Manifolds of zero dimension and $mathbb R^0$?
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Tu Manifolds Section 5.4
Example 5.13 (Manifolds of dimension zero). In a manifold of dimension zero, every singleton subset is homeomorphic to $mathbb R^0$ and so is open. Thus, a zero-dimensional manifold is a discrete set. By second countability, this discrete set must be countable.
Why exactly is the manifold $M$ discrete? I actually proved that the singleton subsets are open in their components but was not able to show they are open in $M$ itself.
Here is what I have done thus far:
Let $M$ be a smooth manifold with dimension zero, which means by definition that all of the connected components of $M$'s topological manifold (see here) ${C_{alpha}}_{alpha in J}$ have dimension zero.
Let $alpha in J$. $C_{alpha}$ has dimension zero, which means by definition (see here) that $forall p in C_{alpha}, exists$ homeomorphism $varphi: U to V$ for some $U$, a neighborhood of $p$ in $C_{alpha}$ and some $V$, an open subset of $mathbb R^0={0}$. $V$ is either ${0}$ or $emptyset$. Since $U$ contains $p$, $U ne emptyset$. Hence, $V ne emptyset$ because from nothing comes nothing, so $V=mathbb R^0={0}$. Sets that are homeomorphic to singletons are singletons. Therefore, $U$ is a singleton containing p, so $U={p}$.
Therefore, we have
$forall p in M, exists$ unique $alpha in J: {p}$ is open in $C_{alpha}$.
I remember the connected components $C_{alpha}$ are:
closed in $M$
not necessarily open in $M$.
open in $M$ if $J$ is finite.
I know ${p}$ is open in one of the connected components of $M$. How do we arrive at the conclusion that ${p}$ is open in $M$ itself?
general-topology differential-geometry manifolds smooth-manifolds connectedness
asked Nov 22 at 9:54
Jack Bauer
1,246531
add a comment |
up vote
0
down vote
favorite
Tu Manifolds Section 5.4
Example 5.13 (Manifolds of dimension zero). In a manifold of dimension zero, every singleton subset is homeomorphic to $mathbb R^0$ and so is open. Thus, a zero-dimensional manifold is a discrete set. By second countability, this discrete set must be countable.
Why exactly is the manifold $M$ discrete? I actually proved that the singleton subsets are open in their components but was not able to show they are open in $M$ itself.
Here is what I have done thus far:
Let $M$ be a smooth manifold with dimension zero, which means by definition that all of the connected components of $M$'s topological manifold (see here) ${C_{alpha}}_{alpha in J}$ have dimension zero.
Let $alpha in J$. $C_{alpha}$ has dimension zero, which means by definition (see here) that $forall p in C_{alpha}, exists$ homeomorphism $varphi: U to V$ for some $U$, a neighborhood of $p$ in $C_{alpha}$ and some $V$, an open subset of $mathbb R^0={0}$. $V$ is either ${0}$ or $emptyset$. Since $U$ contains $p$, $U ne emptyset$. Hence, $V ne emptyset$ because from nothing comes nothing, so $V=mathbb R^0={0}$. Sets that are homeomorphic to singletons are singletons. Therefore, $U$ is a singleton containing p, so $U={p}$.
Therefore, we have
$forall p in M, exists$ unique $alpha in J: {p}$ is open in $C_{alpha}$.
I remember the connected components $C_{alpha}$ are:
closed in $M$
not necessarily open in $M$.
open in $M$ if $J$ is finite.
I know ${p}$ is open in one of the connected components of $M$. How do we arrive at the conclusion that ${p}$ is open in $M$ itself?
general-topology differential-geometry manifolds smooth-manifolds connectedness
asked Nov 22 at 9:54
Jack Bauer
1,246531
1
I don't think you have the right definition of an $n$-dimensional manifold. Being an $n$-dimensional manifold means just that the space is locally homeomorphic to open subsets of ${mathbf R}^n$. In your other question, you have the hypothesis of "locally Euclidean", which implies in particular that the connected components are open, in which case the two notions of $n$-dimensional coincide.
