A question about induced $C^\ast$-algebra

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Recently, I read the book Crossed Products of C*-algebras, and meat a question.

The question is how to prove $\mathrm{Ind}_c(A,\alpha)$ is dence in $\mathrm{Ind}(A,\alpha)$.

On the page 102, the author said that it is easy to see, but it is not easy for me.

Thanks to anyone who can give me a hint!

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The author points out that there is a nondegenerate action of $C_0(G \backslash P)$ on $\text{Ind} \, \alpha$, so $\text{Ind} \, \alpha$ is a $C_0(G \backslash P)$-algebra. The details are straightforward to check (and a good exercise in understanding $C_0(X)$-algebras). This is all you need, since it implies that $$ C_0(G\backslash P) \cdot \text{Ind } \alpha = \text{span}\left\{\varphi \cdot f : \varphi \in C_0(G \backslash P), \; f \in A \right\} $$ is dense in $\text{Ind } \alpha$. Now $C_c(G \backslash P)$ is dense in $C_0(G \backslash P)$, so $C_c(G \backslash P) \cdot \text{Ind} \, \alpha$ is also dense in $\text{Ind} \, \alpha$. But $$ C_c(G \backslash P) \cdot \text{Ind} \, \alpha \subseteq \text{Ind}_c \, \alpha \subseteq \text{Ind} \, \alpha, $$ so $\text{Ind}_c$ is dense in $\text{Ind} \, \alpha$.

This idea of approximating elements of $C_0(X)$-algebras with "compactly supported" ones is extremely useful. You should look at Proposition 3.40 and Corollary 3.42 of http://arxiv.org/pdf/0905.4681v1.pdf for more details.