stationary sets Let $\{S_i: i < \omega\}$ be a strictly decreasing sequence of stationary subsets of $\omega_1$,
is the intersection of all $S_i$'s stationary?
 A: I am answering this as CW, or else it will keep popping up periodically.
Suppose $\{S_n\mid n\lt\omega\}$ is a decreasing collection of stationary subsets of $\omega_1$. If their intersection is non-stationary, then it must be the case that some $S_n\setminus S_{n+1}$ is stationary, or else $S=\bigcap_n S_n$ is stationary, and in fact $S_0\setminus S$ is non-stationary (i.e., modulo the non-stationary ideal, $S$ and $S_0$ are the same). 
The same observation in fact gives us that infinitely often $S_n\setminus S_{n+1}$ must be stationary. By passing to a subsequence, we may as well assume that this holds for all $n$. But then, we may also assume that $\bigcap_nS_n$ is empty, as we can remove $S$ from each $S_n$ and preserve their stationarity.
Hence, if a sequence is as asked, we obtain that, letting $T_n=S_n\setminus S_{n+1}$, the sequence of sets $T_n$ is a pairwise disjoint sequence of stationary sets. Conversely, from any such sequence we obtain a decreasing sequence of stationary sets $S_n$ with empty intersection, by letting $S_n=\bigcup_{m\ge n}T_m$.
What this says is that the question is equivalent to whether there is a sequence of pairwise disjoint stationary sets, and this is well-known to be true.
In the interest of being self-contained, let me add that the usual proof that there are such sequences is due to Ulam, using what we now call a Ulam matrix: Fix injections $f_\alpha:\alpha\to\omega$ for all $\alpha\lt\omega_1$, and set, for $\alpha\lt\omega_1$ and $n\lt\omega$, $$S_{\alpha,n}=\{\beta\lt\omega_1\mid \alpha<\beta\mbox{ and }f_\beta(\alpha)=n\}.$$ Then, for any fixed $\alpha$, $$\bigcup_n S_{\alpha,n}=\omega_1\setminus(\alpha+1),$$ so there is some $n_\alpha$ such that $S_{\alpha,n_\alpha}$ is stationary. 
But then it follows that for some $m$, $n_\alpha=m$ for $\omega_1$ many $\alpha$. Note now that for any $n$, $$S_{\alpha,n}\cap S_{\alpha',n}=\emptyset$$ whenever $\alpha\ne\alpha'$, and we conclude that, considering $n=m$, there are in fact $\omega_1$ pairwise disjoint stationary sets.
An easy elaboration of this argument shows that any stationary set can be split into $\omega_1$ pairwise disjoint stationary subsets. The same argument shows that any stationary subset of $\kappa^+$ can be split into $\kappa^+$ pairwise disjoint stationary subsets, for any infinite $\kappa$, and (considering the ideal of measure zero sets rather than the non-stationary ideal) that no successor cardinal is real-valued measurable. 
That stationary subsets of $\lambda$ can be split into $\lambda$ pairwise disjoint stationary subsets actually holds for all regular (uncountable) cardinals $\lambda$, not just successors (but a different argument is needed, of course). This was proved by Solovay. 
