# R with trivial topology is not locally Euclidean

Question:

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

Attempt:

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.

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.
• 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? – Mathematicing Sep 25 '16 at 10:04
• 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. – Cave Johnson Sep 25 '16 at 10:25