What does an Ulam matrix look like? I'm trying to visualize an Ulam matrix but I"m having trouble. So it has Aleph one rows and aleph null columns? What do elements of a Ulam matrix look like?
 A: It need not have $\aleph_1$ rows and $\aleph_0$ columns: one can define larger Ulam matrices. However, $(\aleph_1,\aleph_0)$-Ulam matrices are a reasonable place to start. Such a matrix is a collection of sets $A_{\alpha,n}$ for $\alpha<\omega_1$ and $n<\omega$ such that

*

*each $A_{\alpha,n}\subseteq\omega_1$;

*if $\alpha<\beta<\omega_1$, then $A_{\alpha,n}\cap A_{\beta,n}=\varnothing$ for each $n<\omega$; and

*for each $\alpha<\omega_1$, $\left|\omega_1\setminus\bigcup_{n<\omega}A_{\alpha,n}\right|\le\aleph_0$.

In other words, the sets in each column are pairwise disjoint, and the union of the sets in each row is almost all of $\omega_1$ — specifically, all but at most countably many elements of $\omega_1$.
It’s hard to say much more about what the sets look like: any collection satisfying all of those conditions is an Ulam matrix. You can perhaps get a little more insight by working through a proof that there actually is an Ulam matrix.

For each $\xi<\omega_1$ let $f_\xi:\omega\to\omega_1$ be a function such that $\xi\subseteq\operatorname{ran}f_\xi$. For example, for each $n<\omega$ we can define $f_n$ by $f_n(k)=k$ for each $k<n$ and let $f(k)=0$ for $n<k<\omega$. If $\omega\le\xi<\omega_1$, then $\xi$ is countably infinite, so we can simply take $f_\xi:\omega\to\xi$ to be any bijection.
For each $\alpha<\omega_1$ and $n<\omega$ let $A_{\alpha,n}=\{\xi<\omega_1:f_\xi(n)=\alpha\}$. This is the key trick, and you’ll probably have to stare at it a bit to get any kind of feel for why it works. Fortunately, merely checking that it works is easier.
First, these sets $A_{\alpha,n}$ are clearly subsets of $\omega_1$.
Now let’s look at two of them in the same column, say $A_{\alpha,n}$ and $A_{\beta,n}$, where $\alpha,\beta<\omega_1$. Suppose that some $\xi\in A_{\alpha,n}\cap A_{\beta,n}$; then by definition $f_\xi(n)=\alpha$ and $f_\xi(n)=\beta$, which is obviously possible only if $\alpha=\beta$. Thus, the sets in any column of the matrix are indeed pairwise disjoint.
Finally, consider the sets in one row, say row $\alpha$: we need to show that $\omega_1\setminus\bigcup_{n<\omega}A_{\alpha,n}$ is countable. Suppose that $\xi<\omega_1$ is not in any of the sets $A_{\alpha,n}$ with $n<\omega$. That means that for each $n<\omega$ we have $f_\xi(n)\ne\alpha$. Recall how we chose the function $f_\xi$: $\xi\subseteq\operatorname{ran}f_\xi$, which means that for each $\gamma<\xi$ there is an $n<\omega$ such that $f_\xi(n)=\gamma$. Thus, if $\xi\notin\bigcup_{n<\omega}A_{\alpha,n}$, then $\alpha\not<\xi$, i.e., $\xi\le\alpha$. Thus, $\omega_1\setminus\bigcup_{n<\omega}A_{\alpha,n}\subseteq\alpha+1=\{\xi:\xi\le\alpha\}$, which is clearly a countable set.

This completes the proof that $\{A_{\alpha,n}:\alpha<\omega_1\text{ and }n<\omega\}$ is an Ulam matrix. $\dashv$
To get any more insight than that, you probably have to look at an argument or two actually using an Ulam matrix to prove a result.
