Show that $\mathbb{R}$ with the trivial topology is not locally Euclidean.


Let $\mathbb{R}$ be the set with the trivial topology $T_{t}=\left \{ \varnothing ,\mathbb{R} \right \}$ Let $\mathbb{R}^{n}$ be the set with the standard topology $T_{R}$.

Recall: that a topological space X is locally Euclidean if $\exists n \in \mathbb{Z}^{+}, \forall x \in X$ there is an open set $U \subseteq X $containing x that is homeomorphic to some open set $V \subseteq \mathbb{R}^{n}$.

By contrapositivity: that a topological space X is locally Euclidean if $\exists n \in \mathbb{Z}^{+}, \forall x \in X$ there is an open set $U \subseteq X $containing x that is not homeomorphic to some open set $V \subseteq \mathbb{R}^{n}$.

We assume a homeomorphism $f:\mathbb{R}\rightarrow \mathbb{R}^{n}$

f is homeomorphic to f is bijective, f and $f^{-1}$ is continuous between the topological spaces.

So we have $f^{-1}: V\rightarrow U$

$f:U\rightarrow V$

So we have that

f maps open sets to open sets. $f^{-1}\left ( \varnothing \right )=\varnothing \in T_{R}$ on $\mathbb{R}^{n}$

$f^{-1}\left ( \mathbb{R} \right ) \in T_{R}$ on $\mathbb{R}^{n}$

I would like some help is seeking a contradiction which would solve the problem.

Thanks in advance.

  • $\begingroup$ @egreg Every locally euclidean space is hausdorff??? You can have non-hausdorff manifolds... $\endgroup$ – orientablesurface Jan 15 at 10:57

Assume there is a homeomorphism $f:\mathbb{R}\to\mathbb{R}^n$, the former $\mathbb{R}$ being equipped with trivial topology. The preimage of any open neighborhood of $f(x_0)$ is non-empty and open, hence is $\mathbb{R}$. This simply implies $f(x)$ is a constant, thus cannot be a bijection.

| cite | improve this answer | |
  • $\begingroup$ The open sets in the topology on $R^{n}$ are open intervals for which there is a point x in that open interval. The pre-image of that open interval maps that open intervals to a point in R, yes? But this violates the definition of a continuous map f that maps an open set to an open set. Did I understood this? $\endgroup$ – Mathematicing Sep 25 '16 at 10:04
  • 1
    $\begingroup$ Your reasoning also works. But what I mean is: take a ball $B(f(x_0),\varepsilon)$ entered at $f(x_0)$ with radius $\varepsilon$. The continuity of $f$ implies $f^{-1}(B(f(x_0),\varepsilon))$ is non-empty(since it contains $x_0$) and open in the trivial topology of $\mathbb{R}$. So it has to be $\mathbb{R}$. This shows the image of $f$ is constrained in an arbitrarily small ball centered at $f(x_0)$. Since $\mathbb{R}^n$ is Hausdorff, we have showed that all $f(x)$ equals $f(x_0)$. Now that $f$ is not even a injection, it can't be a homeomorphism. $\endgroup$ – Cave Johnson Sep 25 '16 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.