If I have a beaker with pure water in it that has a piston right above the surface of the water that maintains a pressure of 1 atm then below 100 °C I would not expect to have any water vapor. In other words, the piston will keep touching the surface of the water. Above 100 °C the piston will go up a lot and I will have only vapor inside the beaker.

The only coexistence region of water+vapor is at 100 °C. This is all in the thermodynamic limit so I am not considering fluctuations.

However, now consider the following. Suppose I have a special gas that does not interact or dissolve in water. I introduce some of this special gas between the water surface and the piston and still maintain 1 atm pressure. Now let's consider the system at 50 °C.

I cannot see any difference between the two configurations (with and without the special gas) as far as the water is concerned and thus at 50 °C I would not expect any of the water to be in vapor form. And yet my gut/experience tells me that there is some vapor in the gap between the water surface and the piston. To see this imagine putting some ice cubes on the piston. One would expect droplets to condense on the piston.

What am I missing?

Edit 1 Adding phase diagrams for clarification and discussion. It's from James Sethna's book on statistical mechanics.

Edit 2 I am adding a diagram of the setup. The point is that I start with the setup at 50 °C and 1 atm with water at the bottom and an ideal gas that does not dissolve in water at the top with the two separated by a partition. What will happen when I remove the partition. Will some of the water in liquid phase convert into vapor phase? If so, how does one figure out how much of the water will convert into vapor? Assume we know the mass of the water and that of the gas.