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In a book I've been reading (focused on combinatorial optimization), the following definition is given

The dimension $\text{dim} \; X$ of a non empty subset $X \in \mathbb{R}^2$ is defined to be $$ n - \max \left\{rank(A) : A \in \mathbb{R}^{n \times n}, Ax = Ay \; \forall x,y \in X\right\} $$

I've a lack of intuition of what geometrically this means so I've tried a simple case.

Consider $n = 2$, $r > 0$ and $X = \left\{(x,y) \in \mathbb{R}^2 : \sqrt{x^2 + y^2} < r \right\}$

What's the dimension in this case?

I've tried to compute it by myself, but I wasn't neither able to set up the problem. The fact that it has to be $Ax = Ay$ for all $x,y \in X$ implies that $A(x-y) = 0$, so I would suspect is related to some kernel dimension of a linear transformation, however I'm sure that explicitly stating that $Ax = Ay$ must mean something more than that. Any hints of both what the dimension in this context means and how to compute my specific case?

Snippet from the book I'm reading through

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    This notion of dimension is only suitable for convex subsets. – 2017-02-17
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    I don't see the restriction in the definition I have. It talks about polyhedra, but when it comes to such definition I don't see any restriction applied. I reported the definition from the book, but I don't see any reference to convex set, although that seems the context. – 2017-02-17
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    I do not know what book you are reading but the [notions of dimension](https://en.wikipedia.org/wiki/Dimension_theory) standard in topology (and applied, say, to arbitrary subsets of $R^n$) are quite different. – 2017-02-17
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    This is the book, section 3.1. https://books.google.co.uk/books?id=X-klBQAAQBAJ&pg=PA65&dq=combinatorial+optimization&hl=it&sa=X&redir_esc=y#v=onepage&q=combinatorial%20optimization&f=false – 2017-02-17

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As already noted in the comments, this notion of dimension is only usefull for convex bodies. I left as an exercise to you, but it should not be difficult to prove that, with the given definition, $\dim\,X$ is the dimension of the smallest affine subspace of $\mathbb{R}^n$ containing $X$.

Just note that $Ax=Ay$ for all $x,y\in X$ only when there is some vector $b$ such that for all $x\in X$, $Ax=b$, i.e. if and only if $X\subseteq \{x\,|\,Ax=b\}$ where the latter is an affine subspace of dimension $n-\mathrm{rank}\,A$ by linear algebra. The reason why we can pick $A$ square of order $n$ is that any affine subspace of $\mathbb{R}^n$ can be described always with $n$ equations, so we don't need more than that.

For an example, in which this definition fails, consider the non-convex set $C:=\{(x,y)\,|\,x^2+y^2=1\}$ for which the dimension of the smallest affine subspace containing it is $2$ as it is not contained in any line and so $\dim\,C=2$ according to the above definition although we don't expec to define the dimension of a circle as $2$.

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    So in my case such space is the whole $\mathbb{R}^2$, with the implication that the dimension is $0$. Is that right? – 2017-02-17
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    The space is right, but the dimension is two not zero. Note that the matrix you have to take is the zero matrix and so it has zero rank, so the dimension will be 2. – 2017-02-17