Consider the GARCH(1,1) process
\begin{align}
r_{t+1} &= \sigma_{t+1} z_{t+1} \\
\sigma^2_{t+1} &= \omega+\alpha r^2_t +\beta \sigma^2_{t}
\end{align}
for the returns $r_t$, with ${z_t} \sim N (0,1)$ IID.
In what follows, let us distinguish the conditional return variance
$$ V [ r_{t+1} \vert \mathcal{F}_t ] = \sigma^2_{t+1} $$
from the unconditional return variance
$$ V [ r_{t+1} ]$$
where $\mathcal{F}_t$ denotes the information available up to time $t$ (filtration), in our case all past values $\{\sigma_t,...,\sigma_0,r_t,...,r_0\}$.
First, notice that if both $\alpha$ and $\beta $ happen to be zero, then if $\omega$ is allowed to reach zero as well there is a possibility that the conditional variance will become zero.
Second, because $z_t$ are zero-mean, unit variance and i.i.d, we have that the unconditional returns' variance is strictly equal to (*)
\begin{align}
V[r_{t+1}] &= E[r_{t+1}^2]-E [r_{t+1}]^2 = E[r_{t+1}^2] \\
&= E [ \sigma_{t+1}^2 z_{t+1}^2] = E [ \sigma_{t+1}^2] E [z_{t+1}^2] \\
&= E[\sigma_{t+1}^2]
\end{align}
We can use this relationship along with the GARCH definition to write
\begin{align}
V[r_{t+1}] &= E[\sigma_{t+1}^2] = E [ (\omega+\alpha r^2_t +\beta \sigma^2_{t}) ] \\
&= \omega + \alpha V [r_t] + \beta V [r_t] \\
&= \omega + V [r_t] (\alpha + \beta)
\end{align}
Now, assuming weak stationarity we should have that the unconditional variance $\sigma^2_\infty$ is such that
$$ \sigma^2_\infty = V [r_{t+1}] = V [r_t] $$
which using the above gives
$$ \sigma^2_\infty = \frac {\omega}{1-\alpha-\beta} $$
From which we see why the condition $\alpha + \beta < 1$ is relevant but also that allowing $\omega $ to reach zero would also imply a possibility that the unconditional variance becomes zero.
At the end of the day, we therefore have that if $\omega=0$ while the other coefficients are positive, you will have (conditional) heteroskedasticity in the sense that conditional variance will evolve through time, but the unconditional stationary variance will be zero, with unrealistic consequence that returns become deterministic at some point.
(*) This could have been anticipated since
\begin{align}
V [r_{t+1}] &= V [r_{t+1} \vert \mathcal{F}_0] \\
&= E [r_{t+1}^2 \vert \mathcal{F}_0] \\
&= E [ E [ r_{t+1}^2 \vert \mathcal{F}_t] \vert \mathcal{F}_0] \\
&= E [ V [ r_{t+1} \vert \mathcal{F}_t] \vert \mathcal{F}_0] \\
&= E [ \sigma_{t+1}^2 \vert \mathcal{F}_0] \\
&= E [ \sigma_{t+1}^2 ]
\end{align}
where we have made use of the tower law (and the fact that $z_t $ are zero-mean and independently distributed so that $E [r_{t+1}]=0$).
