Yesterday night I attended a reception for the international scholars to UCSB. This was the first time that such a celebration took place in Santa Barbara, thanks to an anonymous donor. It was a lively event, and various important people from the UCSB campus showed up. I won’t bore you with the details.
Amongst the interesting facts that I collected yesterday, was that most international students in the US come to study in the science and technology fields, while very few people from the US go out to study and when they do, the statistic is mostly on humanities. I also found out that there are about 640000 students from abroad in US Universities, and that US Universities graduate about 30000 PhD’s annually.
Not surprisingly, current cuts in US University funding (especially state Universities) will hurt the efforts to get the best (graduate) students to the US, with the consequent deterioration of the pool of people with talent contributing to the US economy. An interesting article regarding this issue can be found here.
If you couple all of this progressive lack of investment by the states into the education system, the future starts looking pretty bad, not just for education, but for the US economy.
This year I am working on a University committee in charge of international education. So I have to learn quite a bit about this stuff.
Simple observation: if you lose loose the people from abroad (who are extremely good students), and you lose loose the local people (because they can not afford to go to school any longer), where is the human capital investment in the future development of technology going to come from?
Remember, modern economies depend on having the best and most innovative modern technology in order to compete. And, new ideas for technology come from people who know what the current technologies are, what they can do and how to make them. Ideas are not born from thin air.
Posted in Academia, Santa Barbara, science and society
&bg=ffffff&fg=000000&s=0 "GL_n(\mathbb{F}_q)")
have zeroes in all diagonal entries?![(q-1)^n \sum_{k=0}^n (-1)^k \frac{[n]!}{[k]!}](http://s1.wordpress.com/latex.php?latex=(q-1)%5En+%5Csum_%7Bk=0%7D%5En+(-1)%5Ek+%5Cfrac%7B%5Bn%5D!%7D%7B%5Bk%5D!%7D&bg=ffffff&fg=000000&s=0 "(q-1)^n \sum_{k=0}^n (-1)^k \frac{[n]!}{[k]!}")
doesn’t seem to work for n > 2.
just isn’t the same as the geometry of 
, but a continuous model would be very difficult to design and simulate (or would it?). But perhaps there could be some cleverly chosen simple rule that would tend to protect “clumps” of 1s and allow them to move, and to do complicated things like rotating (whatever that can be made to mean in 
). Or perhaps a complicated ecosystem could develop that was more 
, and put a 1 at every grid point that lies in the annulus. Then in the interior circle of the annulus put a random scattering of not too many 1s. And then slightly move the annulus part, and slightly move all the little particles inside. If the first position of the annulus represents where some very simple structure is at time 1 and the second where it is at time 2, then conservation of mass and momentum would tell us to expect it to continue moving in the same direction (so it would be more sophisticated than a cellular automaton of the kind described earlier because its behaviour would depend not just on how it behaved an instant earlier), and to stay the same size. We might also have “forces” between neighbouring 1s that encouraged them to stay together somewhat, and so on.
be a countable model of a language extending the language of groups, with a universal extension 
. Let 
be a continuous, invariant Keisler measure that is also invariant under translations. Let 
be a symmetric definable subset of +=+1%7D&bg=ffffff&fg=000000&s=0 "{\mu(X) = 1}")
and %7D&bg=ffffff&fg=000000&s=0 "{\mu(X \cdot X)}")
finite, and let 
be the group generated by 
. Let 
be a wide type over 
contained in %7D&bg=ffffff&fg=000000&s=0 "{a \in q(G)}")
, there exists %7D&bg=ffffff&fg=000000&s=0 "{b \in q(G)}")
such that %7D&bg=ffffff&fg=000000&s=0 "{tp(a/M \cup \{b\})}")
and %7D&bg=ffffff&fg=000000&s=0 "{tp(b/M \cup \{a\})}")
are both wide. 
is a normal subgroup of 
. 
and 
seems to be a statement that there exist plenty of pairs %7D&bg=ffffff&fg=000000&s=0 "{(a,b)}")
in %7D&bg=ffffff&fg=000000&s=0 "{q(G)}")
that are in “general position” somehow. In the case of abelian groups, it seems that this hypothesis not necessary. The hypothesis is for all %7D&bg=ffffff&fg=000000&s=0 "{a \in q(G)}")
, but by homogeneity it suffices to verify it for a single 
(since the action of the automorphism group of 
is transitive on %7D&bg=ffffff&fg=000000&s=0 "{q(G)}")
). 
