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The direct sum bornology is defined the following way:
Let $I \neq \emptyset$ be a index set and $(X_i,\mathcal{B}_i)$,$i \in I$ a family of bornological vector spaces, $X = \bigoplus_{i \in I}X_i$ and $\iota_i : X_i \to X$,$i \in I$ the canonical maps. The direct sum bornology $\mathcal{B}_{\bigoplus}$ is defined as the final bornology with respect to $\iota_{i}$,$i \in I$ hence is generated by $\bigcup_{i \in I}\iota_i(\mathcal{B}_i)$ and is therefore the finest vector bornology which makes each $\iota_{i}$,$i \in I$ a bounded linear map. It is claimed that $\mathcal{B} := \lbrace \bigoplus_{i \in I}B_i | B_i \in \mathcal{B}_i,i \in I \rbrace$ is a basis of $\mathcal{B}_{\bigoplus}$.

My first question is: What does $\bigoplus_{i \in I}B_i$ even mean since the $B_i$ are just bounded sets and not vector spaces. In the book "Bornologies and Functional Analysis" from Hogbe Nlend on page 34 it is said that it is a finite sum of bounded sets. My guess is that it is the finite outer direct sum of $B_i$ so a Cartesian product of finitely man bounded sets. In this case $B_1 \times B_2$ is bornological isomorphic to $B_1 + B_2$ by $(x_1,x_2) \mapsto x_1 + x_2$ where $x_1 \in B_1$ and $x_2 \in B_2$.

Anyway, I want to show that $\mathcal{B}$ is closed under scalar multiplication, finite addition, consists of balanced sets and covers $X$. I was able to show everything except that it consists of balanced sets. In the book which was referenced above it is stated that $\mathcal{B}$ consists of balanced sets.

My second question is: How does one see that $\mathcal{B}$ consists of balanced sets?

Edit: It is only necessary to show that $\mathcal{B}$ is closed under balanced hulls and thanks to the comments I was able to show this too.

Orb
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  • But what if the $B_i$ are bounded sets in $(\mathbb{C},\mathcal{B}{can})$ where $\mathcal{B}{can}$ is the canonical bornology so all absolute value bounded set and $\mathbb{C}^{(I)} = \bigoplus_{i \in I}\mathbb{C}$. One can define the direct sum bornology on $\mathbb{C}^{(I)}$ with the canonical maps $\iota_i : \mathbb{C} \to \mathbb{C}^{(I)}$.Then the finite sum of the $B_i \in \mathcal{B}_{can}$ is in $\mathbb{C}$ but as a basis of the direct sum bornology the sum should be in $\mathbb{C}^{(I)}$. – Orb Dec 27 '22 at 21:16
  • The finite dimensional bornology is defined using the identification of a vector space $X$ with $\mathbb{C}^{(I)}$ and the above described direct sum bornology using the canonical maps $\iota_i : \mathbb{C} \to \mathbb{C}^{(I)}$. The sum must somehow be a subset of $\prod_{i \in I}X_i$. – Orb Dec 27 '22 at 21:31
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    I guess that the proposition claims that for every bounded set $B$ there are bounded $B_i\subset E_i$ with ${i\in I: B_i\neq {0}}$ finite such that $B$ is contained in $\prod_{i\in I} B_i$ (this set is contained in the direct sum). – Jochen Dec 28 '22 at 07:49
  • Then the set of all those set of this form is closed under balanced hulls since $\mbox{bal}(B) = \bigcup_{|\lambda| \leq 1}\lambda B \subseteq \bigcup_{|\lambda| \leq 1} \lambda \prod_{i \in I}B_i \subseteq \prod_{i \in I} \bigcup_{|\lambda| \leq 1} \lambda B_i = \prod_{i \in I}\mbox{bal}(B_i)$ and $\mbox{bal}(B_i) $ are bounded since the $\mathcal{B}i$ are vector bornologies. Thus there exists finitely many $A_i \in \mathcal{B}_i$ such that $\mbox{bal}(B) \subseteq \prod{i \in I}A_i$. – Orb Dec 28 '22 at 12:49

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