Is every connected metrizable locally path connected space a length space?

Question

Does every connected metrizable locally path connected topological space $X$ admit a compatible metric $d$ so that $(X,d)$ is a length space?

(Edit to correct definition: Recall that a metric space $(X,d)$ is a length space if for every $x$ and $y$ in $X$ and every $e>0$, there exists a rectifiable path from $x$ to $y$ whose length is less then $d(x,y)+e$.)

The answer is certainly yes for Peano continua, but this is not a trivial fact.

More generally the answer is apparently yes for such locally compact spaces, but local compactness is certainly not necessary: for example, familiar Hilbert space is a length space.

Do the above claims survive without local compactness?

Answer 1

This is not an answer but a long comment with references for the locally compact case. It is proved in [Tominaga-Tanaka, Convexification of locally connected generalized continua. J. Sci. Hiroshima Univ. Ser. A. 19 (1955), 301–306] that

Theorem. Every locally compact, connected, locally connected, separable metrizable space admits a complete length metric.

The proof is similar to the one in the compact case which is due to Bing and Moise (independently). A relative (stronger) version is due to Doodley [Further extending a complete convex metric. Proc. Amer. Math. Soc. 40 (1973), 590–596] and [Extending a complete convex metric. Proc. Amer. Math. Soc. 34 (1972), 553–559] whose electronic versions can be easily found on the web.

Answer 2

I think the answer is no. There is a metrizable version of the Tangent Disc Topology, namely where instead of extending the Euclidean topology in the upper half-plane to all of $\mathbb{R}$ you extend it to a countable subset of $\mathbb{R}$. This will be locally path-connected and connected (see Counterexamples in Topology by Steen/Seebach), but not locally compact, and as far as I can tell there will be no equivalent metric which is a length space.

To see this, pick just a single point $x \in \mathbb{R}$ and adjoin to the Euclidean topology on $\mathbb{H}^2$ neighborhoods of the form $D \cup \lbrace x \rbrace$, where $D$ is any open disc in $\mathbb{H}^2$ tangent to $x$, creating a topological space $(\mathbb{H}^2 \cup \lbrace x \rbrace, \tau)$. Then if $C$ is the boundary of one of these tangent discs, points along $C$ do not converge to $x$ in $\tau$. However, if $V$ is a vertical segment with endpoint $x$, then points along $V$ do converge to $x$, and clearly $V$ is a length-minimizing path.

If $c_n, v_n$ are points with the same $y$-coordinate along $C$ and $V$ respectively (assume all the $c_n$ are either to the left, or to the right, of $V$) with $v_n \rightarrow x$, then $|c_n - v_n| \rightarrow 0$. Notice that the lengths of paths from $c_n$ to $x$ converges to zero (by triangle inequality and moving horizontally, then vertically), yet each $c_n$ lies outside the neighborhood $D \cup \lbrace x \rbrace$, impossible in a length space.

This last bit is the 'topological agreement' criterion, I'm not sure what it's normally called; see p. 27, no. (4) in A Course in Metric Geometry by Burago/Burago/Ivanov:

If $x \in X$ and $U$ is a nbhd of $x$, then $\inf \ell(x, c) > 0$ for $c \in U^c$.

Especially there will be no equivalent metric that's a length space, since an intrinsic length metric is topologically unique and will have the same defect.

Hopefully this argument is right; even if there's a small error, I wanted to bump this question. I'm not an expert in length spaces, perhaps replacing $x$ with $\mathbb{Q}$ is necessary, but those details seemed quite complicated.