Particular Example of a Subsequence: Reasoning to How it Works

Question

Say the function $x: \mathbb{N}\rightarrow \mathcal{P}(\mathbb{R})$ is a sequence of subsets of $\mathbb{R}$ given by the correspondence $x_n=[\frac{1}{n}, 1]$. The following is stated to be true:

$\bigcup_{i=1}^\infty[\frac{1}{i},1]=\bigcup \underbrace{range(x)}_{=\lbrace [\frac{1}{n}, 1]|n\in \mathbb{N}\rbrace}=(0, 1]$.

$\textbf{Question:}$ How does one formally prove that this is true? In other words, how do I prove this set becomes $(0, 1]$?

I understand the $\lim_{n\rightarrow \infty}\frac{1}{n}=0$ and that $\lim_{n\rightarrow \infty}1=1$. But I was just curious as to how the union comes into play to prove something like this.

Answer 1

To prove that two sets $A$ and $B$ are equal, it's usually easiest to break it into two parts:

1. Prove that $A\subseteq B$. That is, take an arbitrary $a\in A$, and show that it must be true that $a\in B$.
2. Prove that $B\subseteq A$. That is, take an arbitrary $b\in B$, and show that it must be true that $b\in A$.

e.g.

1. Let $y\in \bigcup_{i=1}^\infty [\frac{1}{i}, 1]$. Then $y\in [\frac{1}{n},1]$ for some integer $1\leq n<\infty$, by definition of $\cup$. So we must have $y\in (0,1]$.
2. Let $y\in (0,1]$. Then $y\leq 1$; and there must be some large integer $n$ such that $n \geq \frac{1}{y}$, i.e. $\frac{1}{n} \leq y$, so $y\in [\frac{1}{n},1]$, and so by definition of $\cup$, we must have $y\in \bigcup_{i=1}^\infty [\frac{1}{i}, 1]$.

Here, when I say "definition of $\cup$", what I mean is:

• $y\in \bigcup_i C_i$ if and only if $y\in C_n$ for some $n$.

Edit: oops, changed my notation so it doesn't clash with yours.