Need help showing that $\bigcap_{n\in\mathbb{N}}\left(0,1/n\right]=Ø$

Question

So in order to show that $$\bigcap_{n\in\mathbb{N}}\left(0,1/n\right]=Ø$$

Then we must show that $$\bigcap_{n\in\mathbb{N}}\left(0,1/n\right]\subseteq Ø$$ and the opposite

So what I was thinking. Letting $x\inØ$ then $x=Ø$ s.t for some $n\in\mathbb{N}$ so $Ø\in\bigcap_{n\in\mathbb{N}}\left(0,1/n\right]$. Since $x\inØ$ then $x\in\left(0,1/n\right]$. It follows that $Ø\subseteq\bigcap_{n\in\mathbb{N}}\left(0,1/n\right]$. Now for the opposite I don't quite know how to start. I was thinking of going for a contrapositive approach, but I don't know how to execute it after saying that $x\notin Ø$. I have a hard time proving theorems and any proof related problem. Im not asking for "solve this for me" I want to learn so any hints would be greatly appreciated

Answer 1

Asserting that $\bigcap_{n\in\mathbb N}\left(0,\frac1n\right]=\emptyset$ simply means that no number belongs to every interval $\left(0,\frac1n\right]$. To see why, let $x\in\mathbb R$. Then:

• if $x\leqslant0$, then $x$ belongs to no interval $\left(0,\frac1n\right]$;
• otherwise, take $m\in\mathbb N$ such that $m>\frac1x$. Then $\frac1m<x$ and therefore $x\notin\left(0,\frac1m\right]$, from which it follows that $x\notin\bigcap_{n\in\mathbb N}\left(0,\frac1n\right]$.

By the way, it is false that $\emptyset\in\bigcap_{n\in\mathbb N}\left(0,\frac1n\right]$, since each set $\left(0,\frac1n\right]$ is a set of real numbers, and $\emptyset$ is not a real number.

Answer 2

To show $\bigcap_{n\in \mathbb{N}}(0,1/n]=\varnothing,$ it suffices to show simply that no real number is contained in $\bigcap_{n\in \mathbb{N}}(0,1/n]$. Note that you don't need to show $\varnothing \subseteq \bigcap_{n\in \mathbb{N}} (0,1/n]$, because the emptyset is a subset of every set vacuously. Now: suppose $x\in \bigcap_{n\in \mathbb{N}} (0,1/n]$. Then $x\in (0,1/n]$ for all $n$. So, $0<x \le 1/n$ for all $n$. So, $\lvert x\rvert <\varepsilon$ for all $\varepsilon>0$. Hence, $x=0$. But $0\not\in (0,1/n]$ for any $n$. We have a contradiction.