# Simplex is closed

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 $$\lambda_1^{m}x_1 + \dots +\lambda_n^{m}x_n$$.

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 $$a \in C$$