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I want to prove $$\liminf_{n \to \infty} \left( \inf_{x \in X} f_n(x) \right) \leq \inf_{x \in X} \left( \limsup_{n \to \infty} f_n(x)\right)$$

when $X \subset \mathbb{R}$ and $f_n\colon X \rightarrow \mathbb{R}$ so that $\{f_n(x)\>\colon \> n \in \mathbb{N}, x \in X\}$ is bounded.

I know that should be able to use the property that if $x_n \leq y_n$ then $\limsup_{n} x_n \leq \limsup_{n} y_n$ and that $\inf_{n} x_n \leq \liminf_{n} x_n$.

But how am I supposed finish this proof?

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    $\begingroup$ yes thank you, it's supposed to be bounded $\endgroup$ – Hannah Apr 2 '19 at 20:08
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You have that $$ \inf_{x\in X}f_n(x)\le f_n(x)\:\:\:\: \forall n\in \Bbb N\iff \liminf_{n\to\infty}\left( \inf_{x \in X} f_n(x) \right)\le\liminf_{n\to\infty}f_n(x)\:\:\:\:\forall x\in X. $$ But we have also that $$ \liminf_{n\to\infty}f_n(x)\le\limsup_{n\to\infty}f_n(x)\quad \forall x\in X. $$ Now, since the infimum of a bounded from below set in $\Bbb R$ is the largest of all lower bounds, we get $$ \liminf_{n \to \infty} \left( \inf_{x \in X} f_n(x) \right) \leq \inf_{x \in X} \left( \limsup_{n \to \infty} f_n(x)\right), $$ because $\liminf_{n\to\infty}\big( \inf_{x \in X} f_n(x) \big)$ is a lower bound for the set $$ \left\{\limsup_{n\to\infty}f_n(x)\:\big|\:x\in X\right\}, $$ which in turn is bounded and thus bounded from below since $\{f_n(x)\>\colon \> n \in \mathbb{N}, x \in X\}$ is.

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By properties of the infimum, it suffices to prove that for every $x\in X$, we have $$\liminf_n(\inf_{z\in X} f_n(z)) \,\le\, \limsup_n f_n(x)$$ But this easily follows, since $\inf_z f_n(z) \le f_n(x)$, hence $$\liminf_n(\inf_z f_n(z)) \,\le\, \liminf_n f_n(x)\,\le\, \limsup_n f_n(x)$$

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