Uniqueness of tangent space given local injectivity of orthogonal projection onto it Definition. Let $X\subset \mathbb R^n$ be a locally Euclidean subset. Say it has a tangent space at $p\in X$ if there exists a linear subspace $V\leq \mathbb R^n$ satisfying the following conditions.


*

*$\dim V=\dim_pX$.

*There exists a neighborhood $U\subset X$ of $p$ in $X$ for which
$$\lim_{\substack{h\to 0\\\text{in
}U-p}}\frac{\|\pi_{V^\perp}(h)\|}{\|h\|}=0.$$ Here $U-p$ is the
neighborhood of the origin in $X-p$ obtained by translation of $U$
by $(-p)\in  X\subset\mathbb R^n$.

*The orthogonal projection of $X$ onto $V$ is locally injective at $p$.


Question. Is $V\leq \mathbb R^n$ unique?
The usual argument for the uniqueness of the derivative of a map on an open subset of Euclidean spaces relies on (more or less) convexity of open balls, so is not applicable here.
I am hoping the local injectivity at $p$ of the orthogonal projection will eliminate singular cases by preventing $X$ from "wrapping around $V$" about $p$ as in the cubic cusp (a sort of ramification). Without this condition the other two do not suffice. The example I was given is the surface of revolution of a cusp, e.g $\{(x,y,\sqrt[4]{x^2+y^2})\mid x,y\in \mathbb R\}\subset\mathbb R^3$.
 A: By invariance of domain, the condition $\dim V=\dim_pX$ implies that the local injectivity is locally a bijection to a neighborhood of $p.$ So there are open sets $V_0\subseteq V$ and $X_0\subseteq X$ with $V_0$ containing $0,$ and a function $g:V_0\to X_0$ such that $\pi_V(g(v)-p)=v$ and $g(0)=p.$ By condition 2 we have $g(v)=p+v+o(\|v\|)$ with implicit constants depending only on $X,V$ and $p.$
Assume there's another $V'$ satisfying the same conditions. For each $v\in V$ we have $\lim_{\lambda\to 0^+}\frac{\|\pi_{V'^\perp}(g(\lambda v)-p)\|}{\|g(\lambda v)-p\|}=0.$ Using $g(\lambda v)=p+\lambda v+o(\lambda\|v\|)$ we get $\pi_{V'^\perp}(v)=0,$ so $v\in V'.$ Since $\dim V=\dim V'$ we must have $V=V'.$
A: This is a long comment, rather than a complete answer. It shows, for any subset $X$ near $0$ and any dimension of the tangent(s), that if there are two tangent spaces in the sense of the OP then they must have nonzero intersection.
Lemma. Let $\{0\}\neq X\subseteq\mathbb{R}^n$ be any subset that has the origin as an accumulation point. Let $V$ and $V'$ be two nonzero linear subspaces (of arbitrary dimension) of $\mathbb{R}^n$ satisfying the limit conditions
$$\lim_{\stackrel{h\in X}{h\to 0}}\frac{|\pi_{V^\perp}(h)|}{|h|}=0\quad\quad(\star_V)$$
$$\lim_{\stackrel{h\in X}{h\to 0}}\frac{|\pi_{V'^\perp}(h)|}{|h|}=0\quad\quad(\star_{V'}).$$
Then $V\cap V'\neq\{0\}.$
Proof. Assume by contradiction that $V\cap V'=0$. Write $h\in\mathbb{R}^n$ as $h=(x,y)\in V\oplus V^\perp$. Since the $\ell^2$ (usual Euclidean) norm $|h|$ and the norm given by $|(x,y)|_{V,V^\perp}:=\sqrt{|x|^2+|y|^2}$ are equivalent norms on $\mathbb{R}^n$, the condition 
$$\frac{|y|}{\sqrt{|x|^2+|y|^2}}\to0\;,$$
for $h=(x,y)\to0$ in a given subset having $0$ as accumulation point, is equivalent to the condition $|y|/|x|\to 0$ for $h=(x,y)\to 0$ in the same subset.
For $\varepsilon>0$ let 
$$\mathcal{C}_{V,\varepsilon}:=\{ (x,y)\in\mathbb{R}^n\mid |y|\leq \varepsilon |x|\}.$$
This is a closed cone (invariant under dilations) containing $V$. Likewise, let
$$\mathcal{C}_{V',\varepsilon'}:=\{h\in\mathbb{R}^n\mid |\pi_{V'^{\perp}}(h)|\leq\varepsilon' |\pi_{V'}(h)|\}$$
be the analogous cone containing $V'$.
By the remark on equivalence of norms, condition $(\star_V)$ is easily seen to imply the following: for any given $\varepsilon>0$, there is a $\delta<<1$ such that 
$$X\cap B_\delta\subseteq \mathcal{C}_{V,\varepsilon}.$$
Here $B_\delta$ is the ball or radius $\delta$ in $\mathbb{R}^n$ centered at the origin. In the same way, condition $(\star_{V'})$ implies that, for any given $\varepsilon'>0$, there is a $\delta'<<1$ ensuring 
$$X\cap B_{\delta'}\subseteq \mathcal{C}_{V',\varepsilon'}.$$
In particular, given two $\varepsilon,\varepsilon'$, for $\delta<<1$ we have $X\cap B_\delta\subseteq \mathcal{C}_{V,\varepsilon}\cap\mathcal{C}_{V',\varepsilon'}$.
Now, assume by contradiction that neither $V\subseteq V'$ nor $V'\subseteq V$. In what follows we are going to choose $\varepsilon$ and $\varepsilon'$ such that $\mathcal{C}_{V,\varepsilon}\cap\mathcal{C}_{V',\varepsilon'}=\{0\}$. By the previous remarks, for any $\delta<<1$ we would then have $X\cap B_\delta\subseteq\{0\}$ which is a contradiction because the origin is an accumulation point for $X$.
Let $S_{V'}$ be the (compact) Euclidean unit sphere in $V'$, and let $v_0'$ minimize the quantity $|\pi_{V^\perp}(v')|$ for $v'\in S_{V'}$. Since $V'$ is not contained in $V$, $|\pi_{V^\perp}(v_0')|>0$. If we choose $\varepsilon<|\pi_{V^\perp}(v_0')|/|v_0'|$, we have $v_0'\notin\mathcal{C}_{V,\varepsilon}$ and hence
$$V'\cap\mathcal{C}_{V,\varepsilon}=\{0\}.$$
Let $S_V$ be the Euclidean unit sphere in $V$ centered at the origin, and let $v_0$ minimize the quantity $|\pi_{V'^\perp}(v)|$ on the compact set $S_V\cap\mathcal{C}_{V,\varepsilon}$. Again, $|\pi_{V'^\perp}(v_0)|>0$, and we choose an $\varepsilon'<|\pi_{V'^\perp}(v_0)|/|v_0|$. For this choice of $\varepsilon'$, then, we indeed have
$$\mathcal{C}_{V,\varepsilon}\cap\mathcal{C}_{V',\varepsilon'}=\{0\},$$
and this gives the desired contradiction. $\square$

(Edit: I deleted a corollary because it was not correct)
