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

1$\begingroup$ Is there a topological space satisfying these properties? $\endgroup$– Yonatan HarpazFeb 11 '16 at 19:31

1$\begingroup$ @YonatanHarpaz: Remove from the plane those points which have rational coordinates. $\endgroup$– Ramiro de la VegaFeb 11 '16 at 19:57

1$\begingroup$ @YonatanHarpaz for spaces, in dimension 1 there are examples as well. Just take the 1skeleton of a graph not of finite degree, endowed with either the direct limit topology (of its finite subgraphs), or with its distance topology. $\endgroup$– YCorFeb 11 '16 at 22:05
The GleasonMontgomery theorem on the local compactness of locally pathconnected finitedimensional 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 finitedimensional and locally continuumconnected.
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 nonzero 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 $\{\frac1n(pg):n\in\mathbb Z\setminus\{0\},\;g\in \bigcup_{\beta<\alpha}G_{\beta}\}$ 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 $Bp$, 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 nonempty 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 $<n1$).
This is not quite an answer to the question, but it may be of interest: this paper (see Theorem 1) shows that a locally connected, arcwiseconnected finitedimensional topological group is a Lie group.
Of course we also know that a pathconnected Hausdorff space is arcwise connected (see the literature on Peano spaces and the HahnMazurkiewicz theorem, as mentioned for instance here), so we can say that a pathconnected locally connected finitedimensional $T_0$ topological group is a Lie group. That gets you pretty close, if not a cigar.

$\begingroup$ Theorem 3 here: pnas.org/content/36/11/663, says that any locally arcwiseconnected group of finite dimension is locally compact. On the other hand I don´t know any metrizable locally connected topological space (group or otherwise) which is not locally pathconnected. $\endgroup$ Feb 11 '16 at 20:03

$\begingroup$ Of course :P ... and locally compact? $\endgroup$ Feb 11 '16 at 20:22