– tomasz
Nov 22 at 10:47
@tomasz I think in Tu, "space is locally homeomorphic to open subsets" is the definition for topological manifold instead of smooth manifold. By $n$-dimensional manifold, do you refer to smooth manifold or topological?
– Jack Bauer
Nov 22 at 11:15
Yes. For a smooth manifold you need some further restrictions. But every smooth manifold is in particular a topological manifold.
– tomasz
Nov 22 at 12:12
@tomasz I will take it to mean that you refer to a smooth manifold. Tu's definition is different from yours, if I understand correctly.
– Jack Bauer
Nov 24 at 3:28
That is my point: it is not (really) different. By the definition you gave 5.2, a manifold is a locally Euclidean space, which implies that connected components are open (because Euclidean spaces are locally connected). Hence, all connected components being $n$-dimensional is the same as being locally Euclidean of dimension $n$.
– tomasz
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Tu Manifolds Section 5.4
Example 5.13 (Manifolds of dimension zero). In a manifold of dimension zero, every singleton subset is homeomorphic to $mathbb R^0$ and so is open. Thus, a zero-dimensional manifold is a discrete set. By second countability, this discrete set must be countable.
Why exactly is the manifold $M$ discrete? I actually proved that the singleton subsets are open in their components but was not able to show they are open in $M$ itself.
Here is what I have done thus far:
Let $M$ be a smooth manifold with dimension zero, which means by definition that all of the connected components of $M$'s topological manifold (see here) ${C_{alpha}}_{alpha in J}$ have dimension zero.
Let $alpha in J$. $C_{alpha}$ has dimension zero, which means by definition (see here) that $forall p in C_{alpha}, exists$ homeomorphism $varphi: U to V$ for some $U$, a neighborhood of $p$ in $C_{alpha}$ and some $V$, an open subset of $mathbb R^0={0}$. $V$ is either ${0}$ or $emptyset$. Since $U$ contains $p$, $U ne emptyset$. Hence, $V ne emptyset$ because from nothing comes nothing, so $V=mathbb R^0={0}$. Sets that are homeomorphic to singletons are singletons. Therefore, $U$ is a singleton containing p, so $U={p}$.
Therefore, we have
$forall p in M, exists$ unique $alpha in J: {p}$ is open in $C_{alpha}$.
I remember the connected components $C_{alpha}$ are:
closed in $M$
not necessarily open in $M$.
open in $M$ if $J$ is finite.
I know ${p}$ is open in one of the connected components of $M$. How do we arrive at the conclusion that ${p}$ is open in $M$ itself?
general-topology differential-geometry manifolds smooth-manifolds connectedness
asked Nov 22 at 9:54
Jack Bauer
1,246531
Tu Manifolds Section 5.4
Example 5.13 (Manifolds of dimension zero). In a manifold of dimension zero, every singleton subset is homeomorphic to $mathbb R^0$ and so is open. Thus, a zero-dimensional manifold is a discrete set. By second countability, this discrete set must be countable.
Why exactly is the manifold $M$ discrete? I actually proved that the singleton subsets are open in their components but was not able to show they are open in $M$ itself.
Here is what I have done thus far:
Let $M$ be a smooth manifold with dimension zero, which means by definition that all of the connected components of $M$'s topological manifold (see here) ${C_{alpha}}_{alpha in J}$ have dimension zero.
Let $alpha in J$. $C_{alpha}$ has dimension zero, which means by definition (see here) that $forall p in C_{alpha}, exists$ homeomorphism $varphi: U to V$ for some $U$, a neighborhood of $p$ in $C_{alpha}$ and some $V$, an open subset of $mathbb R^0={0}$. $V$ is either ${0}$ or $emptyset$. Since $U$ contains $p$, $U ne emptyset$. Hence, $V ne emptyset$ because from nothing comes nothing, so $V=mathbb R^0={0}$. Sets that are homeomorphic to singletons are singletons. Therefore, $U$ is a singleton containing p, so $U={p}$.
Therefore, we have
$forall p in M, exists$ unique $alpha in J: {p}$ is open in $C_{alpha}$.
