Easy problem (it was hard for me though lol): $\hspace{120mm}$ Let $M_{i}\subseteq{\mathbb{Z}}$, $i=1,2,\dots$, be a sequence of subsets such that: $\;\;\;$ (i) $M_{1}\supseteq{M_{2}}\supseteq{\dots}$ $\hspace{70mm}$ (ii) If $a,b\in{\mathbb{Z}}$ are such that $ab\in{M_{i}}$, then either $a\in{M_{i}}$ or $b\in{M_{i}}$. $\hspace{90mm}$ Prove that, if $ab\in{\cap_{i=1}^{\infty}}M_{i}$, then either $a\in{\cap_{i=1}^{\infty}}M_{i}$ or $b\in{\cap_{i=1}^{\infty}}M_{i}$. $\hspace{100mm}$ $\hspace{100mm}$ Solution: From the hypothesis, we obtain that $ab\in{M_{i}}$ $\forall$ ${i=1,2,\dots}$. If we have $a\in{M_{i}}$ $\forall$ $i$ or $b\in{M_{i}}$ $\forall$ $i$, then we're done. If not, let $r$ the least positive integer such that $a\in{M_{r}}$ but $a\notin{M_{r+1}}$ (it follows that $a\notin{M_{i}}$ $\forall$ $i\geq{r+1}$, since $M_{r+1}\supseteq{M_{r+2}}\supseteq{\dots}$). Similarly, let $s$ the least positive integer such that $b\in{M_{s}}$ but $b\notin{M_{s+1}}$ (it follows again that $b\notin{M_{i}}$ $\forall$ $i\geq{s+1}$). Since $ab\in{M_{r+1}}$ and $a\notin{M_{r+1}}$, we find that $b\in{M_{r+1}}$, and then by the definition of $s$ it follows that $r+1\leq{s}$. Analogously, we obtain that $s+1\leq{r}$. But this implies that $r+1\leq{s}<{s+1}\leq{r}$, which is a contradiction. Now we're done. $\;\;\;\;\;$ $\hspace{100mm}$ (I think...)