0

For each $n \in \mathbb{N}$, define $S_n = \{m \in \mathbb{Z}:m \leq n\}$. Prove that $\bigcup\limits_{n\in \mathbb{N}} S_n = \mathbb{Z}$

So far I have:

Let $x \in \mathbb{Z}$. Assume by contradiction that $x \notin \bigcup\limits_{n\in \mathbb{N}} S_n$. We have three cases.

Case 1: $x$ is positive. Take $S_x = \{m \in \mathbb{Z}:m \leq x\}$. Since $x$

Case 2: $x$ is negative.

Case 3: $x$ is zero.

Am I even on the right track? Do I need to prove that the equality works both ways? Eg. to prove A=B I need to prove A $\subseteq$ B and B $\subseteq$ A. What contradiction would I be looking to find?

Owen Murray
  • 9
  • 4
    You don't need to use contradiction or cases. You certainly have that $S_n \subseteq \mathbb{Z}$ for each $n$. So, you just need to show that every $x \in \mathbb{Z}$ is an element of $S_n$ for some $n$. – J126 Feb 01 '24 at 00:53
  • All you have to do is note that $17$ is in the union, because $17$ is in $S_{18}$ (where $17$ is a variable). – Gerry Myerson Feb 01 '24 at 02:55
  • Any thoughts on the answer and the comments, Owen? – Gerry Myerson Feb 02 '24 at 09:30
  • People are trying to help you, Owen. It's unseemly, not to take any notice of them. – Gerry Myerson Feb 03 '24 at 20:58

1 Answers1

0

Owen. Remember that to establish set equality, you need to prove that $\bigcup_{n \in \mathbb{N}}S_n \subseteq \mathbb{Z}$ and $ \mathbb{Z} \subseteq \bigcup_{n \in \mathbb{N}}S_n$. Regarding the statement $ \mathbb{Z} \subseteq \bigcup_{n \in \mathbb{N}}S_n$, if we know that

$$\bigcup_{n \in \mathbb{N}}S_n = S_1 \cup S_2 \cup S_3 \cdots,$$

When you employ proof by contradiction and assert that $x \notin \bigcup_{n \in \mathbb{N}}S_n$, it is like saying that $x \notin S_1$, $x \notin S_2$, and so on, that is,

$$\forall n \in \mathbb{N}, x \notin S_n.$$

So given that $S_n = \left \{ m \in \mathbb{Z} : m \leq n \right \}$. If you assert that $x \notin S_n$ for all $n \in \mathbb{N}$, then

$$\forall n \in \mathbb{N}, (x \notin \mathbb{Z} \space \vee \space x>n).$$

Clearly, the first statement is false since you've already assumed that $x \in \mathbb{Z}$, so do you think you can you use the second statement to reach a contradiction?

Regarding the statement $\bigcup_{n \in \mathbb{N}}S_n \subseteq \mathbb{Z}$. Assume that $x \in \bigcup_{n \in \mathbb{N}}S_n$. This is similiar to saying, either $x \in S_1$, $x \in S_2$, $x \in S_3$, and so on, but you don't really know which one specifically, but you know there is at least one set where $x$ belongs, in other words

$$\exists k \in \mathbb{N}, x \in S_k.$$

How do you continue from here?

J. J. Gonzalez
  • 87