I am sure that there is a problem with the point 0, but I don't know how to show it rigorously.

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    $\begingroup$ You can define a smooth manifold with boundary, and this is an example of such a thing, with boundary $\{ 0 \}$. $\endgroup$ Sep 30, 2022 at 23:23
  • $\begingroup$ @QiaochuYuan What about manifolds without boundary. Is it possible? $\endgroup$
    – XXX
    Sep 30, 2022 at 23:25
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    $\begingroup$ If $[0,\infty)$ with the usual topology can be made into a smooth $N$-manifold without boundary, there'd exist $\epsilon > 0$ small enough such that $[0,\epsilon)$ is diffeomorphic (and hence homeomorphic) to an open (and necessarily connected) subset of $\mathbb{R}^N$, where $N=1$ by topological invariance of dimension. Why is this a problem? $\endgroup$ Sep 30, 2022 at 23:30
  • $\begingroup$ @BranimirĆaćić The preimage of an open set must be open in the standard topology, but $[0,\epsilon)$ is not. But why can we find such an interval mapped to an open set? $\endgroup$
    – XXX
    Sep 30, 2022 at 23:34
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    $\begingroup$ @BranimirĆaćić answered taking into account that one could try to endow $[0,\infty)$ with an $N$-dimensional smooth manifold's smooth structure in order to induce the standard topology... for any $N$. The first step is determining that since the charts will map open subsets of $\mathbb{R}_{>0}$ to open subsets of $\mathbb{R}^N$ by invariance of domain we are forced into $N$ being equal to $1$. Now that we know that $N=1$ then we can analyze an adequate chart from a neighbourhood of $0$ to an open subset of $\mathbb{R}$ to arrive at a contradiction. It can't happen then for any $N$. $\endgroup$ Sep 30, 2022 at 23:52

1 Answer 1


No. Assume there exists a smooth structure on $M = [0,\infty)$ that induces the standard topology. Then $M$ would be an $n$-dimensional smooth manifold for some $n$. In particular $M$ would be locally Euclidean with "model space" $\mathbb R^n$.

Thus for each $p \in M$ there would exist a homeomorphism $h : U \to B^n$ from an open neigborhood $U$ of $p$ (in the standard topology!) to the open unit ball $B^n$ in $\mathbb R^n$ such that $h(p) = 0$. Note that $U$ must be connected, i.e. an interval.

Let us now prove that we must have $n=1$. Let $x$ be an interior point of the interval. Then $U \setminus \{x\}$ is disconnected, thus also $h(U \setminus \{x\}) = B^n \setminus \{h(x)\}$ is disconnected. But for $n \ge 2$ this cannot happen.

Now take $p = 0$. Then $U$ must be an interval of the form $[0,a)$ with $0 < a \le \infty$. But $U \setminus \{0\}$ is connected, thus also $h(U \setminus \{0\}) = B^1 \setminus \{h(0)\} =(-1,0) \cup (0,1)$ is connected, which is not true. This contradiction shows that the above assumption was false.


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