to use. 
is wide, 
is wide also. If we also assume that every finite product 
has finite measure, this leads to the consequence that the index of 
in 
in cardinality (i.e. has bounded index, see Lemma 1 of 
disjoint cosets 
in 
, and hence in some finite product 
. By Section 4 of 
-definable, and so is the intersection of countably many definable sets 
. For any distinct 
, 
are disjoint, thus by saturation we have %7D+a_i+%5Ccap+S_%7Bn(i,j)%7D+a_j%7D&bg=ffffff&fg=000000&s=0 "{S_{n(i,j)} a_i \cap S_{n(i,j)} a_j}")
non-empty for some integer %7D&bg=ffffff&fg=000000&s=0 "{n(i,j)}")
. By the 
in which the +=+n%7D&bg=ffffff&fg=000000&s=0 "{n(i,j) = n}")
are constant, thus we have a countable set of disjoint translates 
, which live in some finite product 
. But as 
has positive measure, while 
-definable subgroup of 
and the group is the integers; we normalise 
to have measure 
. Taking ultralimits, we end up with a non-standard discrete interval 
in the non-standard integers, with 
being the counting measure normalised to give 
. (This is not yet a saturated model, but yet us ignore this issue.) Note that the subsets 
are definable for any standard natural number 
. Their intersection 
is a wide subgroup; if 
would generate 
be a definable set of positive measure. Then we can refine 
to a wide type 
such that for every 
be a definable set of positive measure, and suppose that the product 
is covered by two definable sets 
(these are sets of pairs, of course). Then one can refine 
of positive measure, such that at least one of the following holds: 
, the set +%5Cin+C+%5C%7D%7D&bg=ffffff&fg=000000&s=0 "{C_a := \{ b: (a,b) \in C \}}")
has positive measure; or 
, the set +%5Cin+D+%5C%7D%7D&bg=ffffff&fg=000000&s=0 "{D^b := \{ a: (a,b) \in D \}}")
has positive measure.
to be contained in =0%7D&bg=ffffff&fg=000000&s=0 "{\mu(C)=0}")
. We can find a definable set 
between +%3E+1/n+%5C%7D%7D&bg=ffffff&fg=000000&s=0 "{\{ a: \mu(C_a) > 1/n \}}")
and +%5Cgeq+1/n+%5C%7D%7D&bg=ffffff&fg=000000&s=0 "{\{ a: \mu(C_a) \geq 1/n \}}")
. If any of these 
, so we may assume instead that =0%7D&bg=ffffff&fg=000000&s=0 "{\mu(Q_n)=0}")
for all 
, which implies that +=+0%7D&bg=ffffff&fg=000000&s=0 "{\mu(C) = 0}")
by Fubini’s theorem (which can be extended to Kiesler measures, at least in the weak form we need here), giving the desired contradiction. 
and refining repeatedly using Lemma 
and pairs of definable sets 
, so that each pair is revisited infinitely often) one can find a type %7D&bg=ffffff&fg=000000&s=0 "{q({\Bbb U})}")
, and 
is covered by two definable sets 
has positive measure for all 
, or 
has positive measure for all 
. In particular (setting 
and 
empty) this implies that %7D&bg=ffffff&fg=000000&s=0 "{a \in q({\Bbb U})}")
, and define 
to be the type 
for definable 
with =0%7D&bg=ffffff&fg=000000&s=0 "{\mu(C_a)=0}")
, and the complement of the sets of the form 
for definable =0%7D&bg=ffffff&fg=000000&s=0 "{\mu(D^a)=0}")
(note the asymmetry here!). We claim that this collection of sentences is still finitely satisfiable. Indeed, if this were not the case, 
for some definable 
with =%5Cmu(D%5Ea)=0%7D&bg=ffffff&fg=000000&s=0 "{\mu(C_a)=\mu(D^a)=0}")
. Write 
, then the set 
is definable and contains 
, we thus see that 
is a definable set containing 
, contradicting the construction of 
.
