Is the infinite-dimensional simplex convex?

Question

Let

$$S = \left\{ x \in \mathbb{R}_{\geq 0}^{\mathbb{N}^+} \mid \displaystyle\sum_{i=1}^\infty x_i = 1 \right\}$$

Is $S$ convex?

Answer 1

Recall the definition of a convex set: a subset $A$ of a real or complex vector space $V$ is said to be convex if for every $x,y\in A$ and every $\lambda\in[0,1]$ we have $\lambda x+(1-\lambda)y\in A$ or equivalently if for every $\lambda\in[0,1]$ we have $\lambda A+(1-\lambda)A\subset A$. In this problem, the vector space is $\mathbb R^{\mathbb Z_+}$ of all real sequences.

Given $x=\{x_n\}$ and $y=\{y_n\}$ in $S$ and $\lambda\in[0,1]$, we have $$\sum_{n=1}^\infty\lambda x_n+(1-\lambda)y_n =\lambda\sum_{n=1}^\infty x_n+(1-\lambda)\sum_{n=1}^\infty y_n=\lambda+(1-\lambda)=1$$ (convergence is a non-issue). Hence $\lambda x+(1-\lambda)y\in S$, and $S$ is convex.