When does the cardinality disappear? Pardon me, if this question sounds stupid. I am learning real analysis on my own and stumbled on this contradiction while reading this -- http://math.kennesaw.edu/~plaval/math4381/setseq.pdf. I appreciate any help or pointers.
Consider this example:  We choose an open interval of $(0, \frac{1}{n})$ for each n from N, the set of natural numbers i.e. N = {1, 2, 3, ...}. Let us call this interval I(n) as in the pdf.
Now, let us form intersections of all the I(n) we have at any point. I(1) is (0, 1), which has the cardinality of R. Then we form I(2) and look for intersection between I(1) and I(2), which is I(2). When I use n = 3, I get I(3) and the intersection of I(1), I(2), and I(3) is I(3).
But, I can transform $(0, \frac{1}{2})$ and then map it back to R. So |I(2)| = |I(1)| -- that is, the cardinalities of I(1) and I(2) are the same.
Now, I can repeat this argument for each n. Thus, |I(n)| = |I(1)| for all n in N. Thus, the intersection of {I(k)}, k in {1, 2, 3, ..., k} has the cardinality of R.
But, as $n \to \infty$, the intersection is the empty set, whose cardinality is 0.
So, I see that the cardinality of intersections of I(n) went from R to 0. 
But at every step, the cardinality does not change. Nor do I see it decrease.

My question then is -- how does the cardinality simply disappear when I don't see it changing at any step? Did I make one or more wrong assumptions?

 A: Your wrong assumption, if you want to call it that, is that infinite sets behave just like finite sets. This simply isn’t true. In your example every real number in $(0,1)$ gets thrown out at some stage when you intersect more and more of the intervals $I(n)$, so in the end nothing is left. You can make an even simpler example using the positive integers. For $n\in\Bbb Z^+$ let $A_n=\{k\in\Bbb Z^+:k\ge n\}$. For each $n\in\Bbb Z^+$ the map
$$f_n:\Bbb Z^+\to A_n:k\mapsto k+n-1$$
is a bijection, so $|A_n|=|\Bbb Z^+|$. On the other hand, for any $k\in\Bbb Z^+$ we can easily see that if $n>k$, then $k\notin A_n$. Thus, no positive integer is in all of the sets $A_n$, and $\bigcap_{n\in\Bbb Z^+}A_n=\varnothing$, even though for each $m\in\Bbb Z^+$ we have
$$\bigcap_{n=1}^mA_n=A_m\;,$$
a whose cardinality is the same as that of $\Bbb Z^+$.
What it boils down to is that when you deal with infinite sets, you have to retrain your intuition, which initially is derived from your experience with finite sets.
A: In essence this boils down to a question about the interchange of the limit and the cardinality function. In general it is not true that
$$
\lim_{n\to\infty} card(C_n) = card\left(\lim_{n\to\infty} C_n\right)
$$
A: No contradiction here.  In general, even if you have sets $C_n$ with $C_{n+1} \subset C_n$ and $|C_n|$ equal for all $n$, you may still have $\bigcap_{n=1}^{\infty} C_n = \emptyset$.  
One way you can think of this is the following characterization of infinite sets: one can say that a set $A$ is infinite if there is some proper subset $B$ of $A$ for which there is a bijection from $A$ onto $B$ (so their cardinalities are the same).  Note that you cannot do this with finite sets.
A: The problem you are having is that you think if something holds true for every term in a sequence, it must hold true for the limit.  This is definitely not always true.
For example, consider the sequence of real numbers $\{ \frac{1}{n} \}_{n = 1}^{\infty}$.  It's true that for each term in this sequence, the term is $>0$, right?  But this is not true of the limit, since the limit equals $0$.
Similarly, you can have a sequence of functions $f_{n}$ that each are continuous, but the limiting function $f(x) := \lim \limits_{n \to \infty} f_{n}(x)$ is not continuous.  As an example, take $f_{n}(x) = x^{n}$.  For each $n$, $f_{n}$ is a continuous function.  But at $x = 1$, $f(x) = 1$, while for each $x \in [\frac{3}{4}, 1)$, $f(x) = 0$.  So $f(x)$ is not continuous at $x = 1$, even though $f_{n}$ is for each $n$.
