If (En) is a sequence of subsets of a fixed set S, we define n →∞lim sup En=⋂n=1^∞(⋃k=n^∞ Ek) a

Step 1: We start by recalling the definitions of u005C(u005Climinfu005C) and u005C(u005Climsupu005C) for the sequence of

Question

If $\left(E_n\right)$ is a sequence of subsets of a fixed set $S$, we define $$ \underset{n \rightarrow \infty}{\limsup } E_n=\bigcap_{n=1}^{\infty}\left(\bigcup_{k=n}^{\infty} E_k\right) \text { and } \liminf _{n \rightarrow \infty} E_n=\bigcup_{n=1}^{\infty}\left(\bigcap_{k=n}^{\infty} E_k\right) \text {. } $$ Show that $$ \liminf _{n \rightarrow \infty} E_n \subset \underset{n \rightarrow \infty}{\limsup } E_n \text { and that } \underset{n \rightarrow \infty}{\liminf }\left(E_n^c\right)=\left(\limsup _{n \rightarrow \infty} E_n\right)^c \text {. } $$

   If $\left(E_n\right)$ is a sequence of subsets of a fixed set $S$, we define
$$
\underset{n \rightarrow \infty}{\limsup } E_n=\bigcap_{n=1}^{\infty}\left(\bigcup_{k=n}^{\infty} E_k\right) \text { and } \liminf _{n \rightarrow \infty} E_n=\bigcup_{n=1}^{\infty}\left(\bigcap_{k=n}^{\infty} E_k\right) \text {. }
$$

Show that
$$
\liminf _{n \rightarrow \infty} E_n \subset \underset{n \rightarrow \infty}{\limsup } E_n \text { and that } \underset{n \rightarrow \infty}{\liminf }\left(E_n^c\right)=\left(\limsup _{n \rightarrow \infty} E_n\right)^c \text {. }
$$

Real analysis

N. L. Carothers 1st Edition

Chapter 1, Problem 37

Instant Answer

Step 1

\[ \liminf_{n \to \infty} E_n = \bigcup_{n=1}^{\infty} \left( \bigcap_{k=n}^{\infty} E_k \right) \] \[ \limsup_{n \to \infty} E_n = \bigcap_{n=1}^{\infty} \left( \bigcup_{k=n}^{\infty} E_k \right) \] Step 2: We need to show that \(\liminf_{n \to \infty} E_n Show more…

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If $\left(E_n\right)$ is a sequence of subsets of a fixed set $S$, we define $$ \underset{n \rightarrow \infty}{\limsup } E_n=\bigcap_{n=1}^{\infty}\left(\bigcup_{k=n}^{\infty} E_k\right) \text { and } \liminf _{n \rightarrow \infty} E_n=\bigcup_{n=1}^{\infty}\left(\bigcap_{k=n}^{\infty} E_k\right) \text {. } $$ Show that $$ \liminf _{n \rightarrow \infty} E_n \subset \underset{n \rightarrow \infty}{\limsup } E_n \text { and that } \underset{n \rightarrow \infty}{\liminf }\left(E_n^c\right)=\left(\limsup _{n \rightarrow \infty} E_n\right)^c \text {. } $$

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