Question about equivalence of infinite sets If there is a one to one correspondence between sets, then they are said to be equivalent. Explicitly, $A$ is equivalent to $B$, denoted $A \sim B$, if there is a one to one and onto function from $A$ to $B$. It can be shown that $\sim$ is an equivalence relation. In particular, this implies that if $A \sim B$ and $B \sim C$, then $A \sim C$.
Here is my question:
Now consider the intervals $I_n=[0,1/n]$. Then $I_1 
\sim I_{2} \sim I_3...$ From this we have that $I_1 \sim I_n$ for all $n$. Now $\lim _{n \to \infty}I_n=\{0\}$. So $I_n \sim \{0\}$. However, this is not the case.
I am thinking that what went wrong here is that $I_1 \sim I_n$ does not imply that $I_1 \sim \lim_{n \to \infty} I_n$. Does that mean that the transitivity property of equivalence relation doesn't extend to infinity?  
 A: You have a good counterexample of $I_1 \sim \lim_{n\to \infty} I_n$ right there, so yes.  It seems natural that this example doesn't work; maybe I can help convince you:
Let $f_n:\mathbb{R} \to \mathbb{R}$ be given by $f_n(x) = x/n$.  Then $f_n$ is a one-to-one and onto map of $[0,1] = I_1$ to $I_n$.  The transitive property holds because if $I_{\ell} \sim I_m \sim I_n$, then we have bijections $g:I_{\ell} \to I_m$ and $h: I_m \to I_n$, so $h\circ g$ is a bijection between $I_{\ell}$ and $I_n$.  In our case, $f_n \circ f_m^{-1}$ gives a bijection of $I_m$ onto $I_n$, demonstrating that.
The problem in taking the limit is that $\displaystyle f(x) := \lim_n f_n(x) = \lim_n x/n = 0$ for all $x$.  Basically, we're trying to `divide by infinity' in some sense, and this operation is not a $1-1$ map.
Because the setting was analytic (multiplication provided the mapping) we were able to talk about limits of these maps, and in this setting we could phrase your statement as 

"Limits of one-to-one functions are not necessarily one-to-one."
