Output State of comparator circuit for AC signal when open?

I am trying to design a circuit which would interface AC signals from pick-up coils. Below is the circuit.

R4, R3 are for DC bias, and R2 is for feeding that bias to inverting input of the OP amp.

R5 and R8 is for hysteresis regarding noise immunity.

V2 is the input signal.

V1 is power rail.

My question is that how should I determine the state of Vout(18V or 0V) when V2 is open?

I simulated it on CircuitLab. and below is the result V+ = 8.946 V V- = 8.955 V Vout = 1.783 mV

My concern is that V- is higher than V+ only by 9 mV, which I believe is leading Vout to near negative rail(0V).

Can I rely on current circuit for the output to be low always?

Or should I add some kind of pull-up or pull-down resistor for circuit stabilization?

I tried adding pull-up or pull-down resistors to V+ or V- but it prevents the signal from making Vout with 18V peak to Peak.

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.