I remember the connected components $C_{alpha}$ are:
closed in $M$
not necessarily open in $M$.
open in $M$ if $J$ is finite.
I know ${p}$ is open in one of the connected components of $M$. How do we arrive at the conclusion that ${p}$ is open in $M$ itself?
general-topology differential-geometry manifolds smooth-manifolds connectedness
general-topology differential-geometry manifolds smooth-manifolds connectedness
asked Nov 22 at 9:54
Jack Bauer
1,246531
asked Nov 22 at 9:54
Jack Bauer
1,246531
edited Nov 22 at 10:37
asked Nov 22 at 9:54
Jack Bauer
1,246531
asked Nov 22 at 9:54
Jack Bauer
1,246531
asked Nov 22 at 9:54
Jack Bauer
1,246531
1,246531
1
I don't think you have the right definition of an $n$-dimensional manifold. Being an $n$-dimensional manifold means just that the space is locally homeomorphic to open subsets of ${mathbf R}^n$. In your other question, you have the hypothesis of "locally Euclidean", which implies in particular that the connected components are open, in which case the two notions of $n$-dimensional coincide.
– tomasz
Nov 22 at 10:47
@tomasz I think in Tu, "space is locally homeomorphic to open subsets" is the definition for topological manifold instead of smooth manifold. By $n$-dimensional manifold, do you refer to smooth manifold or topological?
– Jack Bauer
Nov 22 at 11:15
Yes. For a smooth manifold you need some further restrictions. But every smooth manifold is in particular a topological manifold.
– tomasz
Nov 22 at 12:12
@tomasz I will take it to mean that you refer to a smooth manifold. Tu's definition is different from yours, if I understand correctly.
– Jack Bauer
Nov 24 at 3:28
That is my point: it is not (really) different. By the definition you gave 5.2, a manifold is a locally Euclidean space, which implies that connected components are open (because Euclidean spaces are locally connected). Hence, all connected components being $n$-dimensional is the same as being locally Euclidean of dimension $n$.
– tomasz
yesterday
add a comment |
1
I don't think you have the right definition of an $n$-dimensional manifold. Being an $n$-dimensional manifold means just that the space is locally homeomorphic to open subsets of ${mathbf R}^n$. In your other question, you have the hypothesis of "locally Euclidean", which implies in particular that the connected components are open, in which case the two notions of $n$-dimensional coincide.
– tomasz
Nov 22 at 10:47
@tomasz I think in Tu, "space is locally homeomorphic to open subsets" is the definition for topological manifold instead of smooth manifold. By $n$-dimensional manifold, do you refer to smooth manifold or topological?
– Jack Bauer
Nov 22 at 11:15
Yes. For a smooth manifold you need some further restrictions. But every smooth manifold is in particular a topological manifold.
– tomasz
Nov 22 at 12:12
@tomasz I will take it to mean that you refer to a smooth manifold. Tu's definition is different from yours, if I understand correctly.
– Jack Bauer
Nov 24 at 3:28
That is my point: it is not (really) different. By the definition you gave 5.2, a manifold is a locally Euclidean space, which implies that connected components are open (because Euclidean spaces are locally connected). Hence, all connected components being $n$-dimensional is the same as being locally Euclidean of dimension $n$.
– tomasz
yesterday
1
1
I don't think you have the right definition of an $n$-dimensional manifold. Being an $n$-dimensional manifold means just that the space is locally homeomorphic to open subsets of ${mathbf R}^n$. In your other question, you have the hypothesis of "locally Euclidean", which implies in particular that the connected components are open, in which case the two notions of $n$-dimensional coincide.
– tomasz
Nov 22 at 10:47
I don't think you have the right definition of an $n$-dimensional manifold. Being an $n$-dimensional manifold means just that the space is locally homeomorphic to open subsets of ${mathbf R}^n$. In your other question, you have the hypothesis of "locally Euclidean", which implies in particular that the connected components are open, in which case the two notions of $n$-dimensional coincide.
– tomasz
Nov 22 at 10:47
@tomasz I think in Tu, "space is locally homeomorphic to open subsets" is the definition for topological manifold instead of smooth manifold. By $n$-dimensional manifold, do you refer to smooth manifold or topological?
