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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, while $\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ withit seems that the induced metric is a space withexistence of midpoints, which is not strictly intrinsica rather strong condition. A better exampleSo far I have the following examples: $[-1,1]\times[0,1]\backslash\{0\}$

$\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced $L_{\infty}$ metric is even a locally compact space with midpoints, which is not strictly intrinsic. These spaces however are, but still it is intrinsic.

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

However this is a badthese space are bad: it is 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, nor locally compactand 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?

Local compactness implies local completeness and so local strict intrinsicnessThere were numerous edits, and so local path connectedness but I don't know how to proceed.

EDIT:because I have added an exampletwo counterexamples, one of a locally compact connected locally path connected space with midpoints but not strictly intrinsicwhich later proved to be completely incorrect, and an example ofanother one had a connected space with midpoints but not intrinsic. These counterexamples refined the original two questions into their "maximum"mistake.

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), while $\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced metric is a space with midpoints, which is not strictly intrinsic. A better example: $[-1,1]\times[0,1]\backslash\{0\}$ with the induced $L_{\infty}$ metric is even a locally compact space with midpoints, which is not strictly intrinsic. These spaces however are intrinsic.

On the other hand the space $\{(x,y)\in \mathbb{R}^{2}, x-y\in\mathbb{Q}\}\bigcup\{x=y\}$ with the induced $L_{\infty}$ metric is a connected space with midpoints but not intrinsic. However this is a bad space: it is neither locally path connected, nor locally compact.

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 intrinsic?

Local compactness implies local completeness and so local strict intrinsicness, and so local path connectedness but I don't know how to proceed.

EDIT: I have added an example of a locally compact connected locally path connected space with midpoints but not strictly intrinsic and an example of a connected space with midpoints but not intrinsic. These counterexamples refined the original two questions into their "maximum".

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.

fixed one example
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erz
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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), while $\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced metric is a space with midpoints, which is not strictly intrinsic. A better example: $[-1,1]\times[0,1]\backslash\{0\}$ with the induced $L_{\infty}$ metric is even a locally compact space with midpoints, which is not strictly intrinsic. These spaces however are intrinsic.

On the other hand the space $\{(x,y)\in \mathbb{R}^{2}, x-y\in\mathbb{Q}\}\bigcup\mathbb{R}$$\{(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 this is a bad space: it is neither locally path connected, nor locally compact.

Now let us assume that in connected locally path-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 intrinsic?

Local compactness implies local completeness and so local strict intrinsicness, and so local path connectedness but I don't know how to proceed.

EDIT: I have added an example of a locally compact connected locally path connected space with midpoints but not strictly intrinsic and an example of a connected space with midpoints but not intrinsic. These counterexamples refined the original two questions into their "maximum".

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), while $\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced metric is a space with midpoints, which is not strictly intrinsic. A better example: $[-1,1]\times[0,1]\backslash\{0\}$ with the induced $L_{\infty}$ metric is even a locally compact space with midpoints, which is not strictly intrinsic. These spaces however are intrinsic.

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

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

Local compactness implies local completeness and so local strict intrinsicness, but I don't know how to proceed.

EDIT: I have added an example of a locally compact connected locally path connected space with midpoints but not strictly intrinsic and an example of a connected space with midpoints but not intrinsic. These counterexamples refined the original two questions into their "maximum".

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), while $\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced metric is a space with midpoints, which is not strictly intrinsic. A better example: $[-1,1]\times[0,1]\backslash\{0\}$ with the induced $L_{\infty}$ metric is even a locally compact space with midpoints, which is not strictly intrinsic. These spaces however are intrinsic.

On the other hand the space $\{(x,y)\in \mathbb{R}^{2}, x-y\in\mathbb{Q}\}\bigcup\{x=y\}$ with the induced $L_{\infty}$ metric is a connected space with midpoints but not intrinsic. However this is a bad space: it is neither locally path connected, nor locally compact.

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 intrinsic?

Local compactness implies local completeness and so local strict intrinsicness, and so local path connectedness but I don't know how to proceed.

EDIT: I have added an example of a locally compact connected locally path connected space with midpoints but not strictly intrinsic and an example of a connected space with midpoints but not intrinsic. These counterexamples refined the original two questions into their "maximum".

added a counterexample and changed the question accordingly
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erz
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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), while $\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced metric is a space with midpoints, which is not strictly intrinsic. A better example: $[-1,1]\times[0,1]\backslash\{0\}$ with the induced $L_{\infty}$ metric is even a locally compact space with midpoints, which is not strictly intrinsic. These spaces however are intrinsic.

On the other hand the space $\{(x,y)\in \mathbb{R}^{2}, x+y\bbox{~or~}x-y\in\mathbb{Q}\}$$\{(x,y)\in \mathbb{R}^{2}, x-y\in\mathbb{Q}\}\bigcup\mathbb{R}$ with the induced $L_{\infty}$ metric is a connected locally path connected space with midpoints but not intrinsic.

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

Local compactness implies local completeness and so local strict intrinsicness, but I don't know how to proceed.

EDIT: I have added an example of a locally compact connected locally path connected space with midpoints but not strictly intrinsic and an example of a connected locally path connected space with midpoints but not intrinsic. These counterexamples refined the original two questions into their "maximum".

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), while $\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced metric is a space with midpoints, which is not strictly intrinsic. A better example: $[-1,1]\times[0,1]\backslash\{0\}$ with the induced $L_{\infty}$ metric is even a locally compact space with midpoints, which is not strictly intrinsic. These spaces however are intrinsic.

On the other hand the space $\{(x,y)\in \mathbb{R}^{2}, x+y\bbox{~or~}x-y\in\mathbb{Q}\}$ with the induced $L_{\infty}$ metric is a connected locally path connected space with midpoints but not intrinsic.

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

Local compactness implies local completeness and so local strict intrinsicness, but I don't know how to proceed.

EDIT: I have added an example of a locally compact connected locally path connected space with midpoints but not strictly intrinsic and an example of a connected locally path connected space with midpoints but not intrinsic. These counterexamples refined the original two questions into their "maximum".

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), while $\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced metric is a space with midpoints, which is not strictly intrinsic. A better example: $[-1,1]\times[0,1]\backslash\{0\}$ with the induced $L_{\infty}$ metric is even a locally compact space with midpoints, which is not strictly intrinsic. These spaces however are intrinsic.

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

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

Local compactness implies local completeness and so local strict intrinsicness, but I don't know how to proceed.

EDIT: I have added an example of a locally compact connected locally path connected space with midpoints but not strictly intrinsic and an example of a connected space with midpoints but not intrinsic. These counterexamples refined the original two questions into their "maximum".

added a counterexample and changed the question accordingly
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Found a counterexample to the second question. Modified accordingly.
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