Why $C_0((-\infty ,-1]\cup [1,+\infty ))$ doesn't have projection?

Question

According to projectionless C$^*$-algebras(line 4), since the set $(-\infty ,-1]\cup [1,+\infty )$ is disconnected, $C_0((-\infty ,-1]\cup [1,+\infty ))$ has projection. But I cannot find any projection in it? Is the state in this link false? And is it true only for unital commutative C$^*$-algebras?

Answer 1

The issue here is that your space is non-compact (hence the algebra is non-unital), and there are no compact connected components. You can check for yourself that $f$ being a projection means that $f(x) \in \{0,1\}$ for all $x \in X$, and the requirement that a projection $f$ must vanish at infinity on both of your non-compact connected components means that $f$ must be constantly 0 in this case.

I suppose it should say for a "unital commutative C*-algebra"; in this case the spectrum having no projections is equivalent to being connected.

If your space has compact connected components, you will end up with a non-trivial projection: for example $C_0((-\infty,0] \cup [1,2]) \simeq C_0((-\infty,0]) \oplus C([1,2])$ will have a projection given by $f(x) = 1$ for $x \in [1,2]$ and 0 elsewhere.