Prove $\bigcup_{n=1}^{\infty} [1/n^{2}, \infty) = (0, \infty)$ Hi I am very confused about index sets.
My effort:
Let $x\in (0,\infty)$ since $(0,\infty)$ is open, $\exists y>0 \ni x\in (x-y,x+y)\subset (0,\infty)$. Since $x-y>0, \exists n_1 \in \mathbb{N}$ such that
$\frac{1}{n_1^2}<\frac{1}{n_1}<x-y$, $x\in \big[\frac{1}{n_1^2},\infty\big)$, $x\in \bigcup_{n=1\rightarrow \infty} \big[\frac{1}{n_1^2} , \infty\big)$
$\big[\frac{1}{n_1^2} , \infty\big)\subset (0,\infty), \forall n\in \mathbb{N}$, so $\bigcup_{n=1\rightarrow \infty} \big[\frac{1}{n_1^2} , \infty\big) \subseteq (0,\infty)$
Putting the two pieces together, we have $\bigcup_{n=1\rightarrow \infty} \big[\frac{1}{n_1^2} , \infty\big)=(0,\infty)$
Am I right? Please help, thank you!
 A: Your effort is quite difficult to read/understand, but here's perhaps a clearer explanation: we know that as $n\to\infty$, $\frac 1{n^2} \to 0$ (i.e. $\lim_{n\to\infty} \frac 1{n^2} = 0$). That is to say, for every positive number $\epsilon$, there will be some $N$ (depending on $\epsilon$) for which all $n\geq N \implies \frac 1{n^2} < \epsilon$. Applying this to any $x \in (0,\infty)$, we see that there must be some $N$ (depending on $x$) such that $\frac 1{N^2} < x$, i.e. $x \in [\frac{1}{N^2},\infty)$. This means that $(0,\infty)$ is contained within the infinite union $\bigcup_{n=1}^\infty [\frac{1}{n^2},\infty)$.
For the other direction of containment, consider any $x \leq 0$ -- we show that such $x$ can not be in the infinite union. Note that for any $n\in \mathbb N$, $x$ can not possibly be in $[\frac{1}{n^2},\infty)$ because $[\frac{1}{n^2},\infty)$ contains strictly positive numbers, and we said that $x$ was not a positive number. Thus, $x$ is not in the infinite union, because $x$ is not in any of the $[\frac{1}{n^2},\infty)$, and we are done.
A: Good effort, and thanks for using latex!
In regards to your attempt:
You use the fact that $(0,\infty)$ is an open interval to justify the fact that $\exists y \big(x-y > 0\big)$. This result is indeed true, but you haven't proved it. It's probably fine to assume that result (assuming you've seen it/proved it before). But you go on to say that because $x-y>0$, then $\exists n_1 \in \mathbb{N} \big(\frac{1}{n_1^2} < x-y\big)$. Again, this statement is true, but you have not proved it.
Instead, to prove it we should use an argument such as: Suppose $x\in (0, \infty)$. Suppose $n_1 > \frac{1}{\sqrt{x}}$. This implies that $n_1^2 > \frac{1}{x}$, which implies $\frac{1}{n_1^2} < x$. Hence for any $n_1 > \frac{1}{\sqrt{x}}$ (for example, take $\lceil \frac{1}{x^2}\rceil$), $x \in \big[\frac{1}{n_1^2}, \infty\big] \implies x \in \bigcup_{i=1}^\infty \big[\frac{1}{n_1^2}, \infty\big]$. Note that we don't need to use any result about the set being open in fact.
For the second part of your argument, I am not sure if that's a typo, or if you are trying to take a limit. Even if you did not mean to take a limit, I think your argument that since $\forall n \in \mathbb{N}\big(\big[\frac{1}{n^2},\infty\big]\subset (0,\infty)\big)$ then $\bigcup_{i=1}^\infty \big[\frac{1}{n_1^2}, \infty\big] \subset (0, \infty)$, is unjustified, and perhaps doesn't have details that you would be expected to include when writing these types of proofs.
Remember that we are ultimately trying to show that $\bigcup_{i=1}^\infty \big[\frac{1}{n_1^2}, \infty\big] \subset (0, \infty)$, so we should take an arbitrary element of $\bigcup_{i=1}^\infty \big[\frac{1}{n_1^2}, \infty\big]$ and show this is also an element of $(0,\infty)$.
Here is how we can do it: Suppose $x\in \bigcup_{i=1}^\infty \big[\frac{1}{n_1^2}, \infty\big]$. Then $\exists i \in \mathbb{N}\big(x \in \big[\frac{1}{i^2}, \infty \big]\big)$. But $\frac{1}{i^2} > 0$, so $x \in (0,\infty)$. Note here we use the definition of what the union is to justify the existence of some $i \in \mathbb{N}$