over 
, but by construction %7D&bg=ffffff&fg=000000&s=0 "{q_a({\Bbb U})}")
and hence %7D&bg=ffffff&fg=000000&s=0 "{q'_a({\Bbb U})}")
is contained in the complement of %7D&bg=ffffff&fg=000000&s=0 "{tp(a/M \cup \{b\})}")
is wide. For if not, then 
of measure zero for some definable 
and 
has measure zero also, and so 
is contained in the complement of 
by construction. In particular, 
lying in 
. This proves Lemma 
be positive integers. Then for 
sufficiently close to 
is a finite subset of a group 
with the bounded quintipling property 
, where 
, and for a proportion at least 
-tuples +%5Cin+(X+%5Ccdot+X)%5El%7D&bg=ffffff&fg=000000&s=0 "{(a_1,\ldots,a_l) \in (X \cdot X)^l}")
, one has 
(recall that 
). Then there exists a subgroup 
such that 
is contained in at most %7D&bg=ffffff&fg=000000&s=0 "{k(m+1)}")
cosets of 
from 
rather than 
, then one can find a sequence of groups 
of tripling uniformly bounded by 
, such that the proportion of 
-tuples %7D&bg=ffffff&fg=000000&s=0 "{(a_1,\ldots,a_l)}")
in %5El%7D&bg=ffffff&fg=000000&s=0 "{(X \cdot X)^l}")
with 
approaches %7D&bg=ffffff&fg=000000&s=0 "{k(m+1)}")
cosets.
+=+1%7D&bg=ffffff&fg=000000&s=0 "{\mu(X) = 1}")
, +%5Cleq+k+%5Cmu(X_0)%7D&bg=ffffff&fg=000000&s=0 "{\mu(X_0 X X) \leq k \mu(X_0)}")
, and such that the set +%5Cin+(X+%5Ccdot+X)%5El:+%5Cmu(+a_1%5EX+%5Cldots+a_l%5EX+)+%5Cgeq+%5Cmu(X)/(m+0.5)+%5C%7D%7D&bg=ffffff&fg=000000&s=0 "{Q := \{ (a_1,\ldots,a_l) \in (X \cdot X)^l: \mu( a_1^X \ldots a_l^X ) \geq \mu(X)/(m+0.5) \}}")
(say) has measure %5El%7D&bg=ffffff&fg=000000&s=0 "{\mu(X \cdot X)^l}")
. (Strictly speaking, one has to replace +%5Cgeq+%5Cmu(X)/(m+1)%7D&bg=ffffff&fg=000000&s=0 "{\mu( a_1^X \ldots a_l^X ) \geq \mu(X)/(m+1)}")
by a definable predicate between +%5Cgeq+%5Cmu(X)/(m+1)%7D&bg=ffffff&fg=000000&s=0 "{\mu( a_1^X \ldots a_l^X ) \geq \mu(X)/(m+1)}")
and +%3E+%5Cmu(X)/(m+1)%7D&bg=ffffff&fg=000000&s=0 "{\mu( a_1^X \ldots a_l^X ) > \mu(X)/(m+1)}")
, but let’s ignore this technicality.) We then extend 
, in fact), so by Lemma 
is a group. Note that this set is contained in 
, which have positive measure as 
has full measure in %5El%7D&bg=ffffff&fg=000000&s=0 "{(X \cdot X)^l}")
, we see that 
intersects 
for every 
, by saturation, 
. In particular, there exists 
in +%5Cgeq+%5Cmu(X)/(m+1).&bg=ffffff&fg=000000&s=0 "\displaystyle \mu( a_1^X \ldots a_l^X ) \geq \mu(X)/(m+1).")
lies in +(m+1)+/+%5Cmu(X)+%5Cleq+k(m+1)%7D&bg=ffffff&fg=000000&s=0 "{\mu(X_0 X X) (m+1) / \mu(X) \leq k(m+1)}")
disjoint translates 
, 
. As a consequence, 
can be covered by %5E4%7D&bg=ffffff&fg=000000&s=0 "{k^4 (m+1)^4}")
such translates, including 
-definable (it is the complement of all the other cosets in 
-definable and 
-definable has to be definable. (Proof: we can write 
. By saturation, 
must be empty for at least one 
which is larger than all other natural numbers. 
