Positive feedback and unstability
It all comes down to how you interpret the gain equation.

In an amplifier, things do not happen instantaneously. Extremely quickly - yes - but there is always a small time delay, \$\delta t\$, (or a lag) before the input is operated on to produce the output.

To take account of this, write the gain equation as:
$$y(1-af)=ax$$

where \$x\$ and \$y\$ are the input and output voltages, respectively.

Now, in the passage through the amplifier, \$y\times af\$ and \$a \times x\$ are subject to the delay,  \$\delta t\$, and the equation may therefore be written:
$$y_n-(af\times y_{n-1})=a\times x_{n-1}$$
or
$$y_n=(a\times x_{n-1})\:+\:(af\times y_{n-1})$$

where the \$n\$ subscript means the current value of time, and \$(n-1)\$ means the previous value of time,  \$\delta t\$ earlier.

If, now, you take an input of 1 volt, \$a=10,\: f=0.5, \: af=5\$, and calculate the \$y\$ values as time proceeds, you get:
$$y_1=10$$
$$y_2=10+50=60$$
$$y_3=10+300=310$$
$$y_4=10+1550=1560\:\: ...$$
which is unstable.

If, however, you take an input of 1 volt, \$a=10,\: f=0.01, \: af=0.1\$, and calculate the \$y\$ values as time proceeds, you get:
$$y_1=10$$
$$y_2=10+1=11$$
$$y_3=10+1.1=11.1$$
$$y_4=10+1.11=11.11\:\:...$$

which is stable

Clearly, we can let  \$\delta t\$ be as small as we wish; it just means that the difference equation is executed more rapidly and the output reaches it's final state, be it finite or infinite, more quickly. In practice, the amplifier's characteristics dictate the speed of response.