It's pretty easy with a comparator.
As the resistance of the thermistor goes down, it increases the voltage of the comparator's positive input (pin 3). When this input goes higher than the negative input (pin 2) the output will go high.
To set the threshold, you simply adjust the potentiometer.
The output of the comparator probably won't be able to drive a relay (check the current requirement of the relay, and the output drive of the comparator). So I have added a transistor which switches current to the relay.
Relays are inductive loads which will produce a negative voltage spike when they switch off. The diode will absorb this spike.
Hysteresis: Make sure that the comparator you choose has some built in hysteresis. E.G. LTC6702 or MAX9021. Alternatively, add a resistor (R5) to force some hysteresis.
You have now specified the exact on/off temperatures. These give voltages of 2.710v and 2.688v. Which is a hysteresis of 22mv. If my calculations are correct (probably not) then I think you need to use 100k for R5.
Added 2: How to calculate R5
First, using the formula for parallel resistors, we calculate the effective resistance of RT1 and R2 as about 460 ohms.
Next, we use consider a potential divider formed by R5 on top and (RT1 & R2) at the bottom. For ease of calculation, assume that we have 2.5v at the bottom (generated by RT1 and R2).
Now, the question is, what value of R5 do we need if we want V to change by 0.022v. I.E. +-0.011v.
Using the 0v case, because it's simpler:
2.489v = (R5 / (R5+460)) * 2.5v
R5 = 460 / (2.5/2.489 - 1)
So, about 100k.