Define $(X_n)$ to be a bounded sequence of real numbers such that $∀n∈ℕ, X_n ≥ 0$
Show that $\liminf \sqrt{X_n} = \sqrt{\liminf X_n}$
Side-question: $(X_n) ⊆ ℝ$?
How should I go about on this proof? I don't see a direct way and am wondering if I should first prove $\liminf \sqrt{X_n} \le \sqrt{\liminf X_n}$ and then $\liminf \sqrt{X_n} \ge \sqrt{\liminf X_n}$?
How do I start? If I consider $u ∈ S(\sqrt{X_n})$, a subsequence limit of $\sqrt{X_n}$, how to proceed?
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side question
If I write $(X_n)$ for a sequence of reals, I do not mean the same thing as the set $\{X_n\}$ of reals. This sequence $$ 0,1,1,1,1,1,1,\cdots $$ (continuing with the value $1$) is not the same as this seeuence $$ 0,1,0,1,0,1,0,1,\cdots $$ (continuing to alternate). These two sequences have different $\liminf$. But of course the two sets we get from them are both $\{0,1\}$.