Pressure inside a closed water bottle due to vapour pressure?

Suppose I have a standard, partially filled bottle of distilled pure water (filled up to around 75%) that’s open to the environment at room temperature and pressure. Now suppose I close the lid of the water bottle so that it constitutes a closed system.

At first, the highly energetic water molecules near the surface are able to escape liquid form and evaporate, cooling the water slightly. Heat quickly flows in from the surroundings to equilibrate the temperature of the water with the ambient temperature (around 25°C). Energetic particles continue to evaporate at the same rate (since the evaporation rate is dependent on temperature, which isn’t changing); however, the more water vapour particles that are present in the closed water bottle, the higher the condensation rate, and so eventually we reach an equilibrium whereupon the condensation rate equals the evaporation rate and the amount of water vapour particles in the bottle is effectively constant.

The pressure in the bottle due only to the water vapour ($P_{mathrm{vap,inside}}$) inside the bottle is now equal to the vapour pressure of water at 25°C (which is around 0.03 bar). But the pressure caused by the air inside the bottle ($P_{mathrm{air,inside}}$) is still equal to the atmospheric pressure outside (since it was equal to 1 bar before the bottle was closed and the gas hasn’t changed in temperature, volume or amount and thus must remain at atmospheric pressure). This means that the total pressure inside the partially filled diathermic bottle should be

$$P_{mathrm{total,inside}}=P_{mathrm{air,inside}}+P_{mathrm{vap,inside}}=1+0.03,mathrm{bar}=1.03,mathrm{bar} > P_{mathrm{atmosphere,outside}}=1,mathrm{bar}.$$

Is my thinking correct so far? Now, suppose we did not close the bottle but rather left it open on a table somewhere at STP. The energetic particles would still evaporate; however, not all of them would return to the liquid phase, and so eventually all the water inside the bottle would evaporate, causing an extremely small increase in total atmospheric pressure (due to the new water vapour in the atmosphere), which would be completely undetectable. Once again, is this correct? If so, then would it be correct to surmise that in a hot climate (say, somewhere near the equator) with lots of liquid water present in the location (say, near the ocean or a large lake), the greater presence of water vapour in the atmosphere would lead to a slight and possibly detectable increase in the atmospheric pressure relatively to locations at the same altitude but with colder temperatures?

Any help with this issue would be most appreciated!