. Then 
is also a natural number which is strictly larger than 
beyond the natural numbers, because 
is defined equal to 
. However, the above argument does show that the existence of a largest number is not compatible with the 
cannot be written as 
for any finite collection 
of natural numbers. 
existed. Then consider the number 
; this is a natural number, but is larger than (and hence not equal to) any of the natural numbers 
is enumerated by 
. 
or 
. (For this post, we allow enumerations to contain repetitions.)
cannot be written as 
for any finite collection of prime numbers. 
existed. Then consider the natural number 
; this is a natural number larger than 
which is not divisible by any of the primes 
. But, by the 
is enumerated by 
implies that the primes anti-correlate almost completely with the multiplicative function 
, which is self-defeating because it also implies complete anti-correlation with 
, and the two are incompatible. Thus we see that the 
is contained in 
, one sees that 
if and only if 
, a contradiction. 
which contains all other sets (not including itself), because the set 
would then be universal. 
of 
for some collection _%7Bx+%5Cin+X%7D%7D&bg=ffffff&fg=000000&s=0 "{(A_x)_{x \in X}}")
of subsets of 
could be enumerated as 
for some _%7Bx+%5Cin+X%7D%7D&bg=ffffff&fg=000000&s=0 "{(A_x)_{x \in X}}")
. Using the axiom schema of specification, one can then construct the set 
. As 
for some 
. But then by the definition of 
if and only if 
, a contradiction. 
cannot be written as 
for any countable collection 
of real numbers. 
of real numbers. Consider the decimal expansion of each of these numbers. Note that, due to the well-known “
” issue, the decimal expansion may be non-unique, but each real number 
has at most two decimal expansions. For each 
be a digit which is not equal to the 
digit of any of the decimal expansions of 
; this is always possible because there are ten digits to choose from and at most two decimal expansions of 
. (One can avoid any implicit invocation of the 
to be (say) the least digit which is not equal to the 
digit of any of the decimal expansions of 
differs in the 
into 
. But notice how the two arguments have a slightly different (though closely related) basis; the former argument proceeds by encoding the liar paradox, while the second proceeds by exploiting Cantor’s diagonal argument. The two perspectives are indeed a little different: for instance, Cantor’s diagonal argument can also be modified to establish the 
which consists precisely of all the ordinals and nothing else. But then 
is an ordinal which is not contained in 
(by the 
. But one can simply adjoin an additional element not already in 
(e.g. 
) to 
of the countable ordinals. (Here we consider finite sets and countably infinite sets to both be countable.) 
of the countable ordinals. Then the set 
is also a countable ordinal, as is the set 
. But 
is not equal to any of the 
(by the axiom of foundation), a contradiction. 
consists precisely of all the finite natural numbers, but is itself infinite (Proposition 
map, as the set of all finite natural numbers is itself infinite.) In any event, even once one reaches the first uncountable ordinal, one may not yet have completely exhausted the reals; for instance, using the diagonal argument given in the proof of Proposition ![{[0,1]}](http://s1.wordpress.com/latex.php?latex=%7B%5B0,1%5D%7D&bg=ffffff&fg=000000&s=0 "{[0,1]}")
will ever be enumerated by this procedure. (Also, the question of whether all real numbers in ![{[0,1]}](http://s2.wordpress.com/latex.php?latex=%7B%5B0,1%5D%7D&bg=ffffff&fg=000000&s=0 "{[0,1]}")
can be enumerated by the iterated diagonal algorithm requires the 
in some range. For instance, 
is thirty-two characters long.”, and the range of 
, the statement %7D&bg=ffffff&fg=000000&s=0 "{S(T)}")
(formed by replacing every appearance of 
) is either true or false. For instance, %7D&bg=ffffff&fg=000000&s=0 "{S(``a'')}")
is true, but %7D&bg=ffffff&fg=000000&s=0 "{S(``ab'')}")
is false. Crucially, predicates are themselves strings, and can thus be fed into themselves as input; for instance, %7D&bg=ffffff&fg=000000&s=0 "{S(S)}")
is false. If however 
is the predicate “
is sixty-five characters long.”, observe that %7D&bg=ffffff&fg=000000&s=0 "{U(U)}")
is true.
given by
%7D&bg=ffffff&fg=000000&s=0 "{x(x)}")
is false.”
, %7D&bg=ffffff&fg=000000&s=0 "{Q(T)}")
is the sentence
%7D&bg=ffffff&fg=000000&s=0 "{T(T)}")
is false.”
