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Robert Israel
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In any Banach space $B$, if $S$ is a countable set of bounded linear maps, there is $a \in B$ such that $S\ni T \mapsto Ta \in B$ is injective. This follows easily from the the Baire Category Theorem: $B$ is not the union of the countably many closed, nowhere dense sets $\{x \in B \mid T_i x = T_j x\}$.

In any Banach space $B$, if $S$ is a countable set of bounded linear maps, there is $a \in B$ such that $S\ni T \mapsto Ta \in B$ is injective. This follows easily from the the Baire Category Theorem: $B$ is not the union of the countably many closed, nowhere dense sets $\{x \in B \mid T_i x = T_j x\}$.

In any Banach space $B$, if $S$ is a countable set of bounded linear maps, there is $a \in B$ such that $S\ni T \mapsto Ta \in B$ is injective. This follows easily from the Baire Category Theorem: $B$ is not the union of the countably many closed, nowhere dense sets $\{x \in B \mid T_i x = T_j x\}$.

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David Roberts
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In any Banach space $B$, if $S$ is a countable set of bounded linear maps, there is $a \in B$ such that $T \in S \mapsto Ta \in B$$S\ni T \mapsto Ta \in B$ is injective. This follows easily from the the Baire Category TheoremBaire Category Theorem: $B$ is not the union of the countably many closed, nowhere dense sets $\{x \in B \mid T_i x = T_j x\}$.

In any Banach space $B$, if $S$ is a countable set of bounded linear maps, there is $a \in B$ such that $T \in S \mapsto Ta \in B$ is injective. This follows easily from the the Baire Category Theorem: $B$ is not the union of the countably many closed, nowhere dense sets $\{x \in B \mid T_i x = T_j x\}$.

In any Banach space $B$, if $S$ is a countable set of bounded linear maps, there is $a \in B$ such that $S\ni T \mapsto Ta \in B$ is injective. This follows easily from the the Baire Category Theorem: $B$ is not the union of the countably many closed, nowhere dense sets $\{x \in B \mid T_i x = T_j x\}$.

Source Link
Robert Israel
  • 54.2k
  • 1
  • 76
  • 152

In any Banach space $B$, if $S$ is a countable set of bounded linear maps, there is $a \in B$ such that $T \in S \mapsto Ta \in B$ is injective. This follows easily from the the Baire Category Theorem: $B$ is not the union of the countably many closed, nowhere dense sets $\{x \in B \mid T_i x = T_j x\}$.