How does the empty set work in arithmetic of surreal numbers?

Question

I'm working my way through Surreal Numbers by Knuth, and am finding myself a little hung up on the explanation of how addition works. The rule for addition is given as:

$$ x + y = ((X_L+y)\cup(Y_L+x), (X_R+y)\cup(Y_R+x)) $$

On the next page, one of the protagonists works out $1+1$ to be:

$$ (\{0+1,0+1\},\emptyset) $$ and follows the reasoning on through to prove that $1+1=2$. Where I'm getting a bit lost is that it seems there should be some additions involving $\emptyset$ that are omitted without explanation. If $1$ is defined as $(0,\emptyset)$, then a true following of the addition rule seems like it should be:

$$ 1+1=((0+1)\cup(0+1), (\emptyset+1)\cup(\emptyset+1)) $$ $$ 1+1=(1,1) $$

All of this is obviously wrong, but it's not clear to me why $\emptyset+1$ isn't $1$. I guess the most basic version of my question is: why is $0$ distinct from $\emptyset$ in surreal numbers, and what is the nature of that distinction?

Answer 1

$1$ is not really defined as $(0,\emptyset)$, but as $(\{0\},\emptyset)$. Then $$1+1=\left(\{0+1\}\cup\{0+1\},(\emptyset+1)\cup(\emptyset+1)\right)$$ However, $\emptyset+1$ is empty since there are no surreals in $\emptyset$ to add to $1$ ($\emptyset+1$ is the set of $x+1$ for all $x\in\emptyset$), so $1+1$ simplifies to $$\left(\{0+1\}\cup\{0+1\},\emptyset\cup\emptyset\right)=\left(\{0+1\},\emptyset\right)\text{.}$$

If you know that $0+1$ is $1$, we have $1+1=(\{1\},\emptyset)$.

To answer your other question, $0$ is distinct from $\emptyset$ in the surreals since every surreal is an ordered pair of sets. $0$ is defined to be the ordered pair $(\emptyset,\emptyset)$.

Answer 2

A surreal number is, in particular, a ordered pair of sets of surreal numbers, so $\emptyset$ isn't a surreal number at all. In the formula for addition, the expressions like $X_R + y$ do not represent one value but rather the set of all numbers of the form $z + y$, where $z \in X_R$. But the definition $1 = (\{0\}, \emptyset)$ implies that $1$ has no right options at all. So if $x=1$, then $X_R$ is empty, and there aren't any such numbers. Similar reasoning applies to $Y_R + x$. The conclusion is that $1+1$ has no right options whatsoever.