Is there a topological group which is Hausdorff, first countable,
locally connected and has finite topological dimension, yet fails
to be locally compact?



A non locally compact group of finite topological dimension?
The Gleason-Montgomery theorem on the local compactness of locally path-connected finite-dimensional topological groups was generalized by Banakh and Zdomskyy (<http://topology.auburn.edu/tp/reprints/v36/tp36027.pdf>) who proved that a topological group $G$ is locally compact if it is compactly finite-dimensional and locally continuum-connected.


On the other hand, for every $n\ge 2$ the Euclidean space $\mathbb R^n$ contains an additive subgroup $G$, which is connected, locally connected, and not Borel (so, not locally compact). 


To construct such a group $G$, consider the family $\mathcal B$ of all uncountable Borel subsets of $\mathbb R^n$ and observe that it has cardinality of continuum. So, $\mathcal B$ can be enumerated by ordinals $<\mathfrak c$ as $(B\_\alpha)\_{\alpha<\mathfrak c}$. Fix any non-zero point $p$ in $\mathbb R^n$ and by transfinite induction, for every ordinal $\alpha<\mathfrak c$ choose a point $x\_\alpha\in B\_\alpha$ so that the subgroup $G\_\alpha$ generated by the set $\{x\_\beta\}\_{\beta\le\alpha}$ does not contain the point $p$. Such a choice is always possible since $B\_\alpha$ has cardinality of continuus and $x\_\alpha$ should be chosen to avoid the subset $\{\frac1n(p-g):n\in\mathbb Z\setminus\{0\},\;g\in \bigcup\_{\beta<\alpha}G\_{\beta}\}$ of cardinality $<\mathfrak c$.
After completing the inductive construction, consider the additive subgroup $G$ of $\mathbb R^n$, generated by the set $\{x\_\alpha\}\_{\alpha<\mathfrak c}$. By construction, this subgroup $G$ does not contain the point $p$ and meets every uncountable Borel subset $B$ of $\mathbb R^n$. In particular, it meets the uncountable Borel set $B-p$, which implies that $p+G\subset \mathbb R^n\setminus G$ meets $B$. This means that $G$ is a Bernstein set in $\mathbb R^n$.


It remains to prove that every Bernstein set $G$ in $\mathbb R^n$, $n\ge 2$, is connected and locally connected. This will follow as soon as we check that for every open subset $U\subset\mathbb R^n$, homeomorphic to $\mathbb R^n$ the intersection $U\cap G$ is connected. To derive a contradiction, assume that $U\cap G$ is disconnected and find two disjoint non-empty open sets $U\_1,U\_2\subset U$ such that $U\cap G=(U\_1\cap G)\cup(U\_2\cap G)$. It follows that $F=U\setminus (U\_1\cup U\_2)$ is a closed subset of $U$ separating $U$, which is disjoint with the Bernstein set $G$. So, $F$ is at most countable and hence $\dim(F)=0$.
But this contradicts Mazurkiewicz Theorem (saying that $\mathbb R^n$ cannot be separated by a closed subset of dimension $<n-1$).


