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My question is regarding Theorem 1 (page 579) of Rothenberg (1971). It is associated with four assumptions given on the same page. But, I only have a question about a single step of the proof, so I don't think the entire context is necessary.

Definition 3: A parameter point $\alpha^0$ is said to be locally identifiable if there exists an open neighborhood of $\alpha^0$ containing no other $\alpha$ in $A$ which is observationally equivalent
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Assumption I: The structural parameter space $A$ is an open set in $\mathbb{R}^m$
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Under Assumptions I-V given in the paper, we have

Theorem 1: Let $\alpha^0$ be a regular point of the matrix $R(\alpha)$. Then $\alpha^0$ is locally identifiable if and only if $R(\alpha^0)$ is nonsingular.

In the proof for one direction of the iff statement, the author begins by supposing $\alpha^0$ is not locally identifiable. This means ``there exists an infinite sequence of vectors $\{\alpha^1, \alpha^2, ..., \alpha^k, ...\}$ approaching $\alpha^0$ such that $g(y, \alpha^k) = g(y, \alpha^0)$ for all $y$ and each $k$.'' Then, the author later claims that

$$d_i^k = \frac{\alpha_i^k - \alpha_i^0}{|\alpha^k - \alpha^0|}$$

has a limit point $d$ on the unit sphere, and, ``as $\alpha^k \to \alpha^0$, $d^k$ approaches $d$.''

This does not make sense; I agree that there is a limit point but not that there is a limit.. As a counterexample, suppose $A = \mathbb{R}$, $\alpha^0 = 0$, and the infinite sequence was

$$\{\alpha^k \}_{k=1}^\infty = (-1)^n \cdot (1/n)$$

This sequence is valid because it approaches $\alpha^0 = 0$, while we can simply assume that the $g(y, \alpha^k) = g(y, \alpha^0)$ condition holds for all $y,k$ for the sake of contradiction. Using this sequence, we get

$$d^k = \frac{\alpha_i^k - \alpha_i^0}{|\alpha^k - \alpha^0|} = \frac{(-1)^n \cdot (1/n)}{(1/n)} = (-1)^n$$

which has limit points but no limit.

Is there an explanation for why the author assumes such a limit exists? I did not catch any other conditions on $\alpha$ in the paper that could make this step in the proof work.

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