If $A_n=\{(n+1)k:k\in\mathbb{N}\}$ determine the sets $\bigcup\{A_n:n\in\mathbb{N}\}$ and $\bigcap\{A_{n}:n\in\mathbb{N}\}$?

Question

I'm a beginner. If I made a mistake explain everything as clearly as possible without skipping a step. (My professor insists to not skip a step.)

Question: If $A_n=\{(n+1)k:k\in\mathbb{N}\}$ determine the sets $\bigcup\{A_n:n\in\mathbb{N}\}$ and $\bigcap\{A_{n}:n\in\mathbb{N}\}$

Hypothesis: $\bigcup\{A_n:n\in\mathbb{N}\}=\mathbb{N}-\{1\}$ and $\bigcap\{A_{n}:n\in\mathbb{N}\}=\emptyset$.

Proof: Assume $x\in\bigcup\{A_n:n\in\mathbb{N}\}$. Since $A_n\subseteq \mathbb{N}-\{1\}$, for all $n\in\mathbb{N}$, $\left(\right.$since if $x=(n+1)k$, for $n,k\in\mathbb{N}$, then $x\in\mathbb{N}-\{1\}$ $\!\!\left.\right)$, we have $A_n\subseteq\mathbb{N}-\{1\}$ for some $n\in\mathbb{N}$. Thus $\bigcup\{A_n:n\in\mathbb{N}\}\subseteq\mathbb{N}-\{1\}$. Moreover, if $x\in\mathbb{N}-\{1\}$, then $x$ can be the product of one or more natural numbers greater than one. Hence, there exists some natural numbers in $\mathbb{N}$ whose product is $x\in\mathbb{N}$. Therefore, if $x=(n+1)k$, for some $n,k\in\mathbb{N}$, then $x\in\bigcup\{A_n:n\in\mathbb{N}\}$. Thus, $\mathbb{N}-\{1\}\subseteq\bigcup\{A_n:n\in\mathbb{N}\}$.

Since $\bigcup\{A_n:n\in\mathbb{N}\}\subseteq\mathbb{N}-\{1\}$ and $\mathbb{N}-\{1\}\subseteq\bigcup\{A_n:n\in\mathbb{N}\}$, we have $\mathbb{N}-\{1\}=\bigcup\{A_n:n\in\mathbb{N}\}$.

Suppose $\bigcap\{A_n:n\in\mathbb{N}\}\neq\emptyset$. If $x\in\bigcap\{A_n:n\in\mathbb{N}\}$, then all $A_n$ intersect eachother based on the lowest common multiple of $\{(n+1):n\in\mathbb{N}\}$. This is determined by natural numbers with the most prime factors. Since there are infinite prime factors, their product is $\infty$ which should be the intersection but $\infty$ is not a number. Hence it is not an element of $\mathbb{N}$. Therefore, $\bigcap\{A_n:n\in\mathbb{N}\}$ is empty. This is a contradiction to the first statement that $\bigcap\{A_n:n\in\mathbb{N}\}$ is not empty, thus $\bigcap\{A_n:n\in\mathbb{N}\}=\emptyset$.

Question: Is my proof correct? What is a shorter proof that a begginer like me can write?

We should have that $1\in A_0$ as $1=(n+1)k$ for $n=0$ and $k=1$. — D.Q., May 05 '21 at 18:39
The problem specificallyt states that $n\in N$ so n cannot be 0. — user247327, May 05 '21 at 18:47
@user247327: Every time you say the word $\Bbb N$, you also have to say whether $0$ is included in it. It's one of those nuisances that never got fixed. — TonyK, May 05 '21 at 18:54

user394334 · Answer 1 · 2021-05-05T18:50:37.930

1

A shorter proof might be:

Let $x$ be a natural number, but $x\ne 1$. Then $x\in A_{x-1}$ since $(x-1+1)\cdot1=x.$

$x =1$ is not in any $A_n$ since $y\in A_n \rightarrow y>n$. Hence $\cup A_n = \mathbb{N}\backslash\{1\}$.

Assume now that $x\in \mathbb{N}$. Then $x$ is not an element of $A_{x}$ since we had $y\in A_n \rightarrow y>n$. Hence $\cap A_n =\emptyset$. Because no element $x$ is in all $A_n$.

edited May 05 '21 at 18:50

answered May 05 '21 at 18:41

user394334

1,707

@00GB Yes, I assumed $0\notin \mathbb{N}$. – user394334 May 05 '21 at 18:54
@00GB $2\cdot 3 \in 2\mathbb{N}$ and $3\cdot 2 \in 3\mathbb{N}$. – user394334 May 05 '21 at 18:57
ueser394334, Thank you I did not pay attention when I wrote – Gob May 05 '21 at 19:01

If $A_n=\{(n+1)k:k\in\mathbb{N}\}$ determine the sets $\bigcup\{A_n:n\in\mathbb{N}\}$ and $\bigcap\{A_{n}:n\in\mathbb{N}\}$?

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