## 2 The \(\mu\)-SEIR Model
### Boundaries
It has to be proved that the constraints still hold for \(k+1\). First note that, since \(E_{k}\leq\frac{1}{2\beta_{E}}\), \(I_{k}\leq\frac{1}{2\beta_{I}}\) and \(0\leq S_{k}\), one has:  
\[S_{k+1}=(1-\beta_{I}I_{k}-\beta_{E}E_{k})S_{k}\geq 0 \tag{9}\]  
and, since \(E_{k}\geq 0\) and \(I_{k}\geq 0\), one has that  
\[I_{k+1}=(1-\gamma)I_{k}+\mu E_{k}\geq 0 \tag{10}\]  
On the other hand, since \(E_{k}\leq\frac{1}{2\beta_{E}}\), \(I_{k}\leq\frac{1}{2\beta_{I}}\), \(0\leq S_{k}\leq\frac{\mu}{2\beta_{E}}\), \(\gamma\in(0,1]\) and \(\beta_{I}/\beta_{E}\leq\gamma/\mu\), one has that:  
\[0\leq E_{k+1} =(1-\mu)E_{k}+(\beta_{I}I_{k}+\beta_{E}E_{k})S_{k}\] \[\leq\frac{1-\mu}{2\beta_{E}}+S_{k}\leq\frac{1-\mu}{2\beta_{E}}+ \frac{\mu}{2\beta_{E}}\] \[=\frac{1}{2\beta_{E}}\]  
and  
\[0\leq I_{k+1} =(1-\gamma)I_{k}+\mu\ E_{k}\] \[\leq\frac{1-\gamma}{2\beta_{I}}+\frac{\mu}{2\beta_{E}}\leq\frac{ 1-\gamma}{2\beta_{I}}+\frac{\gamma}{2\beta_{I}}\] \[=\frac{1}{2\beta_{I}}\]  
In addition, since \(S_{k}\leq\frac{\mu}{2\beta_{E}}\) and \(0\leq(1-\beta_{I}I_{k}-\beta_{E}E_{k})\) since \(0\leq E_{k}\leq\frac{1}{2\beta_{E}}\); \(0\leq I_{k}\leq\frac{1}{2\beta_{I}}\), it follows that \[0\leq S_{k+1}=(1-\beta_{I}I_{k}-\beta_{E}E_{k})S_{k}\leq S_{k}\leq\frac{\mu}{2 \beta_{E}}\]  
Finally, since \(R_{k}\geq 0,I_{k}\geq 0\) and \(\gamma\geq 0\) then \(R_{k+1}\geq 0\). It has been proved that \((min(S_{i},E_{i},I_{i},R_{i})\geq 0;i=0,1,2,\ldots,k)\) implies \((min(S_{i},E_{i},I_{i},R_{i})\geq 0;i=0,1,2,\ldots,k+1)\), and \(\left(S_{i}\leq\frac{\mu}{2\beta_{E}},E_{i}\leq\frac{1}{2\beta_{E}},I_{i}\leq \frac{1}{2\beta_{I}};i=0,1,\ldots,k\right)\) implies \((S_{i}\leq\frac{\mu}{2\beta_{E}},E_{i}\leq\frac{1}{2\beta_{E}},I_{i}\leq\frac{ 1}{2\beta_{I}};i=0,1,\ldots,k+1)\). As a result the proof is complete since \(min(S_{0},I_{0},E_{0},R_{0})\geq 0\) implies \(min(S_{i},E_{i},I_{i},R_{i})\geq 0;\forall i\in\mathbf{Z}_{+}\) and \(\left(S_{0}\leq\frac{\mu}{2\beta_{E}},E_{0}\leq\frac{1}{2\beta_{E}},I_{0}\leq \frac{1}{2\beta_{I}}\right)\to S_{i}\leq\frac{\mu}{2\beta_{E}}\), \(E_{i}\leq\frac{1}{2\beta_{E}},I_{i}\leq\frac{1}{2\beta_{I}};\forall i\in \mathbf{Z}_{+}\) which implies also that the sequences of all the subpopulations and that of the total population are bounded. **Remark 1**. : _The given constraints on the upper bounds of the subpopulations which guarantee that Theorem 1 holds are reasonable since the disease transmission rates are typically small in normalized models and much smaller in un-normalized ones. Typically, the normalized disease transmission rate is normalized by the total population in true-mass action epidemic models. On the other hand, the constraint \(\beta_{I}/\beta_{E}\leq\gamma/\mu\) is reasonable, since typically \(\gamma<\mu\) and the transmission rate of the infectious individuals are on average less than that of the exposed ones since the strongest infective period happens at the end period of the disease incubation. In addition, note that if \(\beta_{I}=\beta_{E}=\beta\) then the constraints \(E_{i}\leq\frac{1}{2\beta_{E}},I_{i}\leq\frac{1}{2\beta_{I}}\) of Theorem 1 become a single constraint \(E_{i}+I_{i}\leq\frac{1}{\beta};\forall i\in\mathbf{Z}_{0+}\) and Theorem 1 remains valid. It also remains valid if \(max(E_{i},I_{i})<\frac{1}{\beta_{I}+\beta_{E}};\forall i\in\mathbf{Z}_{0+}\)._  
The subsequent technical result is useful later on:  
**Proposition 3**. : _Assume that the hypotheses of Theorem 1 hold. Thus, if \(\lim_{i\to\infty}E_{i}=0\) then \(\lim_{i\to\infty}I_{i}=0\) and, conversely, if \(\lim_{i\to\infty}I_{i}=0\) then \(\lim_{i\to\infty}E_{i}=0\)._  
Proof.