How was the bottle crushed by air?
You do not need the water vapour for this effect, the air that is contained in the bottle is sufficient.

You may use the ideal gas law to estimate the amount of air in the volume of the bottle of about $V=1\ \mathrm l=0.001\ \mathrm{m^3}$ when the air is heated to about $T=60\ \mathrm{^\circ C}=333.15\ \mathrm K$ (using hot water) at normal pressure $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ (since the bottle is still open):

$$\begin{align}
pV&amp;=nRT\\[3pt]
n&amp;=\frac{pV}{RT}\\[3pt]
&amp;=\frac{100\,000\ \mathrm{Pa}\times0.001\ \mathrm{m^3}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times333.15\ \mathrm K}\\[3pt]
&amp;=0.036\ \mathrm{mol}
\end{align}$$

If you now close the bottle, the amount of air $n$ is trapped.

When this air is cooled to about $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$ (using ice water) at an approximately constant volume of $V=1\ \mathrm l=0.001\ \mathrm{m^3}$, the pressure in the bottle decreases according to the ideal gas law:

$$\begin{align}
pV&amp;=nRT\\[3pt]
p&amp;=\frac{nRT}{V}\\[3pt]
&amp;=\frac{0.036\ \mathrm{mol}\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times273.15\ \mathrm K}{0.001\ \mathrm{m^3}}\\[3pt]
&amp;=8.2\times10^4\ \mathrm{Pa}=0.82\ \mathrm{bar}
\end{align}$$

Therefore, the ambient air pressure of $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ compresses the bottle until the pressures are equalized (or the bottle fails). The volume in the bottle decreases according to the ideal gas law:

$$\begin{align}
pV&amp;=nRT\\[3pt]
V&amp;=\frac{nRT}{p}\\[3pt]
&amp;=\frac{0.036\ \mathrm{mol}\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times273.15\ \mathrm K}{100\,000\ \mathrm{Pa}}\\[3pt]
&amp;=8.2\times10^{-4}\ \mathrm{m^3}=0.82\ \mathrm l
\end{align}$$

Nevertheless, any condensation of water vapour would help to achieve this effect. The resulting changes in pressure and volume due to condensation of water can be even more extreme than the contraction of air. For example, if a volume of $V=1\ \mathrm l$ is filled with steam at $T=120\ \mathrm{^\circ C}$ and $p=1\ \mathrm{bar}$ (without any air) and then cooled until the steam is almost completely condensed to liquid water (at about $T=99.6\ \mathrm{^\circ C}$), the resulting liquid volume is only $V=0.00058\ \mathrm l=0.58\ \mathrm{ml}$.