0

Question copied from here.

In $E^2$, let $X$ be the infinite family of concentric open disks of radius $1 + 1/n$ for all $n \in \mathbb{Z^+}$. Why is $X$ a closed set?

The question in a nutshell , why : $$\bigcap_{n \in \mathbb{N}} \left(-1-\frac{1}{n}, 1+\frac{1}{n}\right)=[-1,1]$$?

This question was clearly solved here, but I have heard the argument that it is a countable intersection (because $n\in\mathbb N$) so then $n$ will never reach $+\infty$. I have a doubt about it.

Someone may confirm or disprove with an argument ?

Community
  • 1
Al Bundy
  • 492
  • Infinite intersection is enough. – Martín-Blas Pérez Pinilla Nov 22 '16 at 07:26
  • 1
    Countability is not the relevant issue. It's the same if you replace $\mathbb N$ with $\mathbb R^+$ which clearly is uncountable. However note that the set ${-1-1/n|n\in\mathbb N}$ has no maximum, nor has ${-1-1/n|n\in\mathbb R^+}$. Note that finite intersections of open sets are always open, so the fact that there are infinitely many sets is important. Also note that finite sets of real numbers always have a maximum and a minimum. – celtschk Nov 22 '16 at 07:28
  • Consider the analogous situation in 1 dimension: $[-1,1]=\cap_{n\in \mathbb N}(-1-1/n,1+1/n)=\cap_{r>0}(-1-r,1+r).$ – DanielWainfleet Nov 23 '16 at 09:56

3 Answers3

2

$\bigcap_{n \in \mathbb{N}} \left(-1-\frac{1}{n}, 1+\frac{1}{n}\right)\supseteq[-1,1]$ is clear, so we only have to prove the other containment:

$$\bigcap_{n \in \mathbb{N}} \left(-1-\frac{1}{n}, 1+\frac{1}{n}\right)\subseteq[-1,1]$$

Let $x\in \bigcap_{n \in \mathbb{N}} \left(-1-\frac{1}{n}, 1+\frac{1}{n}\right)$.

This means that $-1-\frac{1}{n}<x<1+\frac{1}{n}$ for all $n$.

Taking limits as $n\to\infty$, we get $-1\leq x\leq 1$, i.e. $x\in [-1,1]$.

We have proved the other containment.

yoyostein
  • 19,608
1

Hint

$$\bigcap_{x\in \mathbb R}(1-\frac{1}{1+x^2},1+\frac{1}{1+x^2})=\{1\}$$

hamam_Abdallah
  • 62,951
1

Infinity is not a number. $n$ will never reach $+\infty$ in the sense that $+\infty\notin\mathbb{N}$. But there are arbitrarily large $n$ in $\mathbb{N}$, that could be written as $\sup_{n\in\mathbb{N}} n=+\infty$. You can also interprete this as "there is no number greater than all natural numbers".

$\mathbb{N}$ being countable (countably infinite) is not necessary for the intersection to be closed. $$\bigcap_{x \in \mathbb{R}^{+}}\left(-1-\frac{1}{x}, 1+\frac{1}{x}\right) = [-1,1]$$ is true as well.

The important points are that

  1. For $y>1$ you can always find an $n\in\mathbb{N}$ such that $1+\frac{1}{n}<y$.

  2. There is no $n\in\mathbb{N}$ such that $1+\frac{1}{n} \leq 1$.