I am posting this question in the wake of my similar posting here. What I have concluded there is that infinity $\infty$ stands for something without limit, the $\max (\mathbb N)$ and $\min (\mathbb N)$ do not exist, so do that of the $\mathbb R$. Having said that, let's take a look at this problem down here:
$$\begin{align}
A_t &= \{x \mid t^2 \leq x \leq (t+1)^2 \} \tag{1} \\
\bigcup_{t \in \mathbb N} A_t &= [1, 4] \cup [4, 9] \cup [9, 16]\cup [16, 25] \ldots \tag{2}\\
&= \tag{3}\ldots
\end{align}$$
My first reaction is to take $[1, \infty)$ as the answer since the right element moves to the right without limit. However, in the spirit of yesterday's posting, which do you think is the correct answer and what is your explanation:
(a) $[1, \infty)$, or
(b) $\emptyset$, since $\max (\mathbb N)$ does not exist?
And here is another beginner question:
$$\begin{align}
A_t &= \{x \mid -t \leq x \leq t \} \tag{4} \\
\bigcup_{t \in \mathbb N} A_t &= [-1, 1] \cup [-2, 2] \cup [-3, 3] \ldots \tag{5}\\
&= \tag{6}\ldots
\end{align}$$
Again, which do you think is the correct answer and what is your explanation:
(c) $(-\infty, \infty)$, or
(d) $\emptyset$, since $\max (\mathbb N)$ does not exist?
Thank you for your time and effort.
POST SCRIPT: ~~~~~~~~~~~~~~~~~~~~~
Questions edited per request from moderators. Note also that this question is on the first chapter of Measure & Integration, so the concept of Extended Real Number is not there yet.
Best Answer
First, $\Bbb{N}$ is the set of positive integers, so it is bounded below by zero. This means $\min{\Bbb{N}}=0$ or $1$, depending on whether or not you allow $0$ in the natural numbers. Your intuition on your first question is almost right, but you should find $$\bigcup_{t \in \Bbb{N}} \{x \mid t^2 \leq x \leq (t+1)^2 \} = [1,\infty)$$ because $\infty \notin \Bbb{R}$ as $\infty$ is not a number, and does not belong in the set $\Bbb{R}$ of all real numbers. The half open interval $[1,\infty)$ makes that clear. On your second problem, the answer should be $(-\infty, \infty)$ for similar reasons. For all $y\in \Bbb{R}$ you can find $t \in \Bbb{N}$ such that $|y| \leq t$, and hence $y \in [-t,t]$, which means $$\Bbb{R} \subseteq \bigcup_{t \in \Bbb{N}} \{x \mid -t \leq x \leq t \}$$ The reverse inclusion follows easily since a union of subsets of $\Bbb{R}$ is itself a subset of $\Bbb{R}$, so you have $$\bigcup_{t \in \Bbb{N}} \{x \mid -t \leq x \leq t \} = \Bbb{R} = (-\infty, \infty)$$