finding the cardinality

Question

I was trying to show the complementary set is finite .I got the complenentary set as $\{ k\in \mathbb{N}: P_{k+1} - p_k=2 \}$. Don't know know how to show the set is finite. Any help would be appreciated. Thanks in advance.

Answer 1

Consider the numbers $$n!+2,n!+3,\ldots,n!+n $$ for an arbitrary positive integer $n$. These numbers are all composite and successive. So you can by considering $n$ big enough, there is index $k$ in which $p_{k+1}-p_k>n-1>2$. By changing $n$ infinitely many times, we reach the desired quest.

Answer 2

Lemma: Given the three numbers $k-2$ and $k$ and $k+2$ exactly one of them is divisible by $3$.

Pf: Let $r$ be the remainder when you divide $k$ by $3$. Either $r = 0, 1,$ or $2$. There are no other options. If $r=0$ then $k = 3m$ for some integer $m$ and $k$ is divisible by $3$. If $r = 1$ then $k = 3m + 1$ for some integer $m$ and $k +2 = 3m + 3 = 3(m+1)$ and $k+2$ is divisible by $3$. If $r=2$ then $k = 3m +2$ for some integer $m$ and $k -2 = 3m$ and $k-2$ is divisible by $3$.

Do you see how that lemma means that there have to be an infinite number of $k$ where $p_{k+1} - p_k > 2$? Try to think it out.

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I was trying to show the complementary set is finite.

It's possible that both sets are infinite.

If I asked you to prove there were an infinite number of even numbers, it would be a bad idea for you to try to prove that there are finite number of odd numbers. Yes, if there are a finite number of odd numbers then the compliment must be infinite. But if a set is infinite it does not mean the compliment is finite.