%7D&bg=ffffff&fg=000000&s=0 "{Q(Q)}")
captures the essence of the liar paradox: it is true if and only if it is false. This shows that there is no consistent way to interpret sentences in which the sentences are allowed to come from predicates, are allowed to use the concept of a string, and also allowed to use the concept of truth as given by that interpretation.
%7D&bg=ffffff&fg=000000&s=0 "{T(x)}")
with the set +%5Chbox%7B+true%7D+%5C%7D%7D&bg=ffffff&fg=000000&s=0 "{\{x: T(x) \hbox{ true} \}}")
.
%7D&bg=ffffff&fg=000000&s=0 "{x(x)}")
is not provable”
%7D&bg=ffffff&fg=000000&s=0 "{G := Q(Q)}")
. Then 
which takes a string %7D&bg=ffffff&fg=000000&s=0 "{S(S)}")
, defined as the program (with no input) formed by declaring %7D&bg=ffffff&fg=000000&s=0 "{S(S)}")
does not halt, then %7D&bg=ffffff&fg=000000&s=0 "{Q(Q)}")
will halt if and only if it does not halt, a contradiction. 
computable if there exists a computer program which, when given a natural number 
as input, returns %7D&bg=ffffff&fg=000000&s=0 "{f(n)}")
as output in finite time. Define the 
by setting %7D&bg=ffffff&fg=000000&s=0 "{BB(n)}")
to equal the largest output of any program of at most 
such that one has +%5Cleq+f(n)%7D&bg=ffffff&fg=000000&s=0 "{BB(n) \leq f(n)}")
for all %7D&bg=ffffff&fg=000000&s=0 "{BB(n)}")
grows faster than any computable function: thus if 
is computable, then +%3E+f(n)%7D&bg=ffffff&fg=000000&s=0 "{BB(n) > f(n)}")
for all sufficiently large 
problem
problem is not; see 
appear to be self-defeating; this is most strikingly captured in the celebrated work of Razborov and Rudich, who showed (very roughly speaking) that any sufficiently “natural” proof that 
. The first player can then steal that strategy by placing a naught arbitrarily on the board, and then pretending to be the second player and using 
accordingly. Note that occasionally, the 
strategy will compel the naughts player to place a naught on the square that he or she has already occupied, but in such cases the naughts player may simply place the naught somewhere else instead. (It is not possible that the naughts player would run out of places, thus forcing a draw, because this would imply that 
, then 
guarantees a win for the naughts player; playing the 
strategy for the naughts player against the 
in one inertial reference frame, then by the 
which is immovable in another inertial reference frame which is moving with respect to the first; setting the two on a collision course, we obtain the classic contradiction between an irresistible force and an immovable object. Note however that there are several loopholes here which allow one to avoid contradiction; for instance, the two objects 
could simply pass through each other without interacting.
)
) must intersect if they are sufficiently “generic”; the notion of a wide type is designed, in part, to capture this notion of genericity.
of a language 
, and then consider a universal elementary extension 
of 
is a set of formulae (with 
-definable sets; if 
is a type and 
, then either 
lies in 
) or 
lies in 
) but not both.
is small (i.e. has cardinality less than that of 
, which in particular would be true of %7D&bg=ffffff&fg=000000&s=0 "{p({\Bbb U})}")
of 
that they cut out. In particular, these sets are non-empty. Adding more formulae to a partial type corresponds to shrinking the set that they cut out, and vice versa.
– a type defined over the entire model %7D&bg=ffffff&fg=000000&s=0 "{p({\Bbb U})}")
that they cut out, because these sets are usually empty! However, what we can do is restrict 
over 
.
of 
and 
, we see that 
if and only if %7D%7D&bg=ffffff&fg=000000&s=0 "{p \subset A_{\sigma(b)}}")
, where +%5Cin+A+%5C%7D%7D&bg=ffffff&fg=000000&s=0 "{A_b := \{ a \in {\Bbb U}^n: (a,b) \in A \}}")
is a slice of 
(or in 
). This cuts out a singleton set 
. This is in fact the only invariant global type that cuts out anything at all:
(i.e. %7D&bg=ffffff&fg=000000&s=0 "{p({\Bbb U})}")
is non-empty) if and only if it is realisable in %7D&bg=ffffff&fg=000000&s=0 "{a \in p({\Bbb U})}")
. Since the formula 
is definable in 
, i.e. 
