# Proof that mean-variance opportunity set is closed

In the book Financial Economics (2010) by Hens and Rieger, on page 101 we find the following Lemma 3.1: If we have finitely many assets, the minimum-variance opportunity set is closed and connected.

We have: K number of assets, a sequence of points $$x_n = (\mu_n, \sigma_n) (n=1,2,...)$$ in the opportunity set with $$x_n\rightarrow x= (\mu,\sigma)$$, each $$x_n$$ corrsponds to a portfolio characterized by asset weights $$\lambda_1^n,...,\lambda_K^n$$ with $$\lambda_k^n\geq 0$$ for all $$k = 1,...,K$$ and $$\sum_{k=1}^{K}\lambda_k^n=1$$

The proof that the mean-variance-opportunity set is closed, it is stated that the vector of asset weights $$\lambda = (\lambda_1^n, ... ,\lambda_k^n)$$ is for all $$n\in \mathbb{N}$$ in a compact set. Does the proof infer that since $$\lambda$$ is compact (ie closed and bounded), the opportunity set also must be closed? Why does an arbitrary point $$x_n\rightarrow x= (\mu,\sigma)$$?

So the setting is as follows. We have a map $$F: A \to \mathbb{R^2}, \lambda \to (\mu_\lambda, \sigma_\lambda)$$ where $$\mu_\lambda, \sigma_\lambda$$ are the mean and variance of the portfolio with weights $$\lambda$$ and $$A=\{\lambda_1, \ldots, \lambda_K|\lambda_i \geq 0 \text{ and } \sum_i \lambda_i=1\}$$. Consider a sequence $$(\mu_n, \sigma_n)$$ converging to $$(\mu,\sigma)$$. Since they are result of a portfolio they are in the range of $$F$$. Take a point $$\lambda^n \in F^{-1}((\mu_n, \sigma_n))$$. Then $$(\lambda^n)_{n}$$ is a sequence in $$A$$. Since $$A$$ is bounded there exists a converging subsequence $$(\lambda^{n_i})_i$$. Since $$A$$ is closed the limit of this sequence $$\tilde{\lambda}$$ is in $$A$$.Since $$F$$ is continuous we have
$$F(\tilde{\lambda})= F(\lim\limits_{i\to \infty} \lambda^{n_i}) =\lim\limits_{i\to \infty} F(\lambda^{n_i}) = \lim\limits_{i\to \infty} (\mu_{n_i}, \sigma_{n_i}) =\lim\limits_{n\to \infty} (\mu_{n}, \sigma_{n}) = (\mu,\sigma)$$
So we have found a portfolio with weights $$\tilde{\lambda}$$ that has the mean and variance of the limit. Thus the mean-variance opportunity set is closed