Countably infinite union Is it the case that $(a,b) \subseteq\bigcup_{n\in \mathbb{N}} (a+\frac{1}{n}, b-\frac{1}{n})$?
Seeing as we are only indexing by positive integers? i.e we never reach $\infty$
 A: The inclusion does hold (in fact the sets are equal!)
Given $x \in (a,b)$, we have $$x-a = d_1\\b-x = d_2$$where $d_1, d_2 >0$. Hence, we can choose $n \in \mathbb N$ such that $\frac 1n < \min\{d_1, d_2\}$. So $$a+\frac 1n < x < b-\frac1n$$Hence, $x \in (a+\frac 1n, b-\frac 1n) \subset \bigcup_{n \in \mathbb N}(a+\frac1n, b-\frac1n)$.

The point here is that since $(a,b)$ is an open interval, it doesn't matter that $\frac1n$ will never "reach $0$" in finite time.
The answer lies in the way we define the limit: we say $x_n \to x$ if for any arbitrary distance $\epsilon > 0$, after some finite time, the sequence $x_n$ remains closer than $\epsilon$ from $x$. In your case, since $(a,b)$ is an open interval, any $x \in (a,b)$ will be bounded away from $a$ and $b$ by some positive distance. We can use this distance as our $\epsilon$ in the limit and show that after some point, $(a+\frac 1n, b-\frac 1n)$ will contain $x$.
However, if we had the closed interval $[a,b]$ then it would matter and we wouldn't have the inclusion.
A: It is true. For any $x\in (a,b)$, there is an $n$ such that $a+\frac1n < x$, and there is an $m$ such that $b-\frac1m > x$.
A: Here's a constructive proof. Suppose $x \neq (a+b)/2$ satisfies $a < x < b$. Let $C$ be the minimum of $|a-x|$ and $|b-x|$. Then choose a positive integer $N > 1/C$. You can show that $x$ is contained in $(a+1/N,b-1/N)$.
If $x = (a+b)/2$ the above argument doesn't work but clearly if a point different from $(a+b)/2$ is contained in a interval symmetric around $(a+b)/2$, then $(a+b)/2$ is contained in the interval as well.
A: Yes - for each $x$ such that $ a < x < b$, there exists $ N \in \mathbb{N}$ such that $ x \in (a + \frac{1}{N},b - \frac{1}{N})$, which is the definition of the union on the right hand side.
