Output State of comparator circuit for AC signal when open?
To start, eliminate V2, C1, R2 and R5. Then the junction of R3 and R4 will have a Thevenin equivalent voltage of 9 volts, and a resistance of 5k. Adding R5 provides a hysteresis of $$ \Delta v = 18\times \frac{6k}{1.006 M} = .107\text{ volts} =  9 +/- .053\text{ volts} $$ this will also provide a hysteresis at the R2/R3 junction of  $$ \Delta v = 18\times \frac{5k}{1.005 M} = .089\text{ volts}=  9 +/- .045\text{ volts} $$   With V2 still unconnected, the - input will be forced to the value of the R2/R3 junction. Since this is undetermined, the output voltage is undetermined. 

Effectively, with V2 disconnected the comparator can be in either state, and there is no a priori way to determine which state that will be.