Q: definition of diagonal intersection (set theory) Can anyone help me intuitively understand what a diagonal intersection is?
The definition I have is: $$\triangle_{\alpha < \omega_1} C_\alpha = \{\beta < \omega_1 : \forall \alpha < \beta \; \beta \in C_{\alpha} \}$$
So what I did was I tried to compare with the definition of a regular intersection in a similar format, which I understand:
$$\cap_{a < \omega_1} C_\alpha = \{\beta < \omega_1 : \forall \alpha \; \beta \in C_\alpha\} $$
But I'm not sure how to intuitively interpret $\forall \alpha < \beta$ part. Can anyone explain using relatively non-technical words?
Thank you.

A: I can try to help motivate the name diagonal intersection, at least.
Let's say we have a family of sets $(C_\alpha)_{\alpha<\omega_1}$. Let's put these sets into a table:
$$\begin{array}{c|ccccccc}
 & 0 & 1 & 2 & \dots & \omega & \omega+1 & \dots \\
\hline
C_0 & 1 & 1 & 1 & \dots & 1 & 1 & \dots\\
C_1 & 1 & 1 & 1 & \dots & 0& 1 &  \dots\\
C_2 & 1 & 0 & 1 & \dots & 1 & 1 & \dots\\
\vdots &  \vdots & \vdots & \vdots & \ddots& \vdots& \vdots& \\
C_\omega & 1 & 0 & 1 & \dots & 0 &  1 & \dots\\
C_{\omega+1} & 0 & 1 & 1 & \dots & 0 & 1 & \dots \\
\vdots & \vdots& \vdots& \vdots& & \vdots&\vdots & \ddots
\end{array}$$
Here the row labeled $C_\alpha$ is the characteristic function of $C_\alpha$. So $0\in C_0$, $\omega\notin C_1$, etc.
Now $\beta\in \bigcap_{\alpha<\omega_1}C_\alpha$ if and only if $\beta\in C_\alpha$ for all $\alpha$. This corresponds to the column labeled $\beta$ in the table containing all $1$s. From the table above, we can tell that $0$, $1$, and $\omega$ are not in $\bigcap_{\alpha<\omega_1}C_\alpha$, since their columns contain $0$s. The table is finite, hence very incomplete, but based on the partial information we have, it is possible that $2$ and $\omega+1$ are in $\bigcap_{\alpha<\omega_1}C_\alpha$.
On the other hand, to determine whether $\beta\in \triangle_{\alpha<\omega_1}C_\alpha$, we only need to check if $\beta\in C_\alpha$ for $\alpha<\beta$. In the table, this means we get to ignore everything on the diagonal and below:
$$\begin{array}{c|ccccccc}
 & 0 & 1 & 2 & \dots & \omega & \omega+1 & \dots \\
\hline
C_0 &  & 1 & 1 & \dots & 1 & 1 & \dots\\
C_1 &  &  & 1 & \dots & 0& 1 &  \dots\\
C_2 &  &  &  & \dots & 1 & 1 & \dots\\
 &   &  &  & & \vdots& \vdots& \\
C_\omega &  &  &  &  &  &  1 & \dots\\
C_{\omega+1} &  &  &  &  &  &  & \dots \\
\vdots & & & & & & & 
\end{array}$$
This makes it much "easier" for ordinals to end up in the diagonal intersection. For example, we can already tell (from the woefully incomplete table) that $0$, $1$, and $2$ are in $\triangle_{\alpha<\omega_1}C_\alpha$, since their columns contain only $1$s.
