output voltage of an operational amplifier
Ideally, the output will always be saturated.  This is because a ideal opamp has infinite gain.

Back here in the real world, there are some issues with this circuit:

There is always some inevitable noise on any signal.  When the temperature is close to the threshold, then this noise will cause the temperature signal to go back and forth between being above and below the threshold.  This will make the opamp output appear to be at a in-between level.

If the noise is fast enough, the opamp output will actually be at a in-between level most of the time due to the finite output voltage slew rate.  Put another way, the opamp can't go instantly from full high to full low.  If you keep jerking it around faster than it can react to, it will spend most of its time going back and forth without getting to either extreme.

To be a useful circuit, there really should be some hysteresis.  This is a little positive feedback.  It creates a little dead band around the threshold.  For example, if the threshold is set to 0 V, then it might take +100 mV to go high when low, then -100 mV to go low once high.  The size of the hystersis band should be enough to cover most ordinary noise.  That prevents the case described in #1, above.

There should be a resistor in series with the opamp output, or each LED.  If the opamp were perfect, it would drive the output to either +5 V or -5V.  That would blow up the LEDs.  With a real opamp, you don't know what might get damaged.

The right resistor allows for the desired current thru the LEDs.  For example, let's say these are typical green LEDs that drop 2.1 V and you want to run them at 10 mA.  When the opamp puts out 5 V, then the resistor is dropping 2.9 V.  From Ohm's law, (2.9 V)/(10 mA) = 290 &Omega;.  The common value of 300 &Omega; would be a reasonable choice then.