Let $S$ be a bounded subset of $\mathbb{R}^n$, and define $m = (n+1)^2$. Let $Q$ be the subset of $\mathbb{R}^m$, consisting of all points of the form $(\lambda_0,\dots,\lambda_n,x_0,\dots,x_n)$, where $\lambda_i$ are non-negative scalars such that $\lambda_0+\cdots +\lambda_n=1$, and $x_0,\dots, x_n$ are in the closure of $S$. Consider the mapping $\theta: \mathbb{R}^m \rightarrow \mathbb{R}^n$ such that $(\lambda_0,\dots,\lambda_n,x_0,\dots,x_n) \mapsto \lambda_0 x_0 + \cdots + \lambda_n x_n$. The image of $Q$ under $\theta$ is bounded. But why is it also closed? This shows up in the proof of Theorem 17.2 in Rockafellar's Convex Analysis.
a closedness argument in the proof of Theorem 17.2 in Rockafellar's Convex Analysis
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general-topology
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convex-analysis
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1 Answers
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Note that the domain (which is the product of an $n$-dimensional simplex and a bounded closed subset $\operatorname{cl}(S)$) is compact. Since $\theta$ is continuous, the image is compact, and thus closed.
Reference for the last result: a continuous image of a closed subset of a compact space is a closed subset