proving an equality of two sets Suppose $x,y $ are real numbers, then
$$ (a,b) = \bigcup_{n=N}^{\infty} \left(a, b- \frac{1}{n} \right] \; \; $$
for large enough $N$.
TRY:
Say $x$ is in the union, then there is some $n_0 \geq N$ such that $ a < x \leq b - \frac{1}{n_0} $. I wanna show $x \in (a,b)$. Already I have $a < x$. I need to show $x < b$. If $x \geq b$, then $x \geq b \geq x + \frac{1}{n_0}$. this is contradiction since $1/n_0 > 0 $.
IS this correct? Also, how can I show that $(a,b)$ is contained in the union ?

The other direction: if $x \in (a,b) \to a < x < b \to \exists N: N(b-x) \geq 1$ by Archemedian Principle. Thus $\dfrac{1}{N} \leq b-x \to x \leq b - \dfrac{1}{N} \to x \in \left(a, b-\frac{1}{N}\right] \to x \in \displaystyle \bigcup_{n=N}^\infty \left(a,b-\frac{1}{n}\right]$
