup$
Instead of ${emptyset}$, I'll consider ${*}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that ${*}$ is a singleton as a set.
Let $X$ be a topological space. Then, basis for topology of ${*}times X$ is given by ${ {*}times Umid Usubseteq X text{open}}$.
Define $fcolon Xto {*}times X$ by $f(x) = (*,x)$ and $gcolon {*}times Xto X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $Usubseteq X$ open we have
$$f^{-1}({*}times U) = U, g^{-1}(U) = {*}times U.$$
Thus, $X$ and ${*}times X$ are homeomorphic.
If $X = mathbb R$, instead of $*$ choose some real number $a$. Then ${a}timesmathbb R$ is naturally a line embedded in $mathbb R^2$. Here are examples for some different $a$'s:
answered Jan 5 at 20:26
EnnarEnnar
14.4k32443
$endgroup$
add a comment |
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$begingroup$
By definition, the elements of $Atimes B$ (for sets $A$ and $B$) are the ordered pairs $langle a,brangle$ such that $ain A$ and $bin B.$ Since $ainemptyset$ is impossible, then $emptysettimes B$ has no elements, meaning $emptysettimes B=emptyset,$ regardless of the set $B.$
The open sets of $Bbb R^2$ in the product topology can be obtained from the basis ${Atimes B:A,Btext{ are open subsets of }Bbb R}.$ This will include elements of the form $emptysettimes B=emptyset,$ as well as $Atimesemptyset=emptyset.$ However, it will not have any elements of the form ${emptyset}times B,$ since $emptyset$ isn't an element of $Bbb R.$
answered Jan 5 at 19:52
Cameron BuieCameron Buie
85.1k771155
$endgroup$
$begingroup$
The question is asking about ${emptyset } times A$, not $emptyset times A$
$endgroup$
– leibnewtz
Jan 5 at 19:58
$begingroup$
Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
$endgroup$
– J. Dionisio
Jan 5 at 20:00
$begingroup$
Oh, dear! I misunderstood. I'll edit accordingly.
$endgroup$
– Cameron Buie
Jan 5 at 20:03
$begingroup$
Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
$endgroup$
– J. Dionisio
Jan 5 at 20:23
add a comment |
$begingroup$
By definition, the elements of $Atimes B$ (for sets $A$ and $B$) are the ordered pairs $langle a,brangle$ such that $ain A$ and $bin B.$ Since $ainemptyset$ is impossible, then $emptysettimes B$ has no elements, meaning $emptysettimes B=emptyset,$ regardless of the set $B.$
The open sets of $Bbb R^2$ in the product topology can be obtained from the basis ${Atimes B:A,Btext{ are open subsets of }Bbb R}.$ This will include elements of the form $emptysettimes B=emptyset,$ as well as $Atimesemptyset=emptyset.$ However, it will not have any elements of the form ${emptyset}times B,$ since $emptyset$ isn't an element of $Bbb R.$
answered Jan 5 at 19:52
Cameron BuieCameron Buie
85.1k771155
$endgroup$
$begingroup$
The question is asking about ${emptyset } times A$, not $emptyset times A$
$endgroup$
– leibnewtz
Jan 5 at 19:58
$begingroup$
Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
$endgroup$
– J. Dionisio
Jan 5 at 20:00
$begingroup$
Oh, dear! I misunderstood. I'll edit accordingly.
$endgroup$
– Cameron Buie
Jan 5 at 20:03
$begingroup$
Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
$endgroup$
– J. Dionisio
Jan 5 at 20:23
add a comment |
$begingroup$
By definition, the elements of $Atimes B$ (for sets $A$ and $B$) are the ordered pairs $langle a,brangle$ such that $ain A$ and $bin B.$ Since $ainemptyset$ is impossible, then $emptysettimes B$ has no elements, meaning $emptysettimes B=emptyset,$ regardless of the set $B.$
The open sets of $Bbb R^2$ in the product topology can be obtained from the basis ${Atimes B:A,Btext{ are open subsets of }Bbb R}.$ This will include elements of the form $emptysettimes B=emptyset,$ as well as $Atimesemptyset=emptyset.$ However, it will not have any elements of the form ${emptyset}times B,$ since $emptyset$ isn't an element of $Bbb R.$
answered Jan 5 at 19:52
Cameron BuieCameron Buie
85.1k771155
$endgroup$
By definition, the elements of $Atimes B$ (for sets $A$ and $B$) are the ordered pairs $langle a,brangle$ such that $ain A$ and $bin B.$ Since $ainemptyset$ is impossible, then $emptysettimes B$ has no elements, meaning $emptysettimes B=emptyset,$ regardless of the set $B.$
The open sets of $Bbb R^2$ in the product topology can be obtained from the basis ${Atimes B:A,Btext{ are open subsets of }Bbb R}.$ This will include elements of the form $emptysettimes B=emptyset,$ as well as $Atimesemptyset=emptyset.$ However, it will not have any elements of the form ${emptyset}times B,$ since $emptyset$ isn't an element of $Bbb R.$
answered Jan 5 at 19:52
Cameron BuieCameron Buie
85.1k771155
edited Jan 5 at 20:08
answered Jan 5 at 19:52
Cameron BuieCameron Buie
85.1k771155
answered Jan 5 at 19:52
Cameron BuieCameron Buie
85.1k771155
answered Jan 5 at 19:52
Cameron BuieCameron Buie
85.1k771155
85.1k771155
$begingroup$
The question is asking about ${emptyset } times A$, not $emptyset times A$
$endgroup$
– leibnewtz
Jan 5 at 19:58
$begingroup$
Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
$endgroup$
– J. Dionisio
Jan 5 at 20:00
$begingroup$
Oh, dear! I misunderstood. I'll edit accordingly.
