How to understand the meet in the lattice of partitions? By definition a partition of a set $X$ is a collection of nonempty, disjoint sets whose union equals $X$. The meet in the lattice $P(X)$ of partitions of $X$ is given by
\begin{equation*}
\pi\wedge\sigma
=\{R\cap S:R\in\pi,S\in\sigma,R\cap S\neq\varnothing\}
\end{equation*}
for any $\pi$ and $\sigma$ in $P(X)$.
I know how to verify this, but I would like to also have a better understanding of it. So that e.g. I would be able to derive it as though I were not aware of the above expression. Or when considering similar lattices, such as the lattice of noncrossing partitions (where the meet is the same, but why is that?).
I have been thinking of partitions in terms of relations: two sets $R$ and $S$ are related if they intersect. Thus in a partition $\pi=\{R_1,\ldots,R_n\}$ no distinct blocks $R_i$ and $R_j$ are related.
The meet of partitions $\pi$ and $\sigma$ then is the collection of elements $R\cap S$ (i.e., the meet of $R$ and $S$) for those $R$ in $\pi$ and $S$ in $\sigma$ that are related.
The same approach however seems to fail when considering noncrossing partitions, and when calling sets $R$ and $S$ related when they intersect or cross: when $R$ in $\pi$ and $S$ in $\sigma$ cross but not intersect then they are related but $R\cap S = \varnothing$ does not lie in the meet of $\pi$ and $\sigma$.
Should I consider partitions $\pi$ and $\sigma$ of $X$ initially as arbitrary collections of subsets of $X$? Then a lower bound that comes to mind is
\begin{equation*}
\pi\sigma = \{R\cap S:R\in\pi,S\in\sigma\}.
\end{equation*}
This is not a partition of $X$ if $R\cap S$ is the empty set for some $R$ and $S$, but in that case kicking out the empty set seems to just do the trick. Can this be understood? I see that any block in a partition of $X$ is a subset $R$ of $X$ that is bounded from below by a singleton, i.e., $\{x\}\subseteq R$ for some $x$ in $X$. Is this a helpful way to look at it? Or should it be approached altogether differently?
 A: This is best illustrated via picture, but the meet of two partitions is their common refinement. Say this is our set $X$:

We could cut $X$ into pieces like this (call this partition $\pi$):

Or like this (call this partition $\sigma$):

Now there's an obvious question we might ask:

How can we cut up $X$ in such a way that we can reconstruct $\sigma$ and $\pi$ by moving our pieces around?

There's also an obvious answer: why not just give each dot its own block?

Now we can build any partition we like by rearranging these blocks... of course, this is inefficient. In some cases, this is going to be the best we can do, but in our case we don't need to spend quite so much time with the scissors!

If we cut up $X$ like this, then we can still reconstruct $\pi$ and $\sigma$, but now there's no waste. We've made exactly the extra cuts we needed to. This is the common refinement of $\pi$ and $\sigma$, and this is exactly our meet $\pi \wedge \sigma$!
Notice also that we got a new block in $\pi \wedge \sigma$ for each intersection $R \cap S$ for $R \in \pi$ and $S \in \sigma$. This is not an accident! In fact, it's the "obvious" thing to do if you're trying to build a refinement (can you see why? There's a hint under the fold).

 If you had cut out $\pi$ with scissors and then decided you wanted to rearrange it into $\sigma$, how would you do it? Probably you would look to see what pieces in $\pi$ need to be separated in $\sigma$, and cut them apart. But that exactly means you've cut a piece in $\pi$ (call it $R$) into new pieces for each piece in $\sigma$ (call them $S_k$). That is, you've cut $R$ into the pieces $R \cap S_1$, $R \cap S_2$, etc. for each $S_k \in \sigma$. Of course, if some $S_k$ doesn't intersect $R$, then you don't need to make any cuts at all!

It might also be worth staring at this picture of the lattice of partitions from
wikipedia:

Do you see how the meets in this lattice correspond to refinement?

I hope this helps ^_^
