328 
POINT SETS 
point of (5. For 
d=f( c i)=f( c i) = 
For let D x > Z) 2 > ••• be a sequence of superimposed divisions 
of Q, whose norms = 0. Let 
^11’ ^12’ c ^13 
be the cells of D 1 containing no point of 5f within them. Let 
^21’ ^22’ ^23 
(1 
(2 
denote those cells of Z> 2 containing no point of 51 within them and 
not lying in a cell of 1). In this way we may get an infinite se 
quence of cells 3) = \d mn 5, where for each m, the corresponding n 
is finite, and m = oo. Each point a of A lies in some d m ^ n . For 5f 
being complete, Dist (a, 51) > 0. As the norms S n = 0, a must lie 
in some cell of D n , for a sufficiently large n. The truth of the 
theorem is now obvious. 
310. Let 53 be pantactic in 5i. Then there exists an enumerable 
set (5 <53 which is pantactic in 5i. 
For let I) x > _Z> 2 > ••• be a set of superimposed cubical divisions 
of norms d n = 0. In any cell of D 1 containing within it a point 
of 51, there is at least one point of 53. If the point of 5i lies on 
the face of two or more cells, the foregoing statement will hold 
for at least one of the cells. Let us now take one of these points 
in each of these cells; this gives an enumerable set (§). The 
same holds for the cells of JD 2 . Let us take a point in each of 
these cells, taking when possible points of Qjq. Let (5 2 denote the 
points of this set not in (5 1 . Continuing in this way, let 
(5 = ^ + ^+ - 
Then (5 is pantactic in 51, and is enumerable, since each (5 n is. 
Corollary. In any set 51, finite or infinite, there exists an enumer 
able set (5 which is pantactic in 5f. 
For we have only to set 53 = 51 in the above theorem. 
311. 1. The points (5 ivhere the continuous function f(x 1 ••• x m ) 
takes on a given value g in the complete set %,form a complete set. 
For let <q, <? 2 ••• be points of (5 which = c. We show c is a