Construct a set {P_{i}} of people as in the hint. This set has the property that for each
i, either

1) P_{i} is friends with every P_{j} where j>i. (P_{i} is * gregarious.) *

or

2) P_{i} is not friends with any P_{j} where j>i. (P_{i} is a *hermit *.)

Now either the set of gregarious people is infinite or the set of hermits is infinite, because their
union is an infinite set. Suppose the set of gregarious people is infinite. Then for each pair of people
in the set {P_{i},P_{j}} where i < j,
we know that P_{i} is friends with P_{j}, by definition of gregarious.
Similarly, if the set of hermits is infinite, they yield an infinite set of people, no pair of which
are friends.

Perceptive readers will notice that I need to use the axiom of choice here. For them, here is a solution that explicitly uses that axiom.

Given the infinite set of people at the party, there is a choice function, c, defined on the set of nonempty subsets
of people such that c(A) is an element of A. Let P_{1} = c(set of all people at party).
Define F_{1} and N_{1} to be the set of friends and non-friends of P_{1},
respectively. Let S_{1}= F_{1} if that set is infinite. Otherwise let
S_{1}= N_{1}.
In general suppose {P_{1},...,P_{n}} and {S_{1},...,S_{n}} have been
defined. Let P_{n+1}=c(S_{n}), and define S_{n+1} to be the set
of people in S_{n} who are friends with P_{n+1} if that set is infinite. Otherwise
let S_{n+1} be the set of people in S_{n} who are not friends with
P_{n+1}. By the principle of recursive definition (see, e.g., *Topology* by James Munkres),
this gives us a countably infinite set {P_{i}}. Now proceed as above.