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

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?

• There is something wrong with the inequality on $f$. Take $t=0$ for example. May 14, 2013 at 9:54
• Thanks Sergei! My obtuse (and likely incorrect) definition should be ignored. The reader should use any standard definition of length space. For example, for all x and y in X and all e>0, there exists a rectifiable path from x to y whose length is less then d(x,y)+e. May 14, 2013 at 21:28
• @Paul: I have incorporated your comment into the statement of the question, so as to hopefully correct the definition. Jul 8, 2013 at 21:54

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.

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.

• Is there a rectifiable path to $x$ passing through all $c_n$? Apr 4, 2021 at 4:58
• I don't think so; I think if you draw consecutive segments of points approaching $x$ from the left (WLOG) with increasing $x$-coordinates and decreasing $y$-coordinates in the plane, then their angles should have to be bounded away from zero. Here the topological invariance should be due to $c_n$ being on the boundary of a neighborhood (a 'tangent disc topology circle'), and this boundary also belonging to a 'Euclidean topology circle', and they agree except at $x$. Apr 4, 2021 at 6:48
• Oops, that probably needs to happen for every subsequence of the $c_n$, or at least in some limiting sense. Apr 4, 2021 at 7:04
• In fact I am confused now. In the hyperbolic metric, is not $x$ an ideal point? I mean, at infinite distance from any internal point of the open upper halfplane? Apr 4, 2021 at 7:23
• I don't think so; otherwise the space would be disconnected. When I used "$\mathbb{H}$" I mean with the metric induced from the usual Euclidean one (though I suppose that theoretically it shouldn't matter, but then the arguments will be different). Sadly I'm not an expert on this space, either xD Someone will probably point out some technical issue(s) to deal with, then maybe we will have a clearer picture after sorting through them. Apr 4, 2021 at 7:49