– Jack Bauer
Nov 22 at 11:15
@tomasz I think in Tu, "space is locally homeomorphic to open subsets" is the definition for topological manifold instead of smooth manifold. By $n$-dimensional manifold, do you refer to smooth manifold or topological?
– Jack Bauer
Nov 22 at 11:15
Yes. For a smooth manifold you need some further restrictions. But every smooth manifold is in particular a topological manifold.
– tomasz
Nov 22 at 12:12
Yes. For a smooth manifold you need some further restrictions. But every smooth manifold is in particular a topological manifold.
– tomasz
Nov 22 at 12:12
@tomasz I will take it to mean that you refer to a smooth manifold. Tu's definition is different from yours, if I understand correctly.
– Jack Bauer
Nov 24 at 3:28
@tomasz I will take it to mean that you refer to a smooth manifold. Tu's definition is different from yours, if I understand correctly.
– Jack Bauer
Nov 24 at 3:28
That is my point: it is not (really) different. By the definition you gave 5.2, a manifold is a locally Euclidean space, which implies that connected components are open (because Euclidean spaces are locally connected). Hence, all connected components being $n$-dimensional is the same as being locally Euclidean of dimension $n$.
– tomasz
yesterday
That is my point: it is not (really) different. By the definition you gave 5.2, a manifold is a locally Euclidean space, which implies that connected components are open (because Euclidean spaces are locally connected). Hence, all connected components being $n$-dimensional is the same as being locally Euclidean of dimension $n$.
– tomasz
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
You wrote down yourself that you now know that every point in your manifold is clopen (U={p} implies this, since the domain of charts has to be open). But the only clopen subsets of a connected space are the space itself and the empty set, hence ${p}$ is a maximal connected component, and hence all of $C_alpha$
answered Nov 22 at 10:21
Enkidu
67316
Oh, we deduce ${p}$ is open in $C_{alpha}$ and know ${p}$ is closed in $C_{alpha}$ by Hausdorff and so conclude $M$ is totally disconnected, which implies discrete?
– Jack Bauer
Nov 22 at 11:09
1
well, I do not know what totally disconnected means, but we conclude that $C_alpha$ is just a point, and since $alpha$ was arbitrary, all connected components are singletons, which are clopen, hence the union of them (i.e. M) is disrcete.
– Enkidu
Nov 22 at 12:30
Enkdiu, the empty set and the one-point sets are the only connected subsets of a totally disconnected set by the definition. Thank you!
– Jack Bauer
Nov 22 at 14:21
1
Ah, ok thank you, then yes, precisely!
– Enkidu
Nov 22 at 14:31
add a comment |
up vote
0
down vote
I don't know what you know but I would do it like this:
Pick a point $pin M$. It has an open neighborhood $U$ homeomorphic to $mathbf R^0$. So $U={p}$ (it has only one point!). Hence ${p}$ is open.
answered Nov 22 at 10:24
Tom
20917
How do you know $p$ has such an open neighborhood in M? I know $p$ has such an open neighborhood in one of the connected components of M, but I don't know how to extend this to $M$ itself.
– Jack Bauer
Nov 22 at 10:34
Well this is my definition. Do you consider the set of 1/n together with 0 to be a manifold ? Isn't it a counterexample to your question?
– Tom
2 days ago
Tom, are you actually tomasz and mistakenly replied here instead of in the comments on the question? If not, I don't understand your reply here. My problem is I know ${p}$ is open in a connected component of $M$ but don't know how to conclude ${p}$ is open in $M$ itself.
– Jack Bauer
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You wrote down yourself that you now know that every point in your manifold is clopen (U={p} implies this, since the domain of charts has to be open). But the only clopen subsets of a connected space are the space itself and the empty set, hence ${p}$ is a maximal connected component, and hence all of $C_alpha$
answered Nov 22 at 10:21
Enkidu
67316
Oh, we deduce ${p}$ is open in $C_{alpha}$ and know ${p}$ is closed in $C_{alpha}$ by Hausdorff and so conclude $M$ is totally disconnected, which implies discrete?
