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I have a Cauchy sequence $x_n$ and as part of my proof I am trying to prove that $\liminf_{n \to \infty}x_n \leq \limsup_{n \to \infty}x_n$. So far I have shown that $\inf\{x_n : n \geq k\} \leq \sup\{x_n : n \geq k\}$, but I'm not sure how to prove that these limits exist in the first place so I can make the comparison.

I'm thinking somehow showing that they are either decreasing or increasing, and because they are bounded (Cauchy sequence is bounded).

Any hints would be appreciated!

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  • $\begingroup$ Which limits are you having trouble proving exist? $\endgroup$ Oct 21, 2021 at 17:40
  • $\begingroup$ @user6247850 proving the existence of either $\limsup$ or $\liminf$ $\endgroup$
    – user843046
    Oct 21, 2021 at 17:41
  • $\begingroup$ I am guessing that the definition you are working with is $\liminf x_n = \lim_{k \rightarrow \infty} \inf\{x_n : n \ge k\}$, with $\limsup$ defined similarly. Then I agree you should show they are monotone. Specifically, what is the relation between $\inf\{x_n : n \ge k\}$ and $\inf \{x_n : n \ge k+1\}$? $\endgroup$ Oct 21, 2021 at 18:00

1 Answer 1

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Hint: Let $A$ and $B$ be two non-empty bounded above subsets of reals. Suppose that $A\subseteq B$, then $\sup(A)\leq\sup(B)$.

Now, note that if $k_1\leq k_2$, then $\{x_n:n\geq k_1\}\supseteq\{x_n:n\geq k_2\}$. By using the above lemma, the monotonicity of the sequence $(\sup\{x_n:n\geq k\})_{k\in\mathbb{N}}$ follows. You can use similar argument to prove the monotonicity of the sequence $(\inf\{x_n:n\geq k\})_{k\in\mathbb{N}}$.

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