BIBO Stability for system with no poles
Why does a system with no poles have a finite support?


If a system doesn't have finite poles, then its transfer function is of the form:

$$H(z) = \frac{Y(z)}{X(z)} = a_Nz^N+a_{N-1}z^{N-1}+...+a_1z+a_0$$

So if you go back to the time domain:

$$y[n]=a_Nx[n+N]+a_{N-1}x[n+N-1]+...+a_1x[n+1]+a_0x[n]$$

Note that if $x[n]=\delta[n]$, then $y[n]=h[n]$, the impulse response of the filter. 

$$h[n]=a_N\delta[n+N]+a_{N-1}\delta[n+N-1]+...+a_1\delta[n+1]+a_0\delta[n]$$

It's easy to see that the equation above will return a $h[n]$ that is non-zero for a limited range of values of $n$.

This means that the filter is FIR. Therefore, it has a finite support and it is BIBO stable, leading to your second question.


  Why if the system has a finite support it means that it is BIBO stable?


For BIBO stability of an LTI discrete-time system, one must check that the following condition is satisfied:

$$\sum_{n=-\infty}^{\infty}|h[n]|&lt;\infty$$

If $h[n]$ is the impulse response of a FIR filter, then it has some length $L$:

$$\sum_{n=-\infty}^{\infty}|h[n]|=\sum_{n=K}^{L+K-1}|h[n]|$$

Due to this summation being finite, it converges. Ergo, any FIR filter is BIBO stable.