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For a continuous bijective function from $\Bbb R$ to $\Bbb R$ is a homeomorphism, can we generalise it?

José Carlos Santos
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jjj
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  • You might have to be a bit more specific. Take the identity map from $(\Bbb{R},\tau_1)$ to $(\Bbb{R},\tau_2)$, where $\tau_1$ is the usual (Euclidean) topology on $\Bbb{R}$ and $\tau_{2}={\emptyset,\Bbb{R}}$ is the trivial topology on $\Bbb{R}$, then the map is a continuous bijection but its inverse is not continuous. – Teddy38 Sep 26 '17 at 12:26
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    Examples – Hellen Sep 26 '17 at 12:28
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    Maybe I'm misreading it, but isn't the real question here, "under what conditions can we say that a continuous bijection $X\to X$ is guaranteed to be a homeomorphism?" Can we do any better than, "$X$ is $\mathbb{R}$ with the usual topology"? – G Tony Jacobs Sep 26 '17 at 13:05
  • Like, what properties of $\mathbb{R}$ make this theorem hold in that case, when we already know there are cases where it doesn't hold? – G Tony Jacobs Sep 26 '17 at 13:07
  • If $X$ is a manifold then yes, since we have the open mapping theorem) an njective map from $\mathbb{R}^m$ to itself is open. – orangeskid Sep 26 '17 at 14:54
  • So, we know a few sufficient conditions: $X$ is homeomorphic to $\mathbb{R}^n$, or $X$ is compact, or $X$ is a manifold... but no necessary condition. – G Tony Jacobs Sep 27 '17 at 12:47

2 Answers2

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Let $c_{00}$ be the space of all sequences $(x_n)_{n\in\mathbb N}$ of complex numbers such that $n\gg0\implies x_n=0$. Here, consider the distance$$d\bigl((x_n)_{n\in\mathbb N},(y_n)_{n\in\mathbb N}\bigr)=\max_{n\in\mathbb N}|x_n-y_n|.$$Finally, define$$\begin{array}{rccc}f\colon&c_{00}&\longrightarrow&c_{00}\\&(x_n)_{n\in\mathbb N}&\mapsto&\left(\dfrac{x_n}n\right)_{n\in\mathbb N}.\end{array}$$Then $f$ is bijective and continuous. However, $f^{-1}$ is discontinuous, and therefore $f$ is not a homeomorphism.

José Carlos Santos
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  • This is great, but I'm not sure it addresses the question. "Can we generalise it" seems to mean, "under what conditions can we say that a continuous bijection is a homeomorphism", not under what conditions is it not one. Maybe my interpretation is wrong, but that seems like the more interesting question. – G Tony Jacobs Sep 26 '17 at 13:03
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    @GTonyJacobs At least, my example shows that we can't generalize it to normed vector spaces. – José Carlos Santos Sep 26 '17 at 13:10
  • Yeah, that's fair. So, where can we generalize it? Asked another way, suppose we have a space $X$, and we know that all continuous bijections from $X$ to itself are homeomorphisms. What does this tell us about $X$? – G Tony Jacobs Sep 26 '17 at 13:11
  • @GTonyJacobs It tell us that $X$ is not compact. And also that it is not homeomorphic to $\mathbb{R}^n$. – José Carlos Santos Sep 26 '17 at 13:13
  • What? Are the continuous bijections from $\mathbb{R}^n$ to itself that are not homeo? Even when $n=1$? What about from $[0,1]$ to itself? – G Tony Jacobs Sep 26 '17 at 13:16
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    @GTonyJacobs I am sorry. What I should have written is that if there are continuous bijections from $X$ onto itself which are not homeomorphims, then $X$ is not compact and it is not homeomorphic to $\mathbb{R}^n$. – José Carlos Santos Sep 26 '17 at 13:19
  • Oh, I see. That makes more sense. :) – G Tony Jacobs Sep 26 '17 at 13:21
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if we choose the domain R with discrete topology and range also R with standard topology then identity function is a continuous bijection but not a homeomorphism.

jjj
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