Relation between two sets, one with limit inside, the other with limit outside

Question

Consider these two sets:

$$ A\equiv \{x\in X: \forall \xi>0 \text{ }\exists N_{\xi, x} \text{ s.t. } \forall N\geq N_{\xi, x} \text{ } d(p_N, I(x))\leq \xi)\}, $$ $$ B\equiv \bigcap_{\xi>0} \bigcup_{N=1}^\infty\bigcap_{K=N}^\infty \{x\in X: d(p_K, I(x))\leq \xi\}, $$ where

• $A$ and $B$ are non-empty.
• $(p_N)_N$ is a sequence of reals taking values in $[0,1]$.
• $I(x)$ is an interval in $(0,1)$.
• $d\big(p_N, I(x) \big):= \inf \big\{|p_N - y| : y \in I(x) \big\}$.

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Question: Is $A=B$, $A\subseteq B$, $A\supseteq B$?

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Some thoughts:

1. $A$ is equal to: $$ A= \{x\in X: \lim_{N\rightarrow \infty}d(p_N, I(x))=0)\}, $$ where I use the definition of limit.

2. $B$ is equal to: $$ B=\bigcap_{\xi>0} \liminf_{N\rightarrow \infty}\{x\in X: d(p_N, I(x))\leq \xi \}= \liminf_{N\rightarrow \infty}\{x\in X: d(p_N, I(x))=0 \} $$ where the first equality comes from the definition of $\liminf$ applied to sets (see here) and the second equality comes from my understanding of some comments given to another questions of mine. However, from the comments below, it seems that this second equality is wrong (any clarification would be welcome here).

So, $A$ has the limit inside, while $B$ outside. Is there any obvious relation between $A$ and $B$, e.g., $B$ is a superset/subset of $A$?

Answer 1

I will limit myself to your question 1 and your thoughts about it. Maybe you should ask your question 2 later in a new post, more clearly and with some attempts.

• $B=A$ by definition of (infinite) unions and intersections: $$\begin{align}B&=\bigcap_{\xi>0}\bigcup_{N=1}^\infty\{x\in X:\forall K\ge N, d(p_K, I(x))\le\xi\}\\&=\bigcap_{\xi>0}\{x\in X:\exists N\ge1\;\forall K\ge N\;d(p_K, I(x))\le\xi\}\\&=\{x\in X:\forall\xi>0\;\exists N\ge1\;\forall K\ge N\;d(p_K, I(x))\le\xi\}\\&=A. \end{align}$$
• $B=\bigcap_{\xi>0} \liminf_{N\rightarrow \infty}\{x\in X: d(p_N, I(x))\leq \xi \}$ indeed, by definition of $\liminf$ of a sequence of sets.
• This $\liminf$ cannot be pulled out. More precisely:
  • the following inclusion holds: $$ B\supseteq\liminf_{N\rightarrow \infty}\{x\in X: d(p_N, I(x))=0 \} $$ because if $x\in\liminf_{N\rightarrow\infty}\{x\in X:d(p_N,I(x))=0\}$, i.e. if $\exists N\;\forall K\ge N\;d(p_K, I(x))=0$, then obviously $\lim_{N\rightarrow \infty}d(p_N, I(x))=0$, i.e. $x\in A$, or equivalently $x\in B$.
  • the reverse inclusion generally doesn't hold. For instance if $I(x)=(1/2,1/4)$ ($\forall x\in X$) and $p_n=\frac12-\frac1n$, then $$ B=A=X\not\subseteq\varnothing=\liminf_{N\rightarrow \infty}\{x\in X: d(p_N, I(x))=0 \} $$