Could you explain why we can assume infinity in limits? (if at all?)

Question

I am working on a proof of in which I have to assume infinity. Am I allowed to assume the reader is familiar with how limits work?

Answer 1

We have to prove the following.

$ \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6] = [3, 6] $ where $ n \in \mathbb{N}$

This means $x \in \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6] \iff x \in [3,6]$

Proof;

Let $x \in \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6]$

Then $3- \frac{1}{n} \leq x \leq 6$ for all n. Now, let $n \rightarrow \infty$ then it follows , $3 \leq x \leq 6$

Hence $x \in [3,6]$ and $ \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6] \subseteq [3,6]$

For the reverse conditional we have,

let $x \in [3,6]$ then $6 \geq x \geq 3$.

Given that $3-\frac{1}{n} < 3$ for all n, it follows that $x \geq 3 > 3 - \frac{1}{n}$

and hence $[3 - \frac{1}{n}, 6] $ is a bigger interval than [3, 6] for all n.

Therefore [3, 6] $ \subseteq \bigcap_{n=1}^{\infty}A_{n}$ and $x \in \bigcap_{n=1}^{\infty}A_{n}$

$\therefore \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6] = [3, 6] \blacksquare$

Answer 2

In this answer, I would like to focus on your Lemma (since proving this is similar to your case and since I don't like providing direct solutions, you still gotta work for it). Let me also emphasize that if you have proven this lemma, you still are not done by far. If I were you, I would try to prove the statement similar as to the lemma which I will show below.

So, you want to prove that $\bigcup_{n=1}^{\infty} [3-n^{-1},6] = [2,6]$.

Let $x\in\bigcup_{n=1}^{\infty} [3-n^{-1},6]$. Then you claim something which is not very well formulated (since you are not using standard mathematical reasoning), I would do it as follows. Hence, per definition, there exists an $m\in\{1,2,\dots\}$ such that $x\in[3-m^{-1},6]$. Since $m\geq 1$, we have $3-m^{-1}\geq 2$ and thus $[3-m^{-1},6]\subseteq [2,6]$ which implies that $x\in[2,6]$. This is approximately similar to your reasoning, however, this is way more direct and using the definitions.

The other way around is treated wrongly by you. You would have to argue as follows. Let $x\in[2,6] = [3-1^{-1},6]$ and thus $x\in \bigcup_{n=1}^{\infty} [3 - n^{-1},6]$ since this set would only be bigger.