Topology definition: finite intersections vs. infinite unions

Question

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.

Answer 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.

Answer 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

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

Answer 3

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.