Willard's very extreme example of topological space
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In Willard's General Topology, page 32, he says:
"If $y$ is close to $x$ in a metric space, then $x$ is close to $y$; but it can happen in a topological space that $y$ is in every neighborhood of $x$ while $x$ is in no neighborhood of $y$ (a very extreme example; this doesn't happen in useful topological spaces, although many useful spaces lack symmetry in some degree)"
But I do not get how could this be possible. How can such extreme example of topological space exist if the whole space contains itself and then it is a neighborhood of any point, so there cannot be a point that is in no neighborhood of the other?
Any clarification on what he meant would be accepted, or even an explicit example of such very extreme topological space.
Thanks.
general-topology
asked Dec 2 '18 at 3:15
creepyrodentcreepyrodent
48614
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add a comment |
$begingroup$
In Willard's General Topology, page 32, he says:
"If $y$ is close to $x$ in a metric space, then $x$ is close to $y$; but it can happen in a topological space that $y$ is in every neighborhood of $x$ while $x$ is in no neighborhood of $y$ (a very extreme example; this doesn't happen in useful topological spaces, although many useful spaces lack symmetry in some degree)"
But I do not get how could this be possible. How can such extreme example of topological space exist if the whole space contains itself and then it is a neighborhood of any point, so there cannot be a point that is in no neighborhood of the other?
Any clarification on what he meant would be accepted, or even an explicit example of such very extreme topological space.
Thanks.
general-topology
asked Dec 2 '18 at 3:15
creepyrodentcreepyrodent
48614
$endgroup$
$begingroup$
Would they be so extreme that they don't even exist?
$endgroup$
– creepyrodent
Dec 2 '18 at 3:22
1
$begingroup$
Since a neighborhood is just an open set, a neighborhood of $x$ containing $y$ is also a neighborhood of $y$ containing $x.$ It's hard to understand what he means. Still, a terrific textbook, in my view.
$endgroup$
– saulspatz
Dec 2 '18 at 3:24
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I agree, it is a great book. I really like the comments he makes in every section, making abstract concepts more clear and sometimes foreshadowing the usefulness of it for mathematicians. And also there's a lot of nice examples and counterxamples.
$endgroup$
– creepyrodent
Dec 3 '18 at 12:13
add a comment |
$begingroup$
In Willard's General Topology, page 32, he says:
"If $y$ is close to $x$ in a metric space, then $x$ is close to $y$; but it can happen in a topological space that $y$ is in every neighborhood of $x$ while $x$ is in no neighborhood of $y$ (a very extreme example; this doesn't happen in useful topological spaces, although many useful spaces lack symmetry in some degree)"
But I do not get how could this be possible. How can such extreme example of topological space exist if the whole space contains itself and then it is a neighborhood of any point, so there cannot be a point that is in no neighborhood of the other?
Any clarification on what he meant would be accepted, or even an explicit example of such very extreme topological space.
Thanks.
general-topology
asked Dec 2 '18 at 3:15
creepyrodentcreepyrodent
48614
$endgroup$
In Willard's General Topology, page 32, he says:
"If $y$ is close to $x$ in a metric space, then $x$ is close to $y$; but it can happen in a topological space that $y$ is in every neighborhood of $x$ while $x$ is in no neighborhood of $y$ (a very extreme example; this doesn't happen in useful topological spaces, although many useful spaces lack symmetry in some degree)"
But I do not get how could this be possible. How can such extreme example of topological space exist if the whole space contains itself and then it is a neighborhood of any point, so there cannot be a point that is in no neighborhood of the other?
Any clarification on what he meant would be accepted, or even an explicit example of such very extreme topological space.
Thanks.
general-topology
general-topology
asked Dec 2 '18 at 3:15
creepyrodentcreepyrodent
48614
asked Dec 2 '18 at 3:15
creepyrodentcreepyrodent
48614
asked Dec 2 '18 at 3:15
creepyrodentcreepyrodent
48614
asked Dec 2 '18 at 3:15
creepyrodentcreepyrodent
48614
asked Dec 2 '18 at 3:15
creepyrodentcreepyrodent
48614
48614
$begingroup$
Would they be so extreme that they don't even exist?
$endgroup$
– creepyrodent
Dec 2 '18 at 3:22
1
$begingroup$
Since a neighborhood is just an open set, a neighborhood of $x$ containing $y$ is also a neighborhood of $y$ containing $x.$ It's hard to understand what he means. Still, a terrific textbook, in my view.
$endgroup$
– saulspatz
Dec 2 '18 at 3:24
$begingroup$
I agree, it is a great book. I really like the comments he makes in every section, making abstract concepts more clear and sometimes foreshadowing the usefulness of it for mathematicians. And also there's a lot of nice examples and counterxamples.
$endgroup$
– creepyrodent
Dec 3 '18 at 12:13
add a comment |
$begingroup$
Would they be so extreme that they don't even exist?
$endgroup$
– creepyrodent
Dec 2 '18 at 3:22
1
$begingroup$
Since a neighborhood is just an open set, a neighborhood of $x$ containing $y$ is also a neighborhood of $y$ containing $x.$ It's hard to understand what he means. Still, a terrific textbook, in my view.
$endgroup$
– saulspatz
Dec 2 '18 at 3:24
$begingroup$
I agree, it is a great book. I really like the comments he makes in every section, making abstract concepts more clear and sometimes foreshadowing the usefulness of it for mathematicians. And also there's a lot of nice examples and counterxamples.
