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It is well-known, that a complete metric space, where any two points have a midpoints ($\forall x,y~ \exists z:~d(x,z)=d(y,z)=\frac{d(x,y)}{2}$) is strictly intrinsic, in the sense that any $x,y$ can be joined by a path of length $d(x,y)$.

Also, completeness and existence of $\varepsilon$-midpoints ($\forall x,y~ \exists z:~|d(x,z)-\frac{d(x,y)}{2}|+|d(y,z)-\frac{d(x,y)}{2}|\le \varepsilon$) for any $\varepsilon$ imply the space being intrinsic, i.e. any $x,y$ can be joined by a path of length $d(x,y)+\varepsilon$ for any $\varepsilon$.

Can we replace completeness by something else?

It seems that $\varepsilon$-midpoints are completely useless without completeness (a disc without one radius is an example).

On the other hand, it seems that the existence of midpoints is a rather strong condition. So far I have the following examples:

$\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced metric is a space with midpoints, which is not strictly intrinsic, but still it is intrinsic.

$\{(x,y)\in \mathbb{R}^{2}, x-y\in\mathbb{Q}\}\bigcup\{x=y\}$ with the induced $L_{\infty}$ metric is a connected space with midpoints but not intrinsic.

However these space are bad: neither of them is locally compact, and the latter is not even locally path connected (local compactness implies local completeness and so local strict intrinsicness, and so local path connectedness).

Now let us assume that in connected locally compact (or merely locally path connected) metric space $X$ for any two points there is a midpoint. Is it true that this space is (strictly) intrinsic?

There were numerous edits, because I have added two counterexamples, one of which later proved to be completely incorrect, and another one had a mistake.

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  • $\begingroup$ what is the $L_{\infty}$ metric? do you mean $\parallel \; \parallel_{\infty}$ on the plane? If yes, why your example in locally compact(According to your edit) $\endgroup$ – Ali Taghavi Jan 12 '17 at 19:18
  • $\begingroup$ you have identified the metric correctly. The locally compact example is such because it is just a closed rectangle minus a point (the topology in this metric is the same as with the Euclidean); the last example is not locally compact, and there was a slight error in it (which I fixed now). $\endgroup$ – erz Jan 13 '17 at 5:42
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    $\begingroup$ Why is $[-1,1]\times [0,1]\setminus \{(0,0)\}$ with the $L^\infty$ norm not strictly intrinsic? The curves that are piecewise affine with slope at most one have length equal to the distance between their endpoints. Did you want to use either a rotated rectangle or the $L^1$ metric? $\endgroup$ – Benoît Kloeckner Jan 13 '17 at 8:39
  • $\begingroup$ Yes, you are right, the example in invalid. $\endgroup$ – erz Jan 13 '17 at 22:33
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The answer is no.

There is a locally compact, connected metric space, where any two points have a midpoint, that is not intrinsic.

The example I have in mind looks a little bit like this:

enter image description here

[The little circles are meant to indicate holes in my rectangles (not little arcs).]

The metric for this space is the "modified taxicab metric", by which I mean the metric that measures the shortest path between two points, but where I allow paths that jump over holes. (One could imagine a taxicab in New Orleans, which by necessity has to learn to ignore potholes and just run right over them.)

I say that my example looks "a little bit" like this because I have only drawn two iterations of the construction, but I would like to imagine that we have done infinitely many iterations.

A little more precisely, let's call the big rectangle $B_0$. Notice that its bottom left corner has been removed, making a hole. Let's consider this holey rectangle to be stage $0$ of our construction. To get stage $1$, we build an infinite sequence of "bridges" accross the hole, which we'll call $B_1^n$, $n \in \mathbb N$, and remove a point from each bridge, making more holes. The exact manner in which the bridges are constructed is irrelevant, provided we do two things:

(1) Letting $p$ and $q$ denote the points where a bridge $B_1^n$ meets $B_0$, design the bridge so that the distance from $p$ to $q$ is the same whether we take the shortest "path" (path with a hole in it) through $B_0$, or if we choose to take the bridge instead.

(2) With $p$ and $q$ as before, choose the hole in our bridge so that the distance from $p$ (or $q$) to the hole in $B_0$ is different from its distance to the hole in $B_1^n$. (This is true the way I've drawn it, for example, because I used rectangles that are not squares.)

This completes stage $1$ of the construction. Further stages work the same way: for each hole from stage $n$, we construct an infinite sequence of holey bridges at stage $n+1$ (and we do it in such a way that the above two properties are satisfied). Ultimately, the space I have in mind is the union of all these stages.

Let's check that this space has the properties you require.

It is locally compact.

Every point in our space has a neighborhood that is homeomorphic to either an interval or a triod (the letter T). This makes the space locally compact.

It is connected (even better: it is path connected).

A straightforward induction on $n$ shows that if we add a point to our space at stage $n$ of our construction, then there is a path from that point to $B_0$. Since $B_0$ is path connected (because I didn't remove the top right corner!), it follows that the whole space is path connected.

Any two points have a midpoint.

Let $p$ and $q$ be any two points in our space, both occuring in stage $n$ of the construction. If we temporarily ignore the holes in our space, then there is a (not necessarily unique) shortest path connecting $p$ to $q$, and it is contained in stage $n$ of the construction. Because of the metric we've chosen, the midpoint $r$ of this path should be a midpoint between $p$ and $q$ (the only potential problem being that $r$ might not be in our space -- it could be a hole). If $r$ is not a hole then it is a midpoint for $p$ and $q$. If $r$ is a hole, then we can modify our path between $p$ and $q$ using one of the bridges we built around the hole. Because of our first bridge-building condition, this modified path has the same length as the original one. Therefore its midpoint, $s$, should be a midpoint for $p$ and $q$ as well. Our second bridge-building condition guarantees that $s$ is not a hole.

It is not intrinsic.

Let $p$ be a point in (the interior of) the left edge of $B_0$ and let $q$ be a point in its bottom edge. Any path connecting $p$ and $q$ must go through the top right corner of $B_0$ (Why? Because removing this point disconnects the whole space!). It is clear that any such path is significantly longer than the distance from $p$ to $q$. Thus this space is not intrinsic.

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  • $\begingroup$ That is a fantastic example! Please allow me to suggest a partial case, a more concrete construction, which is easier to visualize: Start with $B_0$ and a single holey bridge, which is based exactly in the middles of the corresponding sides. Now we have a rectangle with holes in the upper right and lower left corners. Each time we have such an object, divide it with a vertical and a horizontal "middle" lines in 4 smaller rectangles, and place a hole where these lines cross. Hence we obtain 2 "normal" rectangles and 2 rectangles with holes. Ultimately we will have something like a fractal grid. $\endgroup$ – erz Jan 19 '17 at 21:18
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    $\begingroup$ @erz: If I'm understanding your idea correctly, then I don't think it works. The problem is that you will end up with holes that have points both directly above and directly below them. These pairs of points have a unique shortest path connecting them (and it crosses the hole. Their only chance at having a midpoint is if it's the midpoint of that path. If the points are equidistant from the hole, then you're out of luck. $\endgroup$ – Will Brian Jan 20 '17 at 15:48
  • $\begingroup$ yes, you are right, sorry. Something like this perhaps? postimg.org/image/3q0vavdhp $\endgroup$ – erz Jan 20 '17 at 22:05
  • $\begingroup$ That looks like it should work -- and I like the symmetry of your version! $\endgroup$ – Will Brian Jan 20 '17 at 22:08

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