Page:Cantortransfinite.djvu/139

then $1+\omega$ is not equal to $\omega +1$. For, if $f$ is a new element, we have by (1):

$1+\omega =({\overline {f,E}})$,
$\omega +1=({\overline {E,f}})$.

But the aggregate

$(f,E)=(f,e_{1},e_{2},...,e_{\nu },...)$

is similar to the aggregate E, and consequently

$1+\omega =\omega$.

On the contrary, the aggregates $E$ and $(E,f)$ are not similar, because the first has no term which is highest in rank, but the second has the highest term $f$. Thus $\omega +1$ is different from $\omega =1+\omega$.

Out of two ordered aggregates $M$ and $N$ with the types $\alpha$ and $\beta$ we can set up an ordered aggregate $S$ by substituting for every element $n$ of $N$ an ordered aggregate $M_{n}$ which has the same type $\alpha$ as $M$, so that

and, for the order of precedence in

we make the two rules:

(1) Every two elements of $S$ which belong to one and the same aggregate $M_{n}$ are to retain in $S$ the same order of precedence as in $M_{n}$;

(2) Every two elements of $S$ which belong to two different aggregates $M_{n_{1}}$ and $M_{n_{2}}$ have the same relation of precedence as $n_{1}$ and $n_{2}$ have in $N$.