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Let $F$ a separable space and $T: E \to F$ a linear isometry. Is $E$ a separable space? (E, F are linear spaces)

I was working in the following problem: Showing that $T: l_\infty \to L(l_2,l_2)$ defined by $T(a) (b) = (a_n b_n), \quad a = (a_n) \in l_\infty , b= (b_n) \in l_2$ and then, conclude that $L(l_2,l_2)$ isn't a separable space. Well, we have that $l_\infty$ isn't separable, so if the setence above is correct, we are done.

My attempt:

Let's consider a set $A = ${$ y_n \in F: n \in \mathbb N $} dense in $F$. So, $\forall x \in E$, as $T(x) \in F$, given $n \in \mathbb N$, exists $M(n) \in \mathbb N$ with $|| T(x) - y_{M(n)} || < 1/n$. With this, how can I get a countable set in $E$ that is dense in $E$?

Thank you

user 242964
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2 Answers2

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Hint: you can assume without loss of generality that $E \subseteq F$. As $F$ is separable, we can assume, as you did, that $F$ has a dense subset $\{y_0, y_1, \ldots\}$. Given $m, n \in \Bbb{N}$, if there is an $x \in E$ such that $\|x - y_n\| < 1/(m+1)$, define $x_{mn}$ to be some such $x$, otherwise define $x_{mn}$ to be $0$. Now show that the $x_{mn}$ form a countable dense subset of $E$.

Rob Arthan
  • 48,577
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Hint: a metric space is separable if and only if it is second-countable.

tomasz
  • 35,474