/sci/ - Science & Math

The mapping T : x(t) --> |x(t)| is continuous in both (a) and (b) (right?). So, X\A must be closed since it's the preimage of [1, infty) under T, and [1, infty) is closed in IR. Hence, A is open in both (a) and (b).

Is my reasoning (somewhat) correct? Would appreciate any help or hints.