Why is $\bigcup_{n=1}^{\infty} [a+\frac{1}{n},b-\frac{1}{n} ]=(a,b)$?

Why is $\displaystyle \bigcup_{n=1}^{\infty} \left[a+\frac{1}{n},b-\frac{1}{n} \right]=(a,b)$? Isn't it $\displaystyle \lim_{n \to \infty} \frac{1}{n} = 0$, which implies the union is $[a,b]$?

I am more comfortable with $\displaystyle \bigcup_{n=1}^{\infty} \left(a+\frac{1}{n},b-\frac{1}{n} \right)=(a,b)$, but not convinced by the union of infinite closed intervals.

So they are equal?

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In which of the closed intervals is $a$ contained in the union you mention in the title? –  Mariano Suárez-Alvarez Jan 13 '13 at 5:13
By the way: please make the question bodies self-contained. The title is just a title: what would you think of a book whose title is the first sentence of the main text? –  Mariano Suárez-Alvarez Jan 13 '13 at 5:16

Recall that if $x \in \bigcup_{\alpha \in \Gamma} A_{\alpha} \implies x \in A_{\beta}$ for some $\beta \in \Gamma$. Now consider $$S = \bigcup_{n \in \mathbb{Z}^+}\left[a+\dfrac1n, b- \dfrac1n \right].$$ If $a \in S$, then $a \in \left[a+\dfrac1n, b- \dfrac1n \right]$ for some $n \in \mathbb{Z}^+$, which is incorrect. Hence, $a \notin S$ and similarly $b \notin S$.

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Note that no matter how big you take $n$, the interval $\left[a+\frac1n,b-\frac1n\right]$ does not include either $a$ or $b$. That is, for every $n\in\Bbb Z^+$ we have

$$a,b\notin\left[a+\frac1n,b-\frac1n\right]\;,$$

and if follows immediately from the definition of union that

$$a,b\notin\bigcup_{n\in\Bbb Z^+}\left[a+\frac1n,b-\frac1n\right]\;.$$

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