Understanding metric tangent cones.

A fun, albeit quite tame, example of non-uniqueness of tangents is given by a spiral in the plane that converges very quickly in the radial direction. Consider the spiral $$[0,\infty)\to\mathbb{R}^2,\quad t\mapsto (r(t)cos(t),r(t)sin(t))$$ where $r$ is a decaying to zero faster than exponentially, take for instance $r(t)=e^{-e^t}$. When you zoom in on this curve, you will see every half-line starting from 0 as a possible tangent along some sequence of zooming factors.

But this is just an example where uniqueness of tangents fails only at one point, and all the tangents are even diffeomorphic (as they are all half-lines). However in general, if you put no restrictions on your metric space $(X,d)$, basically anything can happen. See for instance the paper "Locally rich compact sets" by Chen and Rossi, where they show the existence of a compact set that has all compact metric spaces with diameter $\leq 1$ as tangents at all points.

In fact, perhaps more surprisingly, they show that this behavior is typical in the sense of the Baire category theorem. That is, the collection of compact metric spaces that do not have all compact metric spaces of diameter $\leq 1$ as tangents at all points form a meager set. So "most" compact metric spaces have horrendous behavior in their tangent cones. This is the same sort of statement as the fact that "most" functions $\mathbb{R}\to\mathbb{R}$ are not differentiable at any point.

The construction of Chen and Rossi of a space with the above property is essentially based on the following simple idea. Suppose we want to construct a metric space $X\subset\mathbb{R}^n$ that has at the point $0\in X\subset\mathbb{R}^n$ two distinct metric spaces $Y_1\subset\mathbb{R}^n$ and $Y_2\subset\mathbb{R}^n$ as tangents. For simplicity, I will work with the assumption that $Y_1$ and $Y_2$ are both contained in $B(0,1)$.

Pick two sequences of scales $\alpha_j\to 0$ and $\beta_j\to 0$ such that they are asymptotically insignificant with respect to one-another. For instance we can choose decreasing sequences such that $\alpha_j>\beta_j>\alpha_{j+1}$ for all $j$ and $$\lim_{j\to\infty}\frac{\beta_j}{\alpha_j} = \lim_{j\to\infty}\frac{\alpha_{j+1}}{\beta_j} = 0.$$ We construct the space $X$ by scaling down the spaces $Y_1$ to all the scales $\alpha_j$ and scaling down the spaces $Y_2$ to all the scales $\beta_j$. That is we define $$X = \bigcup_{j\in\mathbb{N}}\Big(\alpha_jY_1\setminus B(0,\beta_j)\Big)\cup \bigcup_{j\in\mathbb{N}}\Big(\beta_jY_2\setminus B(0,\alpha_{j+1})\Big)\subset\mathbb{R}^n.$$ Now consider the unit balls of tangents of $X$ at $0$. First look at this set at the scale $\alpha_j$, i.e. dilate $X$ by $1/\alpha_j$. Then outside of the ball $B(0,\beta_j/\alpha_j)$ we see exactly the set $Y_1$. But by the construction of our sequences, the balls $B(0,\beta_j/\alpha_j)$ will become smaller and smaller, so along the sequence $\alpha_j$ we see that the unit ball in the tangent cone is the space $Y_1$. A similar argument shows that along the sequence $\beta_j$, the unit ball in the tangent cone is the space $Y_2$.