1

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.

bof
  • 78,265
X Leo
  • 508
  • 2
    If it were $[a,b]$ then $a$ would have to be an element of $(a+1/n,b-1/n)$ for some $n$. Which $n$? – Angina Seng Aug 11 '18 at 18:47
  • @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. – X Leo Aug 11 '18 at 19:03
  • 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) –  Aug 11 '18 at 19:10
  • 2
    @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. –  Aug 11 '18 at 19:15
  • That make sense. Can you tell me how can we think it for the intersection? Seems that we always have $a-\epsilon$ belong to the result. So the result is always a little bigger than $[a,b]$ for the intersection?@Hurkyl – X Leo Aug 11 '18 at 19:16
  • You seem to be a bit fuzzy about the definitions of infinite union and intersection. For a family of sets $A_i$, with $i \in I$, the union $\cup_{i \in I} A_i$ is just the set of elements that belong to at least one of the $A_i$. Likewise, the intersection $\cap_{i \in I} A_i$ is the set of elements that belong to all of the $A_i$. – Sambo Aug 11 '18 at 19:23
  • @Sambo I agree with you about that. Seems that my question does not confuse with this? Please notice that I change the sign of the lower boundary from subtraction to addition when I change from intersection to union. – X Leo Aug 11 '18 at 19:27
  • Your question doesn't make mistakes about these definitions; however, to figure out the answer to your question, you just need to take a look at the definition. As Lord Shark puts it, if $a$ is in the union of your sets, it must belong to one of them. Which one? – Sambo Aug 11 '18 at 19:49
  • @Hurkyl So why we ask inifinite unions but finite intersections by the definition? The results of two examples you given are the whole set and the empty set, they are both close set and open set. So they dont disgree with the motivation - keep the sets to be open set after the allowed operation. – X Leo Aug 11 '18 at 19:49
  • @XLiu The motivation to "keep the sets open" still works, since an important thing to consider is that a set can be both open and closed. This is what happens in Hurkyl's examples. The resulting sets are closed, yes, but they are also open. The set obtained by infinite intersection that you described in your question, on the other hand, is not open. – Sambo Aug 11 '18 at 19:51
  • 1
    @Sambo This make sense. With others answers I now know clearly the difference.Thank you. – X Leo Aug 11 '18 at 19:51

3 Answers3

2

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

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

Lev Bahn
  • 2,892
  • 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 – X Leo Aug 11 '18 at 19:25
  • Yes $$ \bigcap_{n\in \mathbb{N}}\left(a-\frac{1}{n},b+\frac{1}{n}\right)=[a,b]$$ – Lev Bahn Aug 11 '18 at 19:29
1

\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.

José Carlos Santos
  • 427,504
1

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.

Robert Lewis
  • 71,180