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Currently I am reading the book 'Isometries in Banach Spaces: Vector-valued Function Spaces and Operator Spaces, Volume Two', Chapter $7,$ page $8.$

Definitions: A $T$-set is a subset $S$ of a Banach space $X$ with the property that for any finite collection $x_1,x_2,...,x_n \in S,$ $$\left\Vert \sum_{j=1}^n x_j \right\Vert = \sum_{j=1}^n \|x_j\|,$$ and such that $S$ is maximal with respect to this property.

Definitions: Two $T$-sets $S$ and $R$ in a Banach space $X$ are said to be discrepant if either $R\cap S = \{0\},$ or there exists a $T$-set $L$ such that $R\cap L = S \cap L = \{0\}.$

Question: After reading the second definition, I guess that all $T$-sets contain $0.$ However, I fail to prove this.

Any hint would be appreciated.

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As Jonas mentioned in his comment, the way to go about proving this is to use maximality. Suppose $S$ is a subset of a Banach space $X$ with the property that $$ \tag{1} x_1,\ldots,x_n\in S \implies \left\|\sum_{j=1}^nx_j\right\| = \sum_{j=1}^n\|x_j\|. $$ Then $S\cup\{0\}$ also has property $(1)$. Indeed, if $x_1,\ldots,x_n\in S$, then $$ \left\| \sum_{j=1}^nx_j+0 \right\| = \left\| \sum_{j=1}^nx_j \right\| = \sum_{j=1}^n\|x_j\| = \sum_{j=1}^n\|x_j\| +0 = \sum_{j=1}^n\|x_j\| +\|0\|. $$ Therefore, due to maximality, every $T$-set must contain $0$. (For otherwise, if $S$ is a $T$-set that doesn't contain $0$, then $S\cup\{0\}$ contradicts the maximality of $S$.)

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    $\begingroup$ Clear explanation. Thanks. $\endgroup$ – Idonknow Oct 4 '17 at 5:55

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