Countable unions and the whole real line R

Question

It has been a while since my undergraduate introductory analysis course, so I am a little rusty. I know that to show two sets $B, C$ are equal, you first show that $B\subseteq C$ and then $C\subseteq B$ . To show $B\subseteq C$ you take any element of $B$, say $b \in B$ and show that $b \in C$.

The main issue I have is actually doing this for some inclusions. Like using the Archimedean Property and so on. Now to my question:

How would you show, for $a,b\in \Bbb{R}$ with $a<b$, that $$\bigcup^\infty_{n=1}(a,b-\frac{1}{n}]=(a,b)$$

where $\bigcup^\infty_{n=1}(a,b-\frac{1}{n}]$ is a countable union of half open intervals.

Also how would you show that $$\bigcup^\infty_{n=1}(-n,n)=\Bbb{R}$$ Note $n\in\Bbb{N}$ here.

Edit 1: For the second one, I think I have the direction $\Bbb{R}\subseteq\bigcup^\infty_{n=1}(-n,n)$.

Consider $x\in\Bbb{R}$. We have three cases $x>0$ and $x<0$ and $x=0$. For $x=0$, we have that for all $n\in\Bbb{N}$, $-n<0<n$ so $0\in\bigcup^\infty_{n=1}(-n,n)$.

Now for any positive real number ,there exists an $N\in\Bbb{N}$ such that $N\leq x < N+1$.

Now for $x<0$ we have that $-x>0$. By what I stated previously, there exists $m\in\Bbb{N}$ such that $m\leq -x < m+1$ . So $-m-1 < -x \leq -m <m+1$ and so $x \in(-m-1, m+1)$. Thus $x \in\bigcup^\infty_{n=1}(-n,n)$. The case for $x>0$ is similar.

But I am unsure of how to show the other inclusion $\bigcup^\infty_{n=1}(-n,n)\subseteq\Bbb{R}$

Answer 1

Do you know when two sets are equal?

Say $B=(a,b)$ and $$A=\bigcup^\infty_{n=1}(a,b-\frac{1}{n}]$$ You must prove $A=B$.

So take any $x\in A$ and prove that it is also in $B$. Then do vice versa.

Answer 2

By double inclusion. For the first question, because for any positive integer $n$, we clearly have $(a,b-\frac{1}{n}]\subset(a,b)$, we know that $\bigcup^\infty_{n=1}(a,b-\frac{1}{n}]\subset (a,b)$

Now consider any element $x \in (a,b)$. Then, $x<b$ and because $\lim_{n\rightarrow \infty}b-\frac{1}{n}=b$, by very definition of the limit, we know that there is some integer $n$ such that $x<b-\frac{1}{n}<b$. Hence, $x\in (a,b-\frac{1}{n}]$. It follows that $(a,b)\subset \bigcup^\infty_{n=1}(a,b-\frac{1}{n}]$, so we are done.

With a similar reasoning, one can show your second question concerning $\mathbb{R}$.

Answer 3

Suppose $\exists x \in (a,b)$ s.t. $x \notin\bigcup_{n=1} ^{\infty}(a, b-\frac{1}{n}]$

Then $b-\frac{1}{n}<x$ $\forall n \in \mathbb{N}$

But $$\lim_{n \rightarrow \infty}b-\frac{1}{n}=b$$ which implies that $b \leq x$ by algebra of limits which is a contradiction.

Hence $(a,b) \subset \bigcup_{n=1} ^{\infty}(a, b-\frac{1}{n}]$

Similarly you can prove the other direction.

The second problem is proven simialrly.

Answer 4

Part 2:

Let $n \in \mathbb{Z^+}.$

1) $\bigcup_{n=1}^{\infty} (-n,n) \subset \mathbb{R}.$

Trivial.

Now show that $\mathbb{R} \subset \bigcup_{n=1}^{\infty} (-n,n)$.

2) Let $r$ be real, positive.

Archimedean principle:

There is a $n \in \mathbb{Z^+}$ with $n>r.$

Hence $r \in (-n,n).$

2) Let $r <0.$

Then consider $-r >0.$

Archimedean principle :

There is a $n \in \mathbb{Z^+}$ with

$n >-r .$

Hence $-n < r$ , and $r \in (-n,n)$.

3)Let $r = 0;$ $0 \in (-1,1)$.

Altogether:

$\mathbb{R} \subset \bigcup_{n=1}^{\infty} (-n,n)$.