lim sup of nonoverlapping subintervals

Question

On the interval $A = (0, 1]$, define the subintervals $A_0 = (0, \frac{1}{2}]$, $A_1 = (\frac{1}{2}, \frac{3}{4}]$, $A_2 = (\frac{3}{4}, \frac{7}{8}]$, $\ldots$, $A_n= (\frac{2^n-1}{2^n}, \frac{2^{n+1}-1}{2^{n+1}}]$, $\ldots$.

Am I right to say that $\limsup_{n\rightarrow \infty} A_n = \{1\}$? And if so, how should I reconcile this with the usual interpretation of lim sup that "$\limsup_{n\rightarrow \infty}A_n$ is the set of outcomes that occur infinitely many times within the infinite sequence of events $A_n$", since $\{1\}$ only occurs once?

Answer 1

You are wrong. Since $\limsup_nA_n$ is the set of all elements of $A$ that belong to infinitely many $A_n$'s, $\limsup_nA_n=\emptyset$.

Answer 2

Let's define the limsup of a sequence of sets as follows:

$$\limsup A_n =\bigcap_{n \geq 1}\bigcup_{k \geq n} A_k$$ Using this definition which is the natural generalization of the limsup of a sequence of numbers, you should be able to show that $\limsup A_n = \emptyset$.