. As 
must then be invariant under all %7D&bg=ffffff&fg=000000&s=0 "{tp(a/M)}")
of 
together with the formulae in %7D&bg=ffffff&fg=000000&s=0 "{tp(a/M)}")
is not satisfiable, hence not finitely satisfiable. Thus there is a finite set of formulae in %7D&bg=ffffff&fg=000000&s=0 "{tp(a/M)}")
that cut out 
is definable over 
carefully; the set +%5Csubset+B(%7B%5CBbb+U%7D')%7D&bg=ffffff&fg=000000&s=0 "{p({\Bbb U}') \subset B({\Bbb U}')}")
holds in all extensions 
of 
by adding in the negations of all the formulae definable over %7D&bg=ffffff&fg=000000&s=0 "{\sigma(\phi)}")
is not in %7D&bg=ffffff&fg=000000&s=0 "{\neg \sigma(\phi)}")
must be in %7D&bg=ffffff&fg=000000&s=0 "{\phi \wedge \neg \sigma(\phi)}")
is in 
fixes %7D&bg=ffffff&fg=000000&s=0 "{\sigma(\phi)}")
does lie in 
, and construct recursively a sequence 
by requiring %7D&bg=ffffff&fg=000000&s=0 "{b_n \in p\downharpoonright_{M \cup \{b_1,\ldots,b_{n-1}\}}({\Bbb U})}")
for all 
. (This is always possible since types are satisfiable once restricted to small sets, by saturation, as discussed earlier). Then 
are indiscernible over %7D&bg=ffffff&fg=000000&s=0 "{(b_{i_1},\ldots,b_{i_k})}")
for 
are elementarily indistinguishable (over 
. 
case is clear since the 
all have type 
over 
case. It suffices to show that %7D&bg=ffffff&fg=000000&s=0 "{(b_1,b_2)}")
and %7D&bg=ffffff&fg=000000&s=0 "{(b_i,b_j)}")
are elementarily indistinguishable over 
.
and 
have the same type over 
, and so %7D&bg=ffffff&fg=000000&s=0 "{(b_1,b_2)}")
and %7D&bg=ffffff&fg=000000&s=0 "{(b_1,b_j)}")
are elementarily indistinguishable over %7D&bg=ffffff&fg=000000&s=0 "{(b_1,b_j)}")
and %7D&bg=ffffff&fg=000000&s=0 "{(b_i,b_j)}")
are elementarily indistinguishable.
and 
have the same type over 
of 
realises 
, we see that +%5Cin+A%7D&bg=ffffff&fg=000000&s=0 "{(b_1,x) \in A}")
, hence by invariance +%5Cin+A%7D&bg=ffffff&fg=000000&s=0 "{(b_i,x) \in A}")
also. Since 
, we conclude +%5Cin+A%7D&bg=ffffff&fg=000000&s=0 "{(b_i,b_j) \in A}")
, as required.
case. The higher 
for some set of constants 
be a sequence of indiscernibles (over 
in this sequence, one can find a type 
such that 
has the same type over 
as 
, for all 
. 
be types over %7D&bg=ffffff&fg=000000&s=0 "{a,a' \in p({\Bbb U})}")
, %7D&bg=ffffff&fg=000000&s=0 "{b,b' \in q({\Bbb U})}")
be realisations of 
such that %7D&bg=ffffff&fg=000000&s=0 "{tp(a'/M \cup \{b'\})}")
are wide. Let 
, 
be +%3E+0%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_a \cap B_b) > 0}")
if and only if +%3E+0%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_{a'} \cap B_{b'}) > 0}")
. 
, and maps 
.
+%3E+%5Cdelta+%3E+0%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_a \cap B_b) > \delta > 0}")
and +=+0%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_{a'} \cap B_{b'}) = 0}")
.
