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In the definition of topology, we allow infinite unions but only allow finite intersections. As mentioned by many other answers (see In a topological space, why the intersection only has to be finite?; Why Use Arbitrary Unions and Finite Intersections in Topology?), it is said that we want to keep the sets to be open set after the allowed operation. But the infinite intersection of open sets can be a close set. The example is given: \begin{equation}\bigcap_{n\in\mathbb{N}}(a-\frac{1}{n},b+\frac{1}{n})=[a,b],\quad (b>a)\end{equation} The question is: what is the answer of \begin{equation}\bigcup_{n\in\mathbb{N}}(a+\frac{1}{n},b-\frac{1}{n})=?\quad (b>a+2)\end{equation} is it $[a,b]$ or $(a,b)$?

If it is $(a,b)$ , seems that we use a different rule of limitation between intersection and union; if it is $[a,b]$,seems that we get a close set by operations (allowed by topology) on the open sets.

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    $\begingroup$ If it were $[a,b]$ then $a$ would have to be an element of $(a+1/n,b-1/n)$ for some $n$. Which $n$? $\endgroup$ – Lord Shark the Unknown Aug 11 '18 at 18:47
  • $\begingroup$ @LordSharktheUnknown I am not very clear about your question. For infinite intersection, I guess this is an agreed result. There is not actual $n$ there. It's just like a limitation. For infinite unions, I thought we can use the same argument. $\endgroup$ – X liu Aug 11 '18 at 19:03
  • $\begingroup$ You can get a closed set by operating on open sets. For example, $(-\infty, 1) \cup (-1, \infty)$ and $(1,2) \cap (3,4)$ are both closed sets. (they are also both open sets) $\endgroup$ – Hurkyl Aug 11 '18 at 19:10
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    $\begingroup$ @XLiu: Infinite unions and intersections are not limits. These are operations performed on the entire family of sets. They are not some notion of a "limit" of partial unions or partial intersections of finitely many sets. $\endgroup$ – Hurkyl Aug 11 '18 at 19:15
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    $\begingroup$ @Sambo This make sense. With others answers I now know clearly the difference.Thank you. $\endgroup$ – X liu Aug 11 '18 at 19:51
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For all $n\in \mathbb{N}$, $$\left(a+\frac{1}{n},b-\frac{1}{n} \right)\subset (a,b) $$.

Thus,

$$ \bigcup_{n\in \mathbb{N}}\left(a+\frac{1}{n},b-\frac{1}{n} \right) \subset (a,b)$$.

Conversely, if $x\in (a,b)$, there exists $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subset (a,b)$. Then there exists $N_1,N_2\in \mathbb{N}$ such that

$$b-\epsilon-x > \frac{1}{N_1},\hspace{2mm} \frac{1}{N_2}<x-\epsilon-a $$

Let $N=\max(N_1,N_2)$, then

$$b-\frac{1}{N}>x+\epsilon \text{ and } x-\epsilon > a+\frac{1}{N}$$, so

$$ x\in (x-\epsilon,x+\epsilon)\subset \left(a+\frac{1}{N},b-\frac{1}{N}\right)\subset \bigcup_{n\in \mathbb{N}}\left(a+\frac{1}{n},b-\frac{1}{n} \right)$$.

Thereforem

$$(a,b)= \bigcup_{n\in \mathbb{N}}\left(a+\frac{1}{n},b-\frac{1}{n} \right)$$

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  • $\begingroup$ This make sense. So this is like to say $x\in(a+\epsilon,b-\epsilon)$ for unions. The limitation is $(a,b)$; while $x\in(a-\epsilon,b+\epsilon)$ for intersection, which always includes $[a,b]$, and in limitation should be $[a,b]$? @LevBan $\endgroup$ – X liu Aug 11 '18 at 19:25
  • $\begingroup$ Yes $$ \bigcap_{n\in \mathbb{N}}\left(a-\frac{1}{n},b+\frac{1}{n}\right)=[a,b]$$ $\endgroup$ – Lev Ban Aug 11 '18 at 19:29
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\begin{align}\bigcup_{n\in\mathbb N}\left(a+\frac1n,b-\frac1n\right)&=(a+1,b-1)\cup\left(a+\frac12,b-\frac12\right)\cup\ldots\\&=(a,b).\end{align}In fact:

  1. $(\forall n\in\mathbb{N}):\left(a+\frac1n,b-\frac1n\right)\subset(a,b)$;
  2. if $x\geqslant b$ or $x\leqslant a$, then $x$ belongs to no interval of the form $\left(a+\frac1n,b-\frac1n\right)$;
  3. if $x\in(a,b)$, there's a $n\in\mathbb N$ such that $x\in\left(a+\frac1n,b-\frac1n\right)$, since $\lim_{n\to\infty}a+\frac1n=a$ and $\lim_{n\to\infty}b-\frac1n=b$.

So, the union of all these open sets is again an open set and furthermore it is not a closed set.

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Note that, for every $n \in \Bbb N = \{1, 2, 3, \ldots \}$,

$\left (a + \dfrac{1}{n}, b - \dfrac{1}{n} \right ) = \left \{ x \in \Bbb R \mid a + \dfrac{1}{n} < x < b - \dfrac{1}{n} \right \}; \tag 1$

thus it is easy to see that

$\forall n \in \Bbb N, \; a, b \notin \left (a + \dfrac{1}{n}, b - \dfrac{1}{n} \right ); \tag 2$

it then follows that

$a, b \notin \displaystyle \bigcup_{n \in \Bbb N} \left (a + \dfrac{1}{n}, b - \dfrac{1}{n} \right ); \tag 3$

however, we do have

$y \in (a, b) \Longrightarrow \exists n \in \Bbb N, \; y \in \left \{ x \in \Bbb R \mid a + \dfrac{1}{n} < x < b - \dfrac{1}{n} \right \} = \left (a + \dfrac{1}{n}, b - \dfrac{1}{n} \right ), \tag 4$

from which

$y \in (a, b) \Longrightarrow y \in \displaystyle \bigcup_{n \in \Bbb N} \left (a + \dfrac{1}{n}, b - \dfrac{1}{n} \right ) \Longrightarrow (a, b) \subset \bigcup_{n \in \Bbb N} \left (a + \dfrac{1}{n}, b - \dfrac{1}{n} \right ), \tag 5$

and since for every $n \in \Bbb N$ we clearly have

$\left (a + \dfrac{1}{n}, b - \dfrac{1}{n} \right ) \subset (a, b), \tag 6$

we may infer that

$\displaystyle \bigcup_{n \in \Bbb N} \left (a + \dfrac{1}{n}, b - \dfrac{1}{n} \right ) \subset (a, b); \tag 7$

therefore,

$(a, b) = \displaystyle \bigcup_{n \in \Bbb N} \left (a + \dfrac{1}{n}, b - \dfrac{1}{n} \right ), \tag 8$

and from (3) we see that

$[a, b] \ne \displaystyle \bigcup_{n \in \Bbb N} \left (a + \dfrac{1}{n}, b - \dfrac{1}{n} \right ). \tag 9$

In conclusion, we observe that there is in fact a different "rule of limitation" between intersection and union, since an infinite union of open sets must be open, but an infinite intersection of open sets may indeed be closed.

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