– Jack Bauer
Nov 22 at 11:09
1
well, I do not know what totally disconnected means, but we conclude that $C_alpha$ is just a point, and since $alpha$ was arbitrary, all connected components are singletons, which are clopen, hence the union of them (i.e. M) is disrcete.
– Enkidu
Nov 22 at 12:30
Enkdiu, the empty set and the one-point sets are the only connected subsets of a totally disconnected set by the definition. Thank you!
– Jack Bauer
Nov 22 at 14:21
1
Ah, ok thank you, then yes, precisely!
– Enkidu
Nov 22 at 14:31
add a comment |
up vote
1
down vote
accepted
You wrote down yourself that you now know that every point in your manifold is clopen (U={p} implies this, since the domain of charts has to be open). But the only clopen subsets of a connected space are the space itself and the empty set, hence ${p}$ is a maximal connected component, and hence all of $C_alpha$
answered Nov 22 at 10:21
Enkidu
67316
Oh, we deduce ${p}$ is open in $C_{alpha}$ and know ${p}$ is closed in $C_{alpha}$ by Hausdorff and so conclude $M$ is totally disconnected, which implies discrete?
– Jack Bauer
Nov 22 at 11:09
1
well, I do not know what totally disconnected means, but we conclude that $C_alpha$ is just a point, and since $alpha$ was arbitrary, all connected components are singletons, which are clopen, hence the union of them (i.e. M) is disrcete.
– Enkidu
Nov 22 at 12:30
Enkdiu, the empty set and the one-point sets are the only connected subsets of a totally disconnected set by the definition. Thank you!
– Jack Bauer
Nov 22 at 14:21
1
Ah, ok thank you, then yes, precisely!
– Enkidu
Nov 22 at 14:31
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You wrote down yourself that you now know that every point in your manifold is clopen (U={p} implies this, since the domain of charts has to be open). But the only clopen subsets of a connected space are the space itself and the empty set, hence ${p}$ is a maximal connected component, and hence all of $C_alpha$
answered Nov 22 at 10:21
Enkidu
67316
You wrote down yourself that you now know that every point in your manifold is clopen (U={p} implies this, since the domain of charts has to be open). But the only clopen subsets of a connected space are the space itself and the empty set, hence ${p}$ is a maximal connected component, and hence all of $C_alpha$
answered Nov 22 at 10:21
Enkidu
67316
answered Nov 22 at 10:21
Enkidu
67316
answered Nov 22 at 10:21
Enkidu
67316
answered Nov 22 at 10:21
Enkidu
67316
67316
Oh, we deduce ${p}$ is open in $C_{alpha}$ and know ${p}$ is closed in $C_{alpha}$ by Hausdorff and so conclude $M$ is totally disconnected, which implies discrete?
– Jack Bauer
Nov 22 at 11:09
1
well, I do not know what totally disconnected means, but we conclude that $C_alpha$ is just a point, and since $alpha$ was arbitrary, all connected components are singletons, which are clopen, hence the union of them (i.e. M) is disrcete.
– Enkidu
Nov 22 at 12:30
Enkdiu, the empty set and the one-point sets are the only connected subsets of a totally disconnected set by the definition. Thank you!
– Jack Bauer
Nov 22 at 14:21
1
Ah, ok thank you, then yes, precisely!
– Enkidu
Nov 22 at 14:31
add a comment |
Oh, we deduce ${p}$ is open in $C_{alpha}$ and know ${p}$ is closed in $C_{alpha}$ by Hausdorff and so conclude $M$ is totally disconnected, which implies discrete?
– Jack Bauer
Nov 22 at 11:09
1
well, I do not know what totally disconnected means, but we conclude that $C_alpha$ is just a point, and since $alpha$ was arbitrary, all connected components are singletons, which are clopen, hence the union of them (i.e. M) is disrcete.
– Enkidu
Nov 22 at 12:30
Enkdiu, the empty set and the one-point sets are the only connected subsets of a totally disconnected set by the definition. Thank you!