. Observe that for any 
, either one has +%5Cgeq+%5Cepsilon%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_a \cap B_x) \geq \epsilon}")
for all 
, or one has +%5Cleq+%5Cepsilon%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_a \cap B_x) \leq \epsilon}")
for all 
(since there is a 
-definable set between +%5Cgeq+%5Cepsilon+%5C%7D%7D&bg=ffffff&fg=000000&s=0 "{\{ x: \mu(A_a \cap B_x) \geq \epsilon \}}")
and +%3E+%5Cepsilon+%5C%7D%7D&bg=ffffff&fg=000000&s=0 "{\{ x: \mu(A_a \cap B_x) > \epsilon \}}")
. Suppose first that the former option holds for some 
, thus there is a uniform lower bound +%3E+%5Cepsilon%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_a \cap B_x) > \epsilon}")
. We now define a sequence %7D&bg=ffffff&fg=000000&s=0 "{a_1,a_2,\ldots \in p({\Bbb U})}")
and an indiscernible sequence %7D&bg=ffffff&fg=000000&s=0 "{b_1,b_2,\ldots \in q({\Bbb U})}")
as follows:
and 
to be a realisation of 
. 
have been chosen with 
indiscernible. By Lemma %7D&bg=ffffff&fg=000000&s=0 "{a_{n+1} \in p({\Bbb U})}")
that has the same type over 
for all 
. Since +=+0%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_{a'} \cap B_{b'}) = 0}")
, this implies that +=+0%7D&bg=ffffff&fg=000000&s=0 "{\mu( A_{a_{n+1}} \cap B_{b_i} ) = 0}")
for all =0%7D&bg=ffffff&fg=000000&s=0 "{\mu( A_x \cap B_b )=0}")
is a type-definable formula over 
be a realisation of 
. With this construction and Lemma 
is also indiscernible; now we iterate the procedure.
be the set 
, then observe from the above construction that 
lies in +=+0%7D&bg=ffffff&fg=000000&s=0 "{\mu(C_i \cap C_j) = 0}")
for all %7D&bg=ffffff&fg=000000&s=0 "{\mu(C_i)}")
is uniformly bounded away from zero, this contradicts the finite measure of %7D&bg=ffffff&fg=000000&s=0 "{\mu(X)}")
by the pigeonhole principle.
fixing 
. By hypothesis, +%3E+0%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_{a_1} \cap B_{b_1}) > 0}")
, thus there exists a rational 
such that the predicate that models +%3E+r%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_{a_1} \cap B_x) > r}")
is in 
, hence in +%3E+r%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_{a_i} \cap B_x) > r}")
is in +%3E+r%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_{a_i} \cap B_{b_i}) > r}")
, and the claim follows.
+=+0%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_a \cap B_x) = 0}")
for all 
to be a realisation of 
, then set %7D&bg=ffffff&fg=000000&s=0 "{a_n \in p({\Bbb U})}")
so that +%3E+%5Cdelta%7D&bg=ffffff&fg=000000&s=0 "{\mu(A_{a_n} \cap B_{b_n}) > \delta}")
. (This is possible because for any definable set +%5Cgeq+%5Cdelta+%5C%7D%7D&bg=ffffff&fg=000000&s=0 "{\{ x \in C: \mu(A_x \cap B_b) \geq \delta \}}")
contains 
lie in 
be types in 
such that 
has finite measure. Let 
be such that the type of 
is wide. Assume also that the Keisler measure is translation-invariant. Then 
is also wide. 
has zero measure. (Initially, one would need two different definable sets containing %7D&bg=ffffff&fg=000000&s=0 "{a_1, a_2, \ldots \in q({\Bbb U})}")
with %7D&bg=ffffff&fg=000000&s=0 "{tp(a_n/M \cup \{a_1,\ldots,a_{n-1}\})}")
wide for all +=+0%7D&bg=ffffff&fg=000000&s=0 "{\mu(Aa \cap Ab) = 0}")
, we conclude from Lemma =0%7D&bg=ffffff&fg=000000&s=0 "{\mu(A a_n \cap A a_i)=0}")
for all 
. On the other hand, the %7D&bg=ffffff&fg=000000&s=0 "{\mu(Aa_i)}")
all have the same measure as %7D&bg=ffffff&fg=000000&s=0 "{\mu(A)}")
, which is positive. Finally, the 
are all contained in 
, which has finite measure; this leads to a contradictoin. 
+&bg=ffffff&fg=000000&s=0 "\displaystyle \hbox{det} \begin{pmatrix} 1 & x_1 & \ldots & x_1^{n-1} \\ 1 & x_2 & \ldots & x_2^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & \ldots & x_n^{n-1} \end{pmatrix} = \prod_{1 \leq i < j \leq n} (x_j - x_i)")