$endgroup$
– creepyrodent
Dec 3 '18 at 12:13
$begingroup$
Would they be so extreme that they don't even exist?
$endgroup$
– creepyrodent
Dec 2 '18 at 3:22
$begingroup$
Would they be so extreme that they don't even exist?
$endgroup$
– creepyrodent
Dec 2 '18 at 3:22
1
1
$begingroup$
Since a neighborhood is just an open set, a neighborhood of $x$ containing $y$ is also a neighborhood of $y$ containing $x.$ It's hard to understand what he means. Still, a terrific textbook, in my view.
$endgroup$
– saulspatz
Dec 2 '18 at 3:24
$begingroup$
Since a neighborhood is just an open set, a neighborhood of $x$ containing $y$ is also a neighborhood of $y$ containing $x.$ It's hard to understand what he means. Still, a terrific textbook, in my view.
$endgroup$
– saulspatz
Dec 2 '18 at 3:24
$begingroup$
I agree, it is a great book. I really like the comments he makes in every section, making abstract concepts more clear and sometimes foreshadowing the usefulness of it for mathematicians. And also there's a lot of nice examples and counterxamples.
$endgroup$
– creepyrodent
Dec 3 '18 at 12:13
$begingroup$
I agree, it is a great book. I really like the comments he makes in every section, making abstract concepts more clear and sometimes foreshadowing the usefulness of it for mathematicians. And also there's a lot of nice examples and counterxamples.
$endgroup$
– creepyrodent
Dec 3 '18 at 12:13
add a comment |
1 Answer
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oldest
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You are correct: the entire space will always be a neighborhood of $y$ which contains $x$. (I'm assuming $x neq y$.) I'm guessing he wanted to exclude that neighborhood as a trivial example, and to say that no other neighborhoods of $y$ contain $x$.
In the Sierpiński space, for example, $1$ has a neighborhood that does not contain $0$, but $0$ has a single neighborhood, which does contain $1$. More generally, the "excluded point topology" will give you more examples.
answered Dec 2 '18 at 3:29
Hew WolffHew Wolff
2,245716
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$begingroup$
You are correct: the entire space will always be a neighborhood of $y$ which contains $x$. (I'm assuming $x neq y$.) I'm guessing he wanted to exclude that neighborhood as a trivial example, and to say that no other neighborhoods of $y$ contain $x$.
In the Sierpiński space, for example, $1$ has a neighborhood that does not contain $0$, but $0$ has a single neighborhood, which does contain $1$. More generally, the "excluded point topology" will give you more examples.
answered Dec 2 '18 at 3:29
Hew WolffHew Wolff
2,245716
$endgroup$
add a comment |
$begingroup$
You are correct: the entire space will always be a neighborhood of $y$ which contains $x$. (I'm assuming $x neq y$.) I'm guessing he wanted to exclude that neighborhood as a trivial example, and to say that no other neighborhoods of $y$ contain $x$.
In the Sierpiński space, for example, $1$ has a neighborhood that does not contain $0$, but $0$ has a single neighborhood, which does contain $1$. More generally, the "excluded point topology" will give you more examples.
answered Dec 2 '18 at 3:29
Hew WolffHew Wolff
2,245716
$endgroup$
add a comment |
$begingroup$
You are correct: the entire space will always be a neighborhood of $y$ which contains $x$. (I'm assuming $x neq y$.) I'm guessing he wanted to exclude that neighborhood as a trivial example, and to say that no other neighborhoods of $y$ contain $x$.
In the Sierpiński space, for example, $1$ has a neighborhood that does not contain $0$, but $0$ has a single neighborhood, which does contain $1$. More generally, the "excluded point topology" will give you more examples.
answered Dec 2 '18 at 3:29
Hew WolffHew Wolff
2,245716
$endgroup$
You are correct: the entire space will always be a neighborhood of $y$ which contains $x$. (I'm assuming $x neq y$.) I'm guessing he wanted to exclude that neighborhood as a trivial example, and to say that no other neighborhoods of $y$ contain $x$.
In the Sierpiński space, for example, $1$ has a neighborhood that does not contain $0$, but $0$ has a single neighborhood, which does contain $1$. More generally, the "excluded point topology" will give you more examples.
answered Dec 2 '18 at 3:29
Hew WolffHew Wolff
2,245716
answered Dec 2 '18 at 3:29
Hew WolffHew Wolff
2,245716
answered Dec 2 '18 at 3:29
Hew WolffHew Wolff
2,245716
answered Dec 2 '18 at 3:29
Hew WolffHew Wolff
2,245716
2,245716
add a comment |
add a comment |
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$begingroup$
Would they be so extreme that they don't even exist?
$endgroup$
– creepyrodent
Dec 2 '18 at 3:22
1
$begingroup$
Since a neighborhood is just an open set, a neighborhood of $x$ containing $y$ is also a neighborhood of $y$ containing $x.$ It's hard to understand what he means. Still, a terrific textbook, in my view.
$endgroup$
– saulspatz
Dec 2 '18 at 3:24
$begingroup$
I agree, it is a great book. I really like the comments he makes in every section, making abstract concepts more clear and sometimes foreshadowing the usefulness of it for mathematicians. And also there's a lot of nice examples and counterxamples.
$endgroup$
– creepyrodent
Dec 3 '18 at 12:13