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Could someone please explain why the following hold for a set of a functions $f_n(x)$:

$$\{x : \sup f_n > c \} = \bigcup_{n=1}^\infty \{ x: f_n(x) > c \}$$ $$\{x: \inf f_n < c \} = \bigcup_{n=1}^\infty \{ x: f_n(x) < c \}$$

thank you

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1 Answer 1

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If $\sup f_n (x) > c$, then there exists $n \in \mathbb N$ such that $f_n(x) > c$ (if not, then $f_n(x) \leq c, \forall n$, which implies $\sup f_n(x) \leq c$, contradiction). So we have: $$\sup f_n(x) > c \Leftrightarrow \exists n \in \mathbb N: f_n(x) > c $$ Thus, $$x \in \{\sup f_n > c \} \Leftrightarrow \exists n \in \mathbb N: x \in \{f_n > c \} \Leftrightarrow x \in \cup_n\{f_n > c\}.$$ Similar for $\{\inf f_n < c \}$.

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  • $\begingroup$ ok, so for $\{x: \inf f_n < c\}$ there exists a $n$ such that $f_n(x) < c$ else $f_n(x) \geq c$ but this is not a contradiction as $ inf f_n < c \leq f_n$ $\endgroup$
    – lampj20la
    Apr 18, 2016 at 22:52
  • $\begingroup$ if $f_n(x) \geq c$, then $\inf f_n(x) \geq \inf c = c$. $\endgroup$
    – SiXUlm
    Apr 18, 2016 at 22:55

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