$endgroup$
– Cameron Buie
Jan 5 at 20:03
$begingroup$
Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
$endgroup$
– J. Dionisio
Jan 5 at 20:23
add a comment |
$begingroup$
The question is asking about ${emptyset } times A$, not $emptyset times A$
$endgroup$
– leibnewtz
Jan 5 at 19:58
$begingroup$
Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
$endgroup$
– J. Dionisio
Jan 5 at 20:00
$begingroup$
Oh, dear! I misunderstood. I'll edit accordingly.
$endgroup$
– Cameron Buie
Jan 5 at 20:03
$begingroup$
Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
$endgroup$
– J. Dionisio
Jan 5 at 20:23
$begingroup$
The question is asking about ${emptyset } times A$, not $emptyset times A$
$endgroup$
– leibnewtz
Jan 5 at 19:58
$begingroup$
The question is asking about ${emptyset } times A$, not $emptyset times A$
$endgroup$
– leibnewtz
Jan 5 at 19:58
$begingroup$
Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
$endgroup$
– J. Dionisio
Jan 5 at 20:00
$begingroup$
Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
$endgroup$
– J. Dionisio
Jan 5 at 20:00
$begingroup$
Oh, dear! I misunderstood. I'll edit accordingly.
$endgroup$
– Cameron Buie
Jan 5 at 20:03
$begingroup$
Oh, dear! I misunderstood. I'll edit accordingly.
$endgroup$
– Cameron Buie
Jan 5 at 20:03
$begingroup$
Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
$endgroup$
– J. Dionisio
Jan 5 at 20:23
$begingroup$
Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
$endgroup$
– J. Dionisio
Jan 5 at 20:23
add a comment |
$begingroup$
Instead of ${emptyset}$, I'll consider ${*}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that ${*}$ is a singleton as a set.
Let $X$ be a topological space. Then, basis for topology of ${*}times X$ is given by ${ {*}times Umid Usubseteq X text{open}}$.
Define $fcolon Xto {*}times X$ by $f(x) = (*,x)$ and $gcolon {*}times Xto X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $Usubseteq X$ open we have
$$f^{-1}({*}times U) = U, g^{-1}(U) = {*}times U.$$
Thus, $X$ and ${*}times X$ are homeomorphic.
If $X = mathbb R$, instead of $*$ choose some real number $a$. Then ${a}timesmathbb R$ is naturally a line embedded in $mathbb R^2$. Here are examples for some different $a$'s:
answered Jan 5 at 20:26
EnnarEnnar
14.4k32443
$endgroup$
add a comment |
$begingroup$
Instead of ${emptyset}$, I'll consider ${*}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that ${*}$ is a singleton as a set.
Let $X$ be a topological space. Then, basis for topology of ${*}times X$ is given by ${ {*}times Umid Usubseteq X text{open}}$.
Define $fcolon Xto {*}times X$ by $f(x) = (*,x)$ and $gcolon {*}times Xto X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $Usubseteq X$ open we have
$$f^{-1}({*}times U) = U, g^{-1}(U) = {*}times U.$$
Thus, $X$ and ${*}times X$ are homeomorphic.
If $X = mathbb R$, instead of $*$ choose some real number $a$. Then ${a}timesmathbb R$ is naturally a line embedded in $mathbb R^2$. Here are examples for some different $a$'s:
answered Jan 5 at 20:26
EnnarEnnar
14.4k32443
$endgroup$
add a comment |
$begingroup$
Instead of ${emptyset}$, I'll consider ${*}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that ${*}$ is a singleton as a set.
Let $X$ be a topological space. Then, basis for topology of ${*}times X$ is given by ${ {*}times Umid Usubseteq X text{open}}$.
Define $fcolon Xto {*}times X$ by $f(x) = (*,x)$ and $gcolon {*}times Xto X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $Usubseteq X$ open we have
$$f^{-1}({*}times U) = U, g^{-1}(U) = {*}times U.$$
Thus, $X$ and ${*}times X$ are homeomorphic.
If $X = mathbb R$, instead of $*$ choose some real number $a$. Then ${a}timesmathbb R$ is naturally a line embedded in $mathbb R^2$. Here are examples for some different $a$'s:
answered Jan 5 at 20:26
EnnarEnnar
14.4k32443
$endgroup$
Instead of ${emptyset}$, I'll consider ${*}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that ${*}$ is a singleton as a set.
Let $X$ be a topological space. Then, basis for topology of ${*}times X$ is given by ${ {*}times Umid Usubseteq X text{open}}$.
Define $fcolon Xto {*}times X$ by $f(x) = (*,x)$ and $gcolon {*}times Xto X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $Usubseteq X$ open we have
$$f^{-1}({*}times U) = U, g^{-1}(U) = {*}times U.$$
Thus, $X$ and ${*}times X$ are homeomorphic.
If $X = mathbb R$, instead of $*$ choose some real number $a$. Then ${a}timesmathbb R$ is naturally a line embedded in $mathbb R^2$. Here are examples for some different $a$'s:
answered Jan 5 at 20:26
EnnarEnnar
14.4k32443
edited Jan 5 at 20:40
answered Jan 5 at 20:26
EnnarEnnar
14.4k32443
answered Jan 5 at 20:26
EnnarEnnar
14.4k32443
answered Jan 5 at 20:26
EnnarEnnar
14.4k32443
14.4k32443
add a comment |
add a comment |
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3
$begingroup$
Empty cross anything is empty.
$endgroup$
– Randall
Jan 5 at 19:48
$begingroup$
The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
$endgroup$
– leibnewtz
Jan 5 at 19:59
$begingroup$
Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
$endgroup$
– leibnewtz
Jan 5 at 20:21