2
$\begingroup$

I define the simplex by $C=C(x_1,\dots,x_n)= \{\sum_{i=1}^{n} \lambda_i x_i : \lambda_i \ge 0 \wedge \sum_{i}^{n} \lambda_i = 1\}$. Now assume that $x_1,\dots,x_n$ are linearly independent in some Hilbert space $H$. I know what to show constructively, i.e. without the law of excluded middle, that the simplex is closed. Closed means that every cluster point is in $C$.

Let $a$ be a cluster point of $C$. Hence there exists some sequence $b_n$ in $C$ such that $\lim_{m \rightarrow \infty} b_m = a$.

(1) Since $x_1,\dots,x_n$ are linearly, I may assume that this converging sequence looks like $\lambda_1^{m}x_1 + \dots +\lambda_n^{m}x_n$ where for every $m \in \mathbb{N}$ we have $\sum_{i}^{n} \lambda_i^{m} = 1$ and $\lambda_i^{m} \ge 0$. Hence we get that $a = \sum_{i}^{n}\lambda_i x_i$ where $\lambda_i$ is the limit of $\lambda_i^{m}$. This implies that $a$ is in $C$, hence $C$ is closed.

My problem is step (1). Am I allowed to assume that any converging sequence looks like this? I think I am, but I am not quite sure how to justify this. It has to be closely related to the fact that $x_1,\dots,x_n$ are linearly independent.

In my opinion the reasoning goes like this: For every $m \in \mathbb{N}$ the converging sequence $b_m$ is in the simplex, hence the linear independence of the nodes implies the unique representation $\lambda_1^{m}x_1 + \dots +\lambda_n^{m}x_n$.

$\endgroup$
1
$\begingroup$

First, I think it would be faster to use the property that intersection of a finite number of closed sets is closed itself and the fact that hyperplanes and closed half-spaces are closed.

If you want to stick with your demonstration, if think you have to use linearity properties to make it correct.

As any point in $C$ is, by definition, defined as a linear combination of your $x_i$ points, you are sure the $\lambda_i^m$ exist for all point $b_m$. The uniqueness of those coordinates is due to the linear independence of the $x_i$.

The functions $f_i(b_m)=\lambda_i^m$ is linear as well as the sum of those functions : $f(b_m)=\sum(\lambda_i^m)$, as well as $g(\lambda_1,\lambda_2,...)= \sum{x_i\lambda_i}$.

if $b_m$ converges toward $a$ then, due to linearity, $f(b_m)$ converges toward $f(a)$. But, you can notice that $\forall m , f(b_m)=1$ so converges toward 1. So $f(a)=1$ (1)

As $f_i(b_m)$ converge toward $f_i(a)$ and are always >=0 then $\lambda_i >=0$ (2)

As g is linear, convergence of $(\lambda_1^m,\lambda_2^m,...)$ toward $(\lambda_1,\lambda_2,...)$ implies that the limit of $b_m$ a=$\sum{x_i\lambda_i}$. (3)

1, 2 and 3 prove that $a \in C$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy