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I have read different posts about this subject, all focused on very specific assumptions (compactness, in $\mathbb{R}^N$, etc.). My question aims at a unifying goal.

Let $(X,d_X)$ and $(Y,d_Y)$ metric spaces and $\sigma:X\to Y$ a distance preserving map (isometry). Is it true that $\sigma$ is surjective if (maybe, and only if) $$\mathrm{diam}_X(X)\geq \mathrm{diam}_Y(Y) \, .$$

Update. [the answers below are illuminating]

Add the condition on $\sigma$ that there exists $x\in X$ such that balls in $X$ centered at $x$ are sent by $\sigma$ to balls in $Y$ centered at $\sigma(x)$.

Update 2. Even this condition is not sufficient as the example (of @Dry Bones) in the comments show. Any new guess on how to fix the hypotheses is welcome!

Kosh
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2 Answers2

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No, even if you are working in Hilbert space. Let $H$ be a separable Hilbert space. Fix an orthonormal base $\{e_i\mid i\in\mathbb{N}\}$. Define a unilateral shift operator $U:H\rightarrow H$ by $U(e_i)=e_{i+1}$ (of course, extended by linearity and continuity), then $U$ is an isometry but not surjective.

Danny Pak-Keung Chan
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Complementarily to Danny Pak-Keung Chan's answer, even if the diameters are finite, it doesn't work. For example, the inclusion of a one-dimensional interval in a two-dimensional disk with the usual metrics induced from the plane is an isometry, satisfies the prescribed inequality, but isn't surjective at all.

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Here's another example that must be dealt with in order to find a valid version of your statement.

$X$ is the blue 'ellipse', $x$ is the point at the center of $X$, and $Y=X\cup\{p\}$, where $p$ is the isolated blue point on the right. The isometry is the inclusion $X\rightarrow Y$. All the aforementioned conditions hold, even the stronger '$\sigma$ takes closed balls centered at $x$ to closed balls centered at $\sigma(x)$', but the isometry isn't surjective.

$\hskip4.5cm$no description

Dry Bones
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  • I think your guess is quite interesting, though. Maybe it's possible to adapt it somehow in order to make it true. – Dry Bones Aug 13 '20 at 00:38
  • Actually, I think that your example is simple and illuminating. I think that the condition to add is that there exists $x\in X$ such that balls in $X$ centered at $x$ are sent by $\sigma$ to balls in $Y$ centered at $f(x)$ (something I was giving for granted when I was picturing a possible proof in my mind). – Kosh Aug 13 '20 at 10:32
  • I updated the question – Kosh Aug 13 '20 at 10:48
  • Hm, that's a nice attempt, but I believe you are counting too much on some kind of similarity between metric spaces and, say, topological manifolds, which isn't the case. For instance, you can adapt the example above, but instead of an interval, take the union of an interval and a small disk around the origin. It's an odder subset, but still inherits a metric from the plane. Then such a point as you described exists, but the problem is more or less the same! =] – Dry Bones Aug 13 '20 at 14:46
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    I don't see why the problem is still there in your last example. The images of the balls around the origin are no more mapped to balls when the size increases. – Kosh Aug 13 '20 at 15:59
  • Yeah, you're right! I mistook the ball's quantifier for mere existence. Let me think some more about it! – Dry Bones Aug 13 '20 at 16:36
  • I'm pretty convinced the statement is now true. Would you like me to write down a proof or would you rather try and do it yourself? I believe the Q&A mode is also encouraged here. =] – Dry Bones Aug 13 '20 at 16:59
  • You can write the proof. I already had in mind the proof when I was asking, but with the assumption that I was giving for granted and that now I added. Your example shows that this further assumption is "essential". Please go ahead with the proof so that I can accept the answer ;-) – Kosh Aug 13 '20 at 17:02
  • Wait, I think there's still a small issue here. Consider for example the inclusion of an open interval in a closed interval in the line. Then all assumptions hold, but the inclusion isn't surjective... Maybe the hypothesis on the diameters should be strengthened to ${\rm diam}_X(X)>{\rm diam}_Y(Y)$ ? Or maybe some extra condition should be imposed on the action of $\sigma$ on balls? – Dry Bones Aug 13 '20 at 17:25
  • I see your point. I think that one has to assume that the diameter of $X$ is reached. This is in agreement with the assumptions people usually make on $X$ being compact. – Kosh Aug 13 '20 at 17:42
  • Ok I was trying to write down a proof. I see that it is less easy I was thinking about. I will give another shot later. Thanks for your comments! – Kosh Aug 13 '20 at 18:09