I am struggling to find resources that show how to prove a union/intersection of a family of sets is equal to something. If anybody knows where I can find examples of these types of proofs, it would be greatly appreciated.Can I get a proof verification? Can someone tell me a better way to do this without taking the ceiling? More appreciated would be if someone could direct me to where I can find more practice problems like this.

prove $\bigcup\limits_{n\in \mathbb{N}}(-n,n)=\mathbb{R}$

Prove $\bigcup\limits_{n\in \mathbb{N}}(-n,n)\subset\mathbb{R}$

Since $(-n,n)\subset \mathbb{R}$ for all $n \in \mathbb{N}$ this is proven.

prove $\bigcup\limits_{n\in \mathbb{N}}(-n,n)\supset\mathbb{R}$

let $x \in \mathbb{R}$

If $x \notin \mathbb{Z}$

Take $n =\lceil |x| \rceil$

If $x\in \mathbb{Z}$ Take $n= |x|+1 $

then $\exists n \in \mathbb{N}$ such that $x\in (-n,n)$

so $x\in \bigcup\limits_{n\in \mathbb{N}}(-n,n)$

  • $\begingroup$ Your proof seems to assume that $x$ is positive. For example, if I take $x=-1.5$, then your $n$ would be $-1$, which is not even in $\mathbb{N}$; and the interval $(-n,n)$ would be the interval $(1,-1)$, which makes no sense at all; and even if we assume you mean $(-1,1)$, it does not contain $x$. $\endgroup$ – Arturo Magidin Oct 5 '19 at 3:53
  • $\begingroup$ By the Archemdian property, for every $x \in \mathbb{R}$, there is a natural number $n$ such that $n > \left| x \right|$. So that $\forall x \in \mathbb{R}$, $\exists n \in \mathbb{N}$ such that $x \in \left( -n, n \right)$. This shall complete your proof without using ceil and floor function. $\endgroup$ – Aniruddha Deshmukh Oct 5 '19 at 3:54
  • $\begingroup$ Now corrected.... I would say that you have a step missing. Between your definition of $n$ and your assertion about the existence of an $n$, you should explain why $x\in (-n,n)$ with $n$ defined as you do. However, you will find that a bit difficult, as it is not quite true. $\endgroup$ – Arturo Magidin Oct 5 '19 at 4:00
  • $\begingroup$ Can't one just split into the positive and negative case pretty easily so you don't have to use floor/ceiling/etc. $\endgroup$ – Derek Luna Oct 5 '19 at 4:02
  • $\begingroup$ @ArturoMagidin is that because if $x$ is already a natural number?Would the Archimedean principle take care of that blunder? $\endgroup$ – user707991 Oct 5 '19 at 4:04

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