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I am taking a course on nonlinear optimization and came across this definition of a tangent cone in my lectures:

Let $\emptyset \neq M \subseteq \mathbb{R}^n$ and $x \in M$. Then the tangential cone of $M$ in $x$ is defined as: $T(M,x) = \{d \in \mathbb{R}^n : \exists \eta_k > 0, x_k \in M, \lim_{k \to \infty} x_k = x , \lim_{k \to \infty} \eta_k(x_k - x) = d \}$

I was then asked to show that $T(M,x)$ is always closed but I'm still struggling to see why. So far I have tried the following approach:

Let $(d_k)$ be a sequence within $T(M,x)$ such that $\lim_{k \to \infty} d_k = d$. I will attempt to show that $d \in T(M,x)$. Since $d_k \in T(M,x)$, for each $k$ we have two sequences $\eta_{k,n}$ and $x_{k,n}$ such that for all $k$: $$\lim_{n \to \infty} x_{k,n} = x$$ $$\lim_{n \to \infty} \eta_{k,n}(x_{k,n} - x) = d_k$$

Hence $\lim_{k \to \infty} \lim_{n \to \infty} \eta_{k,n}(x_{k,n} - x) = d$. Now this looks tantalizingly close to the definition of a vector in $T(M,x)$, if I can just "extract out" some sequence $\eta_t$ from $\eta_{k,n}$ and $x_t$ from $x_{k,n}$ and make the above limit works for $\eta_t$ and $x_t$, i.e: $$\lim_{t \to \infty} x_t = x$$ $$\lim_{t \to \infty} \eta_t(x_t - x) = d$$

However I am still stuck here. Is this a correct approach ? If so, how should I proceed ? If not, what should I do instead ? Thank you.

zenman01
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1 Answers1

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Yes, you're following the correct approach. The next step is to use the $\epsilon,K_\epsilon$ and $\epsilon,N_\epsilon$ definitions of the limits. For each $\epsilon>0$, you can choose some $K_\epsilon$ such that $\|d_k-d\|\le \epsilon$ for all $k \ge K_\epsilon$. For the next step, you choose an arbitrary $k \ge K_\epsilon$ - you might as well choose $k:=K_\epsilon$. For the same $\epsilon$ as before, you can also choose some $N_\epsilon$ such that $\|\eta_{K_\epsilon,n}(x_{K_\epsilon,n}-x)-d_{K_\epsilon}\|\le \epsilon$, for all $n \ge N_\epsilon$. Have a go at finishing the proof from here on your own - good luck!

1Rock
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