Why is $\beta(\mathbb{N}\times\mathbb{N})\neq\beta(\mathbb{N})\times\beta(\mathbb{N})$?

Question

In my course on topology, we proved that the Cech-Stone compactification of $\mathbb{N}$ is homeomorphic to $(U(\mathbb{N}), \mathcal{T})$ where $U(\mathbb{N})$ are the ultrafilters on $\mathbb{N}$ and $\mathcal{T}$ is the topology generated by $\{A^*\mid A\subset\mathbb{N}\}$ for $A^*=\{\mathcal{U}\in U(\mathbb{N})\mid A\notin \mathcal{U}\}$.

Afterwards, my professor showed that $\beta(\mathbb{N}\times\mathbb{N})\neq\beta(\mathbb{N})\times\beta(\mathbb{N})$, but I didn't fully understand the proof. The whole point is that, apparently, $\mathbb{N}\times\mathbb{N}$ isn't $\mathcal{C}^*$-embedded in $\beta(\mathbb{N})\times\beta(\mathbb{N})$. In order to prove this, it is sufficient to show that the continuous characteristic function on $\mathbb{N}\times\mathbb{N}$ (which sends all points on the diagonal to $1$ and all others to $0$) doesn't have an extention to $\beta(\mathbb{N})\times\beta(\mathbb{N})$. It is the proof of this no-extention statement that has been bothering me.

We took a point $(p, p)$ on the diagonal outside $\mathbb{N}\times\mathbb{N}$. If $V\in \mathcal{V}(p, p)$, then there is an $A\subset \mathbb{N}$ such that $p\in A^*$ and $A^*\times A^*\subset V$. Also, $(A^*\times A^*)\cap(\mathbb{N}\times\mathbb{N})=(\mathbb{N}\setminus A)\times(\mathbb{N}\setminus A)$. I don't get how we can conclude at this point that the characteristic function cannot be extended to $\beta(\mathbb{N})\times\beta(\mathbb{N})$.

What is going wrong? What am I missing?

Answer 1

Any function from the discrete space $S = \mathbb N \times \mathbb N$ to a compact set $K$ has a unique continuous extension to $\beta(S)$. So, does the diagonal function ($f(x,y) = 1$ if $x = y$, $0$ otherwise) have a continuous extension to $\beta(\mathbb N) \times \beta(\mathbb N)$?

Suppose $g$ is such an extension. For any $x \in \mathbb N$, $g(x, \cdot)$ is a continuous function on $\beta(\mathbb N)$. Now for $m,n \in \mathbb N$, $g(m,n) = 0$ if $m \ne n$. Since $\mathbb N \backslash \{n\}$ is dense in $\beta(\mathbb N)$, we have $g(x, n) = 0$ for all $x \in \beta(\mathbb N) \backslash \mathbb N$ and all $n \in \mathbb N$. Since $\mathbb N$ is dense in $\beta(N)$, that implies $g(x, y) = 0$ for all $x,y \in \beta(\mathbb N)$ with $x \notin \mathbb N$.
Thus $g^{-1}(\{1\}) = \{(n,n): n \in \mathbb N\}$. But that is not compact, therefore not closed in $\beta(\mathbb N) \times \beta(\mathbb N)$, so it is impossible for $g$ to be continuous.

Answer 2

I believe the idea is that $(p,p)$ would have to map to $1$ by density of the diagonal of $\mathbb N\times\mathbb N$ in the diagonal of $\beta \mathbb N\times \beta\mathbb N$. But every neighborhood of $(p,p)$ in $\beta \mathbb N\times \beta\mathbb N$ contains points which map to $0$.

I do not think the specific topology of $\beta\mathbb N$ is relevant. You just need to prove two general exercises:

If $X$ is dense in $Y$ then

(1) the diagonal of $X\times X$ is dense in the diagonal of $Y\times Y$.

(2) $X\times X$ minus its diagonal is dense in $Y\times Y$.