When is the Minkowski sum of weighted compact sets $w_1 B_1 + w_2 B_2 + \ldots$ (with $w \in L^1$) closed?

Question

Let $B_1,B_2,\ldots,$ be compact subsets of $\mathbb R^d$ and $w_1,w_2,\ldots$ be nonnegative numbers summing to $1$. Consider the set

$$ A := w_1 B_1 + w_2 B_2 + \ldots = \left\{\sum_{n=1}^\infty w_n b_n \mid b_1 \in B_1,\,b_2 \in B_2,\,\ldots\right\}. $$

Question. Is $A$ closed in $\mathbb R^d$ ?

Note that if $w_n = 0$ for all $n \ge N$ (where $N$ is some finite integer), then $A$ is compact, and therefore closed. Indeed, the the function $f:B_1 \times \ldots \times B_N \to A$ defined by $f(b_1,\ldots,b_N) := \sum_{n=1}^N w_n b_n$ is continuous and the set $B_1 \times \ldots \times B_N$ is compact.

Answer 1

Another case.

Let $B_1, B_2, \dots$ be compact sets in $\mathbb R^n$, all bounded by $R > 0$. Let $w_n$ be nonnegative numbers with sum $1$. The product topological space $X = B_1 \times B_2 \times \dots$ is compact. The function $F : X \to \mathbb R^n$ defined by $$ F(b_1,b_2,\dots) = \sum_{n=1}^\infty w_n b_n $$ is defined everywhere on $X$ (because of the common bound $R$) and continuous. Therefore $A = F(X)$ is compact.

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remark. If there is no bound on the diameters of $B_n$, then we cannot expect $\sum_{n=1}^\infty w_n b_n$ to converge. If we let $A$ be those series that do converge, then we cannot expect $A$ to be compact. But would it still be closed?

Answer 2

Yes, the set $A$ is closed whenever $A$ is well-defined. Your definition of $A$ as an infinite sum is actually a definition of $A$ as the closure of all finite subsums. Compare with the one-dimensional case $\sum_{i=0}^\infty w_i=:\lim_{N\to +\infty} \sum_{i=0}^N w_i$ where infinite sums are defined as limits. Likewise your definition of $A$ already implies $A$ is the limit of all finitely subsums. I hope its obvious what i mean by the term "subsum". So we just need recall that "derived sets" in point-set topology are closed. I.e. the set of all accumulation points of any set is closed.