Does there exist a smooth structure on $[0,\infty)$ that induces the standard topology? [closed]

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

• You can define a smooth manifold with boundary, and this is an example of such a thing, with boundary $\{ 0 \}$. Sep 30, 2022 at 23:23
• @QiaochuYuan What about manifolds without boundary. Is it possible?
– XXX
Sep 30, 2022 at 23:25
• 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? Sep 30, 2022 at 23:30
• @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?
– XXX
Sep 30, 2022 at 23:34
• @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 $N$. Sep 30, 2022 at 23:52

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.