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Theorem The range $\mathcal{R}(T)$ of a compact linear operator $T\colon X \to Y$ is separable; here, $X$ and $Y$ are normed space.

Proof. Consider the ball $B_n=B(0,n)\subset X$. Since $T$ is compact, the image $C_n= T(B_n)$ is relatively compact. $C_n$ is separable by a precedent result. The norm of any $x\in X$ is finite, so that $\lVert x \rVert < n$, hence $x\in B_n$ with $n$ sufficiently large. Consequently, $$X = \bigcup_{n=1}^\infty B_n \implies T(X)=\bigcup_{n=1}^\infty T(B_n)=\bigcup_{n=1}^\infty C_n.$$

Since $C_n$ is separable, it has a countable dense subset $D_n$ and the union $$D=\bigcup_{n=1}^\infty D_n\subseteq T(X)$$ is countable. I can't see the point that $D$ is dense in $T(X)$, that is $cl(D)=T(X)$.

We have that $$cl(D)=cl\left(\bigcup_{n=1}^\infty D_n\right)\supseteq \bigcup_{n=1}^\infty cl(D_n)=\bigcup_{n=1}^\infty C_n = T(X).$$

And for the other inclusion? Is this correct? Or can the fact that $D$ is dense in $T(X)$ be observed differently?

Jack J.
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1 Answers1

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If $y \in T(X)$, $y = T(x)$ for some $x$. Then $y \in T(B_n)$ if $n \ge \|x\|$. Given $\epsilon> 0$, since $D_n$ is dense in $T(B_n)$ there is $z \in D_n$ with $\|z - y\| < \epsilon$. Since $D_n \subset D$, this says there is $z \in D$ with $\|z - y\| < \epsilon$, i.e. $D$ is dense in $T(X)$.

Robert Israel
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