The following question is from Real Analysis by Folland 2nd edition in the prologue. The author denotes families of sets denoted by $\mathbb{N}$ in the following manner: $$\{E_{n}\}_{n=1}^{\infty}$$ In this situation the notions of limit superior and limit inferior are sometimes useful: $$\limsup E_n = \cap_{k=1}^{\infty} \cup_{n=k}^{\infty} E_n, \ \ \liminf E_n = \cup_{k=1}^{\infty} \cap_{n=k}^{\infty} E_n$$ Verify that, $$\limsup E_n = \{x:x\in E_n \ \ \text{for infinitely many} \ \ n \}$$ and $$\liminf E_n = \{x:x\in E_n \ \ \text{for all but finitely many} \ \ n \}$$

Proof of first case (scratch work): I believe we need to prove both cases in which $x\in E_n$ and when $x\notin E_n$. I want to take the assumption in which $E_n$ is open and an assumption in which we suppose it is closed and see where that goes, but I am not sure if this is the correct approach.

Proof of second case (scratch work): I would use the same approach as above if it is correct.

Any suggestions would be greatly appreciated


If you want to show equality of two sets $A$ and $B$, it suffices to show $A \subset B$ and $B \subset A$. You don't need the notion of open or closed sets. I'll outline the first one...

We first show $\limsup_{n \to \infty} E_n \supset \{x : \text{$x \in E_n$ for infinitely many $n$}\}$. If $x$ is contained in infinitely many $E_n$, then it is in $\bigcup_{n=k}^\infty E_n$ for any $k\ge 1$. Therefore, $x$ is in $\limsup_{n \to \infty}E_n$.

To show the reverse inclusion $\limsup_{n \to \infty} E_n \subset \{x : \text{$x \in E_n$for infinitely many $n$}\}$, consider a point $x$ belonging to $\limsup_{n \to \infty} E_n$. Suppose for sake of contradiction that $x$ belonged to only finitely many of the $E_n$. Then for all large enough $k$, the point $x$ would not belong to $\bigcup_{n = k}^\infty E_n$, and therefore $x$ would not belong to $\limsup_{n \to \infty} E_n$, a contradiction. So $x$ belongs to infinitely many of the $E_n$.

A similar approach works for the second case. I think you are missing the word "but" in "all but finitely many sets."

  • $\begingroup$ For the first part, just want to be clear so if $x$ infinitely many $E_n$ then it must also be in the union. That essentially shows that $x$ is in $\limsup E_n$? I believe for the second part you simply showed a contradiction by assuming that $x$ only belongs to finitely many $E_n$? $\endgroup$ – Wolfy Aug 9 '15 at 0:03
  • $\begingroup$ @MorganWeiss For the first part, if $x$ is in infinitely many $E_n$, then it must be in every union of the form $\bigcup_{n=k}^\infty E_n$ so that therefore $x$ lies in the intersection of all these unions. For the second part yes; another way to look at it is that I showed $A \subset B$ by showing that $A^c \supset B^c$ (where the superscript $c$ denotes set complement). $\endgroup$ – angryavian Aug 9 '15 at 0:45
  • $\begingroup$ Ok, thanks I got it now $\endgroup$ – Wolfy Aug 9 '15 at 16:43

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