Infinite unions and intersections of intervals

Question

Given two real numbers $a

Can someone please explain, because I keep staring at it and don't know where to even start.

Answer 1

Intuition

The left bounds of the intervals $A_n$ $$ \frac{1}{2}-\frac{1}{2n} $$ increase from $0$ (when $n=1$) as $n$ grows and, at the limit, tend to $1/2$.

The right bounds of the intervals $A_n$ $$ \frac{1}{2}+\frac{1}{3n} $$ decrease from $5/6$ (when $n=1$) as $n$ grows and, at the limit, tend to $1/2$.

Noting that $A_{n}\supset A_{n+1}$, we naturally conjecture that the reunion of all $A_n$ is simply $A_1$, since $A_1$ contains all other intervals.

Intuitively, as $n$ grows, $A_n$ "shrinks" to the singleton $\{1/2\}$. This leads us to conjecture that the intersection of all $A_n$ is that singleton.

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Proofs

To prove that two sets $A$ and $B$ are equal according to the definition of equality of sets, we must show that they are subsets of each others. That is, we must show that $A\subseteq B$ and $B\subseteq A$.

To show that a set $A$ is a subset of a set $B$, we must consider an arbitrary element in $A$ and show that it also belongs to $B$.

$\displaystyle\bigcup_{n\in\mathbb{N}}A_n=A_1$

$[\subseteq]$ Let $x$ be an arbitrary element in $\bigcup_{n\in\mathbb{N}}A_n$. By definition of $\bigcup_{n\in\mathbb{N}}A_n$, this means that $x$ is in $A_{n_0}$ for some $n_0 \in \mathbb{N}$. In other words, $x$ is in (at least) one of the intervals $A_n$, say $A_{n_0}$. Now by definition of $A_{n_0}$, this means that $$ \frac{1}{2}-\frac{1}{2n_0} < x < \frac{1}{2}+\frac{1}{3n_0}\tag{1} $$ Our goal is to show that $x$ is also in $A_{\color{#090}{1}}$. Hence we need to deduce $$ \frac{1}{2}-\frac{1}{2\cdot\color{#090}{1}} < x < \frac{1}{2}+\frac{1}{3\cdot\color{#090}{1}} $$ which is equivalent to $$ 0 < x < \frac{5}{6}\tag{2} $$ from the hypothesis $(1)$. How to do so? Well, we'll have to use some properties (obviously taken for granted unless you're in a class dealing with an axiomatic construction of the real numbers) of the order relation $<$ on $\mathbb{R}$. You should know (from highschool, I guess) that if $n_0\in\mathbb{N}$ then $$ \frac{1}{2n_0}<\frac{1}{2} $$ so that $$ -\frac{1}{2}<-\frac{1}{2n_0} $$ and $$ \underbrace{\frac{1}{2}-\frac{1}{2}}_{0}<\frac{1}{2}-\frac{1}{2n_0} $$ Hence $$ 0<\frac{1}{2}-\frac{1}{2n_0} $$ and in view of $(1)$ $$ 0<\frac{1}{2}-\frac{1}{2n_0}

I let you prove similarly that from $$ x < \frac{1}{2}+\frac{1}{3n_0} $$ we may deduce $$ x < \frac{5}{6} $$ All in all, we have shown that the $x$ in $\bigcup_{n\in\mathbb{N}}A_n$ is also in $A_1$. Since $x$ was arbitrary, this shows the desired inclusion $\bigcup_{n\in\mathbb{N}}A_n\subseteq A_1$.

$[\supseteq]$ If $x\in A_1$ then obviously $x\in\bigcup_{n\in\mathbb{N}}A_n$. There is no need to add anything more, really. $\blacksquare$

Note 1: It is unclear how detailed you should provide your proof. Surely, the above proof of $[\subseteq]$, meant to be pedagogical, is a little long and could (should?) be shortened for better readability.

It is important to be rigorous, but it is equally important to be sure that the reader who's trying to read your mind doesn't get lost in trivial details he could supply by himself very easily.

The next proof will be given more succinctly (and will be absolutely as correct).

$\displaystyle\bigcap_{n\in\mathbb{N}}A_n=\{1/2\}$

$[\subseteq]$ Let $x$ in $\bigcap_{n\in\mathbb{N}}A_n$. Then $$ \frac{1}{2}-\frac{1}{2n} < x < \frac{1}{2}+\frac{1}{3n}\quad(\text{for all }n\in\mathbb{N})\tag{3} $$

First, let's show that $x<1/2$ is impossible. Suppose that $x<1/2$. Then $x=1/2-\epsilon$ for some $\epsilon>0$. In view of $(3)$, it follows that $$ \frac{1}{2n} > \epsilon\quad(\text{for all }n\in\mathbb{N}) $$

This contradicts the Archimedean property which assures us that we can make $\frac{1}{2n}$ as small as we want (eventually smaller than $\epsilon$) by taking $n$ sufficiently large. Hence it is not the case that $x<1/2$.

Similarly, we see that $x>1/2$ is impossible.

By trichotomy, it must then be the case that $x=1/2$. Hence $x\in\{1/2\}$.

Since $x$ was arbitrary, this shows the desired inclusion.

Note 2: If you have the right to use the framework of limits, then $x=1/2$ is a direct consequence of $(3)$ by using the squeeze theorem. But I doubt you are allowed that.

Note 3: The Archimedean property of $\mathbb{R}$ is generally taken for granted, even though it can be shown that it is a consequence of more fundamental axioms. Same thing for the trichotomy of the order relation on $\mathbb{R}$.

$[\supseteq]$ Since $$ \frac{1}{2}-\frac{1}{2n} < \frac{1}{2} < \frac{1}{2}+\frac{1}{3n} $$ for all $n\in\mathbb{N}$, it follows that $1/2\in A_n$ for all $n\in\mathbb{N}$. This means $1/2\in\bigcap_{n\in\mathbb{N}}A_n$. The inclusion is thus proved. $\blacksquare$