– Jack Bauer
Nov 22 at 14:21
1
Ah, ok thank you, then yes, precisely!
– Enkidu
Nov 22 at 14:31
Oh, we deduce ${p}$ is open in $C_{alpha}$ and know ${p}$ is closed in $C_{alpha}$ by Hausdorff and so conclude $M$ is totally disconnected, which implies discrete?
– Jack Bauer
Nov 22 at 11:09
Oh, we deduce ${p}$ is open in $C_{alpha}$ and know ${p}$ is closed in $C_{alpha}$ by Hausdorff and so conclude $M$ is totally disconnected, which implies discrete?
– Jack Bauer
Nov 22 at 11:09
1
1
well, I do not know what totally disconnected means, but we conclude that $C_alpha$ is just a point, and since $alpha$ was arbitrary, all connected components are singletons, which are clopen, hence the union of them (i.e. M) is disrcete.
– Enkidu
Nov 22 at 12:30
well, I do not know what totally disconnected means, but we conclude that $C_alpha$ is just a point, and since $alpha$ was arbitrary, all connected components are singletons, which are clopen, hence the union of them (i.e. M) is disrcete.
– Enkidu
Nov 22 at 12:30
Enkdiu, the empty set and the one-point sets are the only connected subsets of a totally disconnected set by the definition. Thank you!
– Jack Bauer
Nov 22 at 14:21
Enkdiu, the empty set and the one-point sets are the only connected subsets of a totally disconnected set by the definition. Thank you!
– Jack Bauer
Nov 22 at 14:21
1
1
Ah, ok thank you, then yes, precisely!
– Enkidu
Nov 22 at 14:31
Ah, ok thank you, then yes, precisely!
– Enkidu
Nov 22 at 14:31
add a comment |
up vote
0
down vote
I don't know what you know but I would do it like this:
Pick a point $pin M$. It has an open neighborhood $U$ homeomorphic to $mathbf R^0$. So $U={p}$ (it has only one point!). Hence ${p}$ is open.
answered Nov 22 at 10:24
Tom
20917
How do you know $p$ has such an open neighborhood in M? I know $p$ has such an open neighborhood in one of the connected components of M, but I don't know how to extend this to $M$ itself.
– Jack Bauer
Nov 22 at 10:34
Well this is my definition. Do you consider the set of 1/n together with 0 to be a manifold ? Isn't it a counterexample to your question?
– Tom
2 days ago
Tom, are you actually tomasz and mistakenly replied here instead of in the comments on the question? If not, I don't understand your reply here. My problem is I know ${p}$ is open in a connected component of $M$ but don't know how to conclude ${p}$ is open in $M$ itself.
– Jack Bauer
yesterday
add a comment |
up vote
0
down vote
I don't know what you know but I would do it like this:
Pick a point $pin M$. It has an open neighborhood $U$ homeomorphic to $mathbf R^0$. So $U={p}$ (it has only one point!). Hence ${p}$ is open.
answered Nov 22 at 10:24
Tom
20917
How do you know $p$ has such an open neighborhood in M? I know $p$ has such an open neighborhood in one of the connected components of M, but I don't know how to extend this to $M$ itself.
– Jack Bauer
Nov 22 at 10:34
Well this is my definition. Do you consider the set of 1/n together with 0 to be a manifold ? Isn't it a counterexample to your question?
– Tom
2 days ago
Tom, are you actually tomasz and mistakenly replied here instead of in the comments on the question? If not, I don't understand your reply here. My problem is I know ${p}$ is open in a connected component of $M$ but don't know how to conclude ${p}$ is open in $M$ itself.
– Jack Bauer
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
I don't know what you know but I would do it like this:
Pick a point $pin M$. It has an open neighborhood $U$ homeomorphic to $mathbf R^0$. So $U={p}$ (it has only one point!). Hence ${p}$ is open.
answered Nov 22 at 10:24
Tom
20917
I don't know what you know but I would do it like this:
Pick a point $pin M$. It has an open neighborhood $U$ homeomorphic to $mathbf R^0$. So $U={p}$ (it has only one point!). Hence ${p}$ is open.
answered Nov 22 at 10:24
Tom
20917
answered Nov 22 at 10:24
Tom
20917
answered Nov 22 at 10:24
Tom
20917
answered Nov 22 at 10:24
Tom
20917
20917
How do you know $p$ has such an open neighborhood in M? I know $p$ has such an open neighborhood in one of the connected components of M, but I don't know how to extend this to $M$ itself.
