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While studying Metric Spaces, I came across the following exercise:

Problem:

Let $X \subset \mathbb{R}^{2}$ such that the Euclidean Metric induces in $X$ the discrete metric. Prove that $X$ has at most three elements.

Where am I struggling?

I just cannot understand how it is possible for $X$ equipped with the standard Euclidean metric to induce in $X$ the discrete metric. With that in mind, I went back and checked again what it means to be an induced metric and it goes like this:

Suppose that $(X, d)$ is a metric space and that $A$ is a non-empty subset of $X .$ Let $d_{A}: A \times A \rightarrow \mathbb{R}$ be the restriction of $d$ to $A \times A\left(\right.$ recall that this means $d_{A}(x, y)=d(x, y)$ for any $x, y$ in $A$ ). The metric space axioms hold for $d_{A}$ since they hold for $d$. The metric space $\left(A, d_{A}\right)$ is called a (metric) subspace of $(X, d)$ and $d_{A}$ is called the metric on A induced by $d$.

How can $X$ equipped with the Euclidean Metric induce the discrete metric in itself? Is the problem wrong? If so, what would be the correct way to restate it?

Thanks In advance, Lucas

Lucas
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    It is saying that when you restrict the usual metric $d:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ to $X\times X$, all of the points in $X$ are either a distance $1$ from each other, or they are a distance $0$ from each other, and in that case, they are the same point. The largest type of configuration which satisfies these conditions is three points at the vertices of an equilateral triangle with side length $1$. – C Squared Apr 26 '21 at 19:50
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    Interesting! Just to confirm: So you are still measuring distance in $X \times X$ using the euclidean metric, right? However, it turns out that in that subset the points satisfy what you just said. Is that correct? – Lucas Apr 26 '21 at 20:00
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    yes thats correct. the set is either empty, in which case its vacuously true, it has one element, it has two elements separated by distance $1$, or its has three elements in the configuration described above – C Squared Apr 26 '21 at 20:00
  • https://math.stackexchange.com/a/4128513/803927 see here for a more general proof – C Squared May 05 '21 at 20:19

1 Answers1

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Suppose $X\subset\mathbb{R}^2$ is induced with the discrete metric, i.e., $d:X\times X\to\mathbb{R}$ where $$d(x,y) = \begin{cases}0&\text{ if }x=y\\1&\text{ if else }\end{cases} $$ when the Euclidean metric is restricted to $X\times X$.

Suppose, for contradiction, that $|X|\geq 4$. Consider any three points $a,b,c\in X$. We must have that $$d(a,b)=d(a,c)=d(b,c)=1$$

We can immediately conclude that these three points form an equilateral triangle $\triangle abc$.

For any fourth, distinct point $e=(e_1,e_2)\in X$, we also require that $$d(a,e)=d(b,e)=d(c,e)=1$$

so we must have, for example, equilateral triangle $\triangle abe$. But since $e$ is distinct from point $c$, then $\triangle abe$ is $\triangle abc$ reflected over the line segment $\overline{ab}$. But then $d(c,e)=\sqrt{3}>1=d(c,e)$, which is a contradiction.

This proof assumes some knowledge of equilateral triangles and may not be completely rigorous, but it expresses the main idea behind a proof of this statement.

C Squared
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  • Incredible! What would you say that it is not 100% rigorous the proof? I think it would be a nice exercise for me to try and fill those gaps! – Lucas Apr 26 '21 at 21:47
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    @Lucas i think it would also be a good idea for me to fill in those gaps, as i don’t have a full idea on how to either. and thank you, glad i could help – C Squared Apr 26 '21 at 21:55
  • Nice! So are you going to try to fill the gaps and come back here later? – Lucas Apr 26 '21 at 22:41
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    @Lucas If I do end up filling in the gaps, I will edit my answer. If you happen to fill them in yourself, post them as an answer. I'd love to see how you filled them in. – C Squared Apr 26 '21 at 22:43
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    Quick note: Maybe a good way to try and fill the gaps is to think of the more general case where $X$ is a subset of $\mathbb{R}^{n}$. Probably X must have at most $n - 1$ elements! – Lucas Apr 26 '21 at 22:50
  • That is probably going to make us think in a more general way and leave the geometrical reasoning behind – Lucas Apr 26 '21 at 22:53
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    @Lucas in that case, the upper bound on $|X|$ increases. For example, in $\mathbb{R}^1$, $|X|$ can be at most two. In $\mathbb{R}^3$, you would need a tetrahedron, so $|X|$ is at most four. I think that lends it self to a proof by induction, and i guess for $X\subset \mathbb{R}^n$, $|X|$ should be at most $n+1$. – C Squared Apr 26 '21 at 22:54
  • You are completely right – Lucas Apr 26 '21 at 22:56
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    Let us continue this discussion in chat.Edit: My bad, I was going to suggest that if you want to work on this together, we could move into a chat. – C Squared Apr 26 '21 at 22:58