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Proposition: $\mathbb{R}$ with the discrete topology is not locally euclidean.

The proof begins with the conditional statement that if $U \subseteq \mathbb{R}^{n}$ is a non-empty open set in the standard topology and $V \subseteq \mathbb{R}$ is a non-empty set in the discrete topology, U and V are not homeomorphic.

I am unable to understand why U and V are not homeomorphic even from the definition of "Homeomorphism". Is it because the map between the two topological spaces are not continuous?

Thanks in advance for any clarification.

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  • $\begingroup$ Blast! typo. You're right. My doubt still holds. $\endgroup$ Commented Sep 24, 2016 at 7:59
  • $\begingroup$ It's not clear what you mean by "Euclidean". ${\mathbb R}$, or ${\mathbb R}^n$, with the standard scalar product remains a euclidean vector space even if you decide to adopt a topology of your own choice. – That the discrete topology is different from the "euclidean topology" is obvious. $\endgroup$ Commented Sep 24, 2016 at 8:10
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    $\begingroup$ $\Bbb R$ with discrete topology is locally just a point, which is homeomorphic to the zero-dimensional euclidean vector space $\Bbb R^0$ $\endgroup$ Commented Sep 24, 2016 at 8:16
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    $\begingroup$ I don't think the proposition as stated is true. A discrete space is (pretty much by definition) a zero-dimensional manifold. What is true is that it is not of positive dimension. You could restate being locally Euclidean to mean that the dimension is positive, but that would make statements of lots of theorems in differential topology terribly awkward. $\endgroup$
    – tomasz
    Commented Sep 24, 2016 at 8:18
  • $\begingroup$ To show that, you can just note that no Euclidean space of positive dimension is locally discrete. $\endgroup$
    – tomasz
    Commented Sep 24, 2016 at 8:20

2 Answers 2

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Suppose $f : U \to V$ is a homeomorphism. Then $f$ is a continuous bijection. Now for any $x \in V$, the set $\{x\}$ is open in the discrete topology. Since $f^{-1}[\{x\}]$ is open (by continuity) and is a single element of $U$ (since $f$ injective).

However, single points are not open in the Euclidean topology, so we get a contradiction. Therefore there does not exist such a homeomorphism $f$.

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Let $U$ be a non-empty open set in $\Bbb R^n$. Suppose that there is a homeomorphism from $U$ to some $D\subseteq\Bbb R$ with the discrete topology.

Fix $p\in U$. There is an $\epsilon>0$ such that $B(p,\epsilon)\subseteq U$. $B(p,\epsilon)$ is a connected set with more than one point, so $h[B(p,\epsilon)]$ must be a connected subset of $D$ with more than one point. But $D$ is discrete, so no subset of $D$ with more than one point is connected. Thus, there cannot be such a homeomorphism $h$.

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