(b) Suppose is countably
infinite. Suppose
, be a finite collection. Then to show
.
Case -1 Assume
, then
is finite then each is finite then for
all i, for all
i=1,...n. Hence
, then we have
.
Case -2 Assume
, then
is finite, where are pairwise
disjoint. Then there exists (say) , such that is finite.
Now note that since are pairwise
disjoint, this gives us
, for all , as if
,
for some i>1, then
, this is a contradiction hence each is finite for all
Hence
, hence
.
Not additive part.
Since is countable we
can take
. Consider ,
then for all i, , now note
that
, which does not match with
. Hence not additive.
(c) Assume is
uncountable.
Case -1 Assume
, then each
is finite then each 's are finite. Then
we have , for all
i, hence
, hence
.
Case-2 Assume
, as there are only countably many , this can not
happen each is finite but the
union is countable infinite. Hence there exists say such that
is
uncountable and since
, is finite.
Then by the same argument note that
, for all . , then note
that there will exists only finitely many as is finite.
Then we have for
all Hence
. Hence
.
Feel free to comment if you have any doubts. Cheers!
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