Is $\cup_{n=1}^{\infty}\{x:f(x) > \frac{1}{n}\}=\cup_{n=1}^{\infty}\{x:f(x) \geq \frac{1}{n}\}$

Question

I am learning real analysis recently, I know that $$\{x:f(x) > \frac{1}{n}\} = \bigcup_{n=1}^{\infty}\{x:f(x) > 0\}$$ but does it equal to $$\bigcup_{n=1}^{\infty}\{x:f(x)\geq \frac{1}{n}\}$$ and why?

Answer 1

First, it should be$$ \{x \mid f(x) > 0\} = \bigcup_{n=1}^{\infty} \left\{x \,\middle|\,f(x) > \frac{1}{n}\right\}. $$

Now, it is easy to see that$$ \{x \mid f(x) > 0\} \supseteq \bigcup_{n=1}^{\infty} \left\{x \,\middle|\,f(x) \geqslant \frac{1}{n}\right\}. $$ And for any $x \in \{x \mid f(x) > 0\}$, suppose $n \in \mathbb{N}_+$ satisfies $n \geqslant \dfrac{1}{f(x)}$, then$$ f(x) \geqslant \frac{1}{n} \Longrightarrow x \in \left\{x' \,\middle|\,f(x') \geqslant \frac{1}{n}\right\}. $$ Thus$$ \{x \mid f(x) > 0\} \subseteq \bigcup_{n=1}^{\infty} \left\{x \,\middle|\,f(x) \geqslant \frac{1}{n}\right\}. $$

Answer 2

Let $A_n =\{x : f(x) > \frac{1}{n}\}$ and $B_n = \{x : f(x) \geq \frac{1}{n}\}$.

Then its easy to see that $A_n \subset B_n$ so we have that $\bigcup_{n=1}^{\infty} A_n \subset \bigcup_{n=1}^{\infty} B_n(1)$.

Now let $x \in \bigcup_{n=1}^{\infty} B_n$ then there exist $n_0$ such that $x\in B_{n_0}$. So , $f(x) \geq \frac{1}{n_0}$ now set

$n' = 2n_0 > n_0$. Then you have that $\frac{1}{n'} < \frac{1}{n_0}$ but since $f(x) \geq \frac{1}{n_0}$ you also have that $f(x) >\frac{1}{n'}$.

Which means that $x \in A_{n'} \subset \bigcup_{n=1}^{\infty} A_n$. So $\bigcup_{n=1}^{\infty} B_n \subset \bigcup_{n=1}^{\infty} A_n$ (2).

And from $(1),(2)$ you have the desired equality !