Consider the standard topology on \(\Bbb R\) generated by the open intervals.

Consider the class of sets \(\{A_n\}_{n\geq 0}\) where \(A_n\subseteq\Bbb R\) is defined by, \[A_n=\left\{\sum\limits_{k=1}^{n}\frac{1}{p_k}\mid p_k\in\Bbb N~\forall~1\leq k\leq n\right\}~\forall~n\geq 1\] and \(A_0=\{0\}\)

Which one of the following options is correct?

1) \[(A_n)^\prime=\bigcap_{i=0}^{n-1} A_i\]

2) \[(A_n)^\prime=\left(\bigcup_{i\textrm{ odd}}A_i\right)\cap\left(\bigcup_{i\textrm{ even}}A_i\right)\]

3) \[(A_n)^\prime=\bigcup_{i=0}^{n-1} A_i\]

4) \[(A_n)^\prime=\left(\bigcap_{i\textrm{ odd}}A_i\right)\cup\left(\bigcap_{i\textrm{ even}}A_i\right)\]

**Details and Assumptions:**

For a subset \(X\) of a topological space, \(X^\prime\) denotes the derived set (set of all limit points) of \(X\).

\(\cup\) and \(\cap\) denote set union and intersection respectively.

\(\Bbb N\) and \(\Bbb R\) denotes the set of natural numbers and real number, respectively.

×

Problem Loading...

Note Loading...

Set Loading...