I witness a lemma in my class note and I think the proof is not quite clear. Could anybody give me some ideas about argument technique to prove the lemma? The lemma 3 in the beginning of the text: $X_n \to X$ in P iff for every subsequence $n_{k}$ there exists a further subsequence $n'_{k}$ such that $X_{n'_{k}} \to X$ almost surely.
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1How to use "every" in the problem? – sincostancot Jan 28 '15 at 21:29
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"$\Rightarrow$": Suppose that $X_n \to X$ in probability. For any $k \in \mathbb{N}$ there exists $N=N(k)$ such that $$\mathbb{P} \left( |X_n-X|> \frac{1}{2^k} \right) < \frac{1}{2^k} \qquad \text{for all} \, \, n \geq N(k).$$ Without loss of generality, $N(1)< N(2)<\dots$ If we set $n_k := N(k)$, then $$\sum_{k \geq 0} \mathbb{P} \left( |X_{n(k)}-X|>\frac{1}{2^k} \right)< \infty$$ and therefore it follows from the Borel-Cantelli lemma that $X_{n(k)} \to X$ almost surely as $k \to \infty$.
"$\Leftarrow$": Assume that $X_n$ does not converge to $X$ in probability, i.e. there exist $\epsilon>0$,$\delta>0$ such that for all $k \in \mathbb{N}$ there exists $n(k) \geq k$ such that $$\mathbb{P}(|X_{n(k)}-X|>\epsilon)>\delta.$$ By construction, the sequence $(X_{n(k)})_{k \in \mathbb{N}}$ does not have a convergent subsequence. This contradicts our assumption.
saz
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