Page:Cantortransfinite.djvu/114

This page has been proofread, but needs to be validated.

OF TRANSFINITE NUMBERS

95

$f(n_{0})=m_{0}$

$f(n)=m_{1}$

for all $n$ 's which are different from $n_{0}$ .

The totality of different coverings of N with M forms a definite aggregate with the elements $f(N)$ ; we call it the "covering-aggregate (Belegungsmenge) of $N$ with $M$ " and denote it by $(N|M)$ . Thus:

(2)

(N|M)={f(N)}

.

If $M\sim M'$ and $N\sim N'$ , we easily find that

(3)

(N|M)\sim (N'|M')

.

Thus the cardinal number of $(N|M)$ depends only on the cardinal numbers ${\overline {\overline {M}}}={\mathfrak {a}}$ and ${\overline {\overline {N}}}={\mathfrak {b}}$ ; it serves us for the definition of ${\mathfrak {a}}^{\mathfrak {b}}$ :

(4)

{\mathfrak {a}}^{\mathfrak {b}}=({\overline {\overline {N|M}}})

.

For any three aggregates, $M,N,P$ , we easily prove the theorems:

(5)

((N|M).(P|M))\sim ((N.P)|M)

,

(6)

((P|M).(P|N))\sim (P|(M.N))

,

(7)

(P|(N|M))\sim ((P.N)|M)

,

from which, if we put ${\overline {\overline {P}}}={\mathfrak {c}}$ , we have, by (4) and by paying attention to § 3, the theorems for any three cardinal numbers, ${\mathfrak {a}}$ , ${\mathfrak {b}}$ , and ${\mathfrak {c}}$ :

(8)

{\mathfrak {a}}^{\mathfrak {b}}.{\mathfrak {a}}^{\mathfrak {c}}={\mathfrak {a}}^{\mathfrak {b+c}}

,

(9)

{\mathfrak {a}}^{\mathfrak {c}}.{\mathfrak {b}}^{\mathfrak {c}}=({\mathfrak {a}}.{\mathfrak {b}})^{\mathfrak {c}}

,

(10)

({\mathfrak {a}}^{\mathfrak {b}})^{\mathfrak {c}}={\mathfrak {a}}^{\mathfrak {b.c}}

.