$ \{ x : f(x) > 0 \} = \bigcup_n f^{-1}([\frac{1}{n}, \infty )) $ $ \{ x : f(x) > 0 \} = \bigcup_n f^{-1}([\frac{1}{n}, \infty )) $
My try: Pick $$x \in \bigcup_n f^{-1}([\frac{1}{n}, \infty )) \implies x \in f^{-1}([\frac{1}{n}, \infty )) \text{ for some $n$ } $$
$$ \therefore f(x) \geq \frac{1}{n} \text{for some $n$ } \implies f > 0$$
Other direction, if $x \in  \{ x : f(x) > 0 \} \implies f(x) > 0 $. We want to show $f \geq \frac{1}{n} $. If not, then $f < \frac{1}{n}  \implies f = 0$ contradiction and hence $f \geq \frac{1}{n} \implies x \in f^{-1}([\frac{1}{n}, \infty )) \text{ for some $n$ } $ therefore, $$x \in \bigcup_n f^{-1}([\frac{1}{n}, \infty )) $$
Is this correct? Im a little unsure about last picture. I would greatly appreciate any feedback.
 A: Yes, its correct (as far as I can see). But personally - and this is just a stylistic thing - I would rather break it into smaller problems. Consider the following argument.
$$\bigcup_n f^{-1}([\frac{1}{n}, \infty )) = f^{-1}\left(\bigcup_n [\frac{1}{n}, \infty )\right)=f^{-1}(0,\infty) = \{ x : f(x) > 0 \}$$
All that remains is to verify the intermediate steps, by proving a few lemmas. Namely:


*

*$\bigcup_n f^{-1}(A_n) = f^{-1}(\bigcup_n A_n)$ for an arbitrary function $f$

*$\bigcup_n [\frac{1}{n}, \infty ) = (0,\infty)$

*$f^{-1}(0,\infty) = \{ x : f(x) > 0 \}$


Since 1 is a standard result, you can probably omit the proof. Also, 3 is just definition chasing. So 2 is what you want to show. I suggest you break it up into the following results:


*

*The sequence $a$ defined by $a_n = 1/n$ is strictly decreasing.

*If $a$ is a strictly decreasing sequence, then $a$ has no least element.

*If $a$ has no least element, then $\bigcup_n [a_n, \infty ) = (\mathrm{inf}(a),\infty).$

*The sequence $a$ defined by $a_n = 1/n$ has the property that $\mathrm{inf}(a) = 0$
Hope that was helpful!
