 uplakanimkk

2022-07-07

Prove that $\bigcup _{n\in \mathbb{N}}\left(-n,\frac{1}{n}\right]=\left(-\mathrm{\infty },1\right]$
I understand that for this to be true, I have to show that $\bigcup _{n\in \mathbb{N}}\left(-n,\frac{1}{n}\right]\subset \left(-\mathrm{\infty },1\right]$ and $\left(-\mathrm{\infty },1\right]\subset \bigcup _{n\in \mathbb{N}}\left(-n,\frac{1}{n}\right]$, however, I'm stuck on figuring it out. I think the latter can be proven by contradiction somehow. Tanner Hamilton

Expert

Step 1
Show mutual inclusion.
First suppose that $x\in \bigcup _{n\in \mathbb{N}}\left(-n,\frac{1}{n}\right]$. Choose $n\in \mathbb{N}$ such that $x\in \left(-n,\frac{1}{n}\right]$. Since $x\le \frac{1}{n}\le 1$, $x\in \left(-\mathrm{\infty },1\right]$. This shows that $\bigcup _{n\in \mathbb{N}}\left(-n,\frac{1}{n}\right]\subseteq \left(-\mathrm{\infty },1\right]$.
Step 2
Conversely, suppose that $x\in \left(-\mathrm{\infty },1\right]$. If $0\le x\le 1$, we may choose $n\in \mathbb{N}$ such that $x\le \frac{1}{n}$. Then $x\in \left(-n,\frac{1}{n}\right]$, so $x\in \bigcup _{n\in \mathbb{N}}\left(-n,\frac{1}{n}\right]$. On the other hand, if $x<0$, there exists $n\in \mathbb{N}$ such that $x>-n$. $x<0<\frac{1}{n}$, so $x\in \left(-n,\frac{1}{n}\right]$, so $x\bigcup _{n\in \mathbb{N}}\left(-n,\frac{1}{n}\right]$. This shows that $\bigcup _{n\in \mathbb{N}}\left(-n,\frac{1}{n}\right]\subseteq \left(-\mathrm{\infty },1\right]$. Desirae Washington

Expert

Step 1

as $n\in \mathbb{N}$ and as $-n>-\mathrm{\infty }$ then $x\in \left(-\mathrm{\infty },1\right]$
Step 2
For the second, we know that for every number a there exists number $b>a$ such that $b\in \mathbb{N}$, in the same way it works backwards, so that for every a there exists $b such that $b\in \mathbb{Z}$.
Step 3
If $0\le x\le 1$ we take $n=1$, else we take $n=y$ where y is a smaller number than x that is in $\mathbb{N}$ and satisfies that $y, so we get that again $x\in \left(-y,\frac{1}{y}\right]$.

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