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A unital $C^*$-Algebra $B$ is called stably finite, if, for all $n\in\mathbb{N}$, $1_{M_n}\in M_n(B)$ is a finite projection.

Claim: If $B$ is a unital, simple, stably finite $C^*$-Algebra, then $B\otimes \mathcal{K}$ contains no infinite projections.

I want to prove this, but I don't really know, how ( I only know the definition)

Here, $\mathcal{K}$ denotes the compact operators on a separable, infinite dimensional Hilbert space. Write $\mathcal{K}=\varinjlim M_n$ and $M_n(B)=B\otimes M_n$, then $B\otimes \mathbb{K}=B\otimes \varinjlim M_n=\varinjlim (B\otimes M_n)$ and every projection in $B\otimes \mathbb{K}$ can be approximated by projections in $M_n(B)$. But now, I only know that $1_{M_n}\in M_n(B)$ is finite for all $n$, so that at this point I don't really know how to continue.

I have seen this question Stably Finite C$^\ast$-algebras (but I don't know how to check $(2)\Rightarrow (1)$). In the recommended book, I just found proposition V.2.1.8: That if $(A_i,\phi_{i,j})$ is an inductive system of unital $C^*$-algebras (with the $\phi_{i,j}$ unital) and $A=\varinjlim (A_i,\phi_{i,j})$ and each $A_i$ is stably finite, then $A$ is stably finite. (Note: A nonunital $C^*$-Algebra $E$ is stably finite, if it's minimal unitization $E^+$ is stably finite) However, in the inductive limit for $\mathcal{K}$ the connecting maps are not necessarily unital and, if you want to apply this proposition, the question is, how to prove: $1\in M_n((B\otimes \mathcal{K})^+)$ is finite for all $n$, $\Rightarrow$ $B\otimes \mathcal{K}$ contains no infinite projections.

I appreciate any further hints or help to prove the claim.

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Thinking again about this question I am not anymore sure if there is a direct argument via projections. For the argument in your reference question I was thinking that infiniteness of a projection is invariant under Murray von Neumann equivalence. If I will find an argument involving this this idea, I will post it here.

However, using scaling elements (V.2.2.8) one can prove this result quite fast.

By V.2.3.6 for a simple C*-algebra containing an infinite projection is equivalent to containing a scaling element. However, if $B \otimes \mathbb K$ contains a scaling element, then so does $B \otimes M_n$ for some $n$.

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  • $\begingroup$ I understand, thank you very much! $\endgroup$ – Sabrina G. Aug 3 '17 at 13:37

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