– Jack Bauer
Nov 22 at 10:34
Well this is my definition. Do you consider the set of 1/n together with 0 to be a manifold ? Isn't it a counterexample to your question?
– Tom
2 days ago
Tom, are you actually tomasz and mistakenly replied here instead of in the comments on the question? If not, I don't understand your reply here. My problem is I know ${p}$ is open in a connected component of $M$ but don't know how to conclude ${p}$ is open in $M$ itself.
– Jack Bauer
yesterday
add a comment |
How do you know $p$ has such an open neighborhood in M? I know $p$ has such an open neighborhood in one of the connected components of M, but I don't know how to extend this to $M$ itself.
– Jack Bauer
Nov 22 at 10:34
Well this is my definition. Do you consider the set of 1/n together with 0 to be a manifold ? Isn't it a counterexample to your question?
– Tom
2 days ago
Tom, are you actually tomasz and mistakenly replied here instead of in the comments on the question? If not, I don't understand your reply here. My problem is I know ${p}$ is open in a connected component of $M$ but don't know how to conclude ${p}$ is open in $M$ itself.
– Jack Bauer
yesterday
How do you know $p$ has such an open neighborhood in M? I know $p$ has such an open neighborhood in one of the connected components of M, but I don't know how to extend this to $M$ itself.
– Jack Bauer
Nov 22 at 10:34
How do you know $p$ has such an open neighborhood in M? I know $p$ has such an open neighborhood in one of the connected components of M, but I don't know how to extend this to $M$ itself.
– Jack Bauer
Nov 22 at 10:34
Well this is my definition. Do you consider the set of 1/n together with 0 to be a manifold ? Isn't it a counterexample to your question?
– Tom
2 days ago
Well this is my definition. Do you consider the set of 1/n together with 0 to be a manifold ? Isn't it a counterexample to your question?
– Tom
2 days ago
Tom, are you actually tomasz and mistakenly replied here instead of in the comments on the question? If not, I don't understand your reply here. My problem is I know ${p}$ is open in a connected component of $M$ but don't know how to conclude ${p}$ is open in $M$ itself.
– Jack Bauer
yesterday
Tom, are you actually tomasz and mistakenly replied here instead of in the comments on the question? If not, I don't understand your reply here. My problem is I know ${p}$ is open in a connected component of $M$ but don't know how to conclude ${p}$ is open in $M$ itself.
– Jack Bauer
yesterday
add a comment |
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1
I don't think you have the right definition of an $n$-dimensional manifold. Being an $n$-dimensional manifold means just that the space is locally homeomorphic to open subsets of ${mathbf R}^n$. In your other question, you have the hypothesis of "locally Euclidean", which implies in particular that the connected components are open, in which case the two notions of $n$-dimensional coincide.
– tomasz
Nov 22 at 10:47
@tomasz I think in Tu, "space is locally homeomorphic to open subsets" is the definition for topological manifold instead of smooth manifold. By $n$-dimensional manifold, do you refer to smooth manifold or topological?
– Jack Bauer
Nov 22 at 11:15
Yes. For a smooth manifold you need some further restrictions. But every smooth manifold is in particular a topological manifold.
– tomasz
Nov 22 at 12:12
@tomasz I will take it to mean that you refer to a smooth manifold. Tu's definition is different from yours, if I understand correctly.
– Jack Bauer
Nov 24 at 3:28
That is my point: it is not (really) different. By the definition you gave 5.2, a manifold is a locally Euclidean space, which implies that connected components are open (because Euclidean spaces are locally connected). Hence, all connected components being $n$-dimensional is the same as being locally Euclidean of dimension $n$.
– tomasz
yesterday