Pigeonhole Principle for infinite case  Suppose $X_n$ are finite sets for any natural integer $n$. let $Y$ be an infinite subset of $\prod_n X_n$. Do there exist $y$ and $y'$ in $Y$  and an infinite subset $S$ of $\mathbb N$ such that $y_n=y'_n$ for all $n$ in $S$?
 A: First, let me improve upon the countable counterexample of
Darij Grinberg by giving an uncountable counterexample Y.
Indeed, I shall give finite sets Xn and a subset Y of the product ΠXn having size continuum (that is, as large as possible), such that any two distinct y, y' in Y
have only finitely many common values.
Let Xn have 2n elements, consisting
of the binary sequences of length n. Now, for each infinite
binary sequence s, let ys be sequence in the
product ΠXn whose nth value is
s|n, the length n initial segment of s. Let Y consist of
all these ys. Since there are continuum many s,
it follows that Y has size continuum.
Note that if s and t are distinct binary sequences, then eventually
the initial segments of s and t disagree. Thus, eventually,
the values of ys and yt are
different. Thus, ys and yt have only
finitely many common values. So Y is very large counterexample, as
desired.
A similar argument works still if the Xn grow
more slowly in size, as long as liminf|Xn| =
infinity. One simply spreads the construction out a bit
further, until the size of the Xi is large
enough to accommodate the same idea. That is, if the liminf
of the sizes of the Xn's is infinite, then one
can again make a counterexample set Y of size continuum.
In contrast, in the remaining case, there are no infinite
counterexamples. I claim that if infinitely many
Xn have size at most k and Y is a
subset of ΠXn having k+1 many elements, then
there are distinct y,y' in Y having infinitely many common
values. To see this, suppose that Y has the property that
distinct y, y' in Y have only finitely many common values.
In this case, any two y, y' must eventually have different
values. So if Y has k+1 many elements, then eventually for
sufficiently large n, these k+1 many sequences in Y must be
taking on different values in every Xn. But
since unboundedly often there are only k possible values in
Xn, this is impossible.
In summary, the situation is as follows:
Theorem. Suppose that Xn is finite and
nonempty.


*

*If liminf |Xn| is infinite, then there is Y
subset ΠXn of size continuum, such that
distinct y, y' in Y have only finitely many values in
common.

*Otherwise, infinitely many Xn have size at most k for some k, and in this case, every Y subset
ΠXn of size k+1 has distinct y,y' in Y with infinitely
many common values.


In particular, if the Xn become increasingly
large in size, then there are very bad counterexamples to the
question, and if the Xn are infinitely often
bounded in size, then there is a very strong positive answer to
the question.
A: What about $X_n=\left\lbrace 1,2,3,...,n\right\rbrace$, and $Y$ consisting of the sequences $\left(1,1,1,...,1,2,3,4,5,6,...\right)$ with $n$ ones for all $n\in\mathbb N$?
The first true thing that comes into my mind when I hear "Pigeonhole Principle for infinite case" are some theorems in infinite Ramsey theory, such as this one.
A: I think that what you are asking for is impossible.  Given any element of $\prod_n X_n$ the element is uniquely determined by its image in each of the individual $X_n$'s.  So if two elements of $Y$ agree on each $X_n$ then they must be the same element.
In language similar to yours, what you probably want for the finite case is "If $X$ and $Y$ are finite sets such that $|X| < |Y|$ and $f:Y\rightarrow X$ is any map then there exists an element $x\in X$ such that $|f^{-1}(x)| > x$."  More generally, given any finite sequence $X_1,\ldots,X_n$ of finite sets and any set $Y$ such that $|Y| > |X_1|\cdot|X_2|\cdots|X_n|$ and any sequence of maps $f_i:Y\rightarrow X_i$ then there exists a sequence of elements $x_1,\ldots,x_n$ and two elements $y,y'\in Y$ such that $f_i(y) = f_i(y')$ for any $i$.
The problem with the infinite case is that there are injective but not surjective maps between infinite sets with the same cardinality. However, it is true that given a sequence of finite sets $X_1,X_2,\ldots$ and a set $Y$ with cardinality greater than that of $\prod X_n$, if you have any sequence of maps $f_i:Y\rightarrow X_i$ then there exists an uncountable subset $Z\subseteq Y$ such that for any two elements $z,z'$ of $Z$ you have $f_i(z) = f_i(z')$ for all $i$.
In even more generality, I believe that if you have any set of sets $\{X_\alpha\}$ and any set $Y$ such that the cardinality of $Y$ is larger than the cardinality of $\prod_\alpha X_\alpha$ then you have a similar statement.
