For clarification: Given a projection $p$ in a C$^{*}$-algebra $A$, we say $p$ is infinite if there is a projection $q\in A$ satisfying $q\lneq p \sim q$; we say $p$ is properly infinite if there are projections $e,f\in A$, which are mutually orthogonal and satisfy $e\leq p$, $f\leq p$, and $e\sim p\sim f$.
If $p$ is non-zero, it is easy to see from the definitions that $p$ is infinite provided it is properly infinite. Since these properties are not equivalent (I know, for instance, that if $1$ is properly infinite in $A$, then $K_{0}(A)=\{[p]_{0}:0\not=p\text{ is a projection in }A\}$), I was wondering if there was an (easy) example of a non-zero projection in a C$^{*}$-algebra that is infinite but not properly infinite.
Let $A=B(H)\oplus 1$. Take $p=I\oplus 1$. Then $p$ is infinite, since $p\sim (I-E_{11})\oplus 1$. The partial isometry being $S\oplus 1$, where $S$ is the unilateral shift.
But it is not properly infinite. If $p=e+f$, we write $e=E\oplus\lambda$, $f=F\oplus \mu$. Being projections, $\lambda,\mu\in\{0,1\}$. So $\lambda+\mu=1$, and one of them is $1$, the other $0$. Say $\lambda=1$, $\mu=0$. But then it is not possible to have $e\sim f$: because if $v=A\oplus b$ is a partial isometry with $v^*v=e$ and $vv^*=f$. This would require $|b|=1$ and $|b|=0$, which is impossible.