1

Let $I \subseteq \mathbb{R}$ and $f_n : I \to \mathbb{R}$ be a sequence of continuous functions. Let $a \in \mathbb{R}$ and define:

$$ A = \{ x \in I : \limsup \limits_{n \to \infty} |f_n(x)| \leq a\}$$

I'm a bit confused about how can I write the set $A$ as unions and intersections of sets. For instance, I know that: $$ \limsup \limits_{n \to \infty} |f_n(x)| = \inf_{j \in \mathbb{N}} \sup_{n \geq j} |f_n(x)|$$ And I've studied that this implies:

\begin{equation} A = \bigcap_{j \in \mathbb{N}} \bigcup_{n=j}^{\infty} \{ x \in I : |f_n(x)| \leq a\} ... (1) \end{equation} (I understand the reason for this equality in terms of set inequalities with $\subseteq$ as partial order).

But why couldn't it be the other way around?, I mean, Why couldn't we have: $$ A = \bigcup_{j \in \mathbb{N}} \bigcap_{n=j}^{\infty} \{ x \in I : |f_n(x)| \leq a\}$$

I'm asking this because if one thinks of a sequence of functions $F_n: I \to \mathbb{R}$ and if we consider the set: $$ B = \{ x \in I : \inf_{n \in \mathbb{N}}|F_n(x)| \leq a \}$$

Then by the definition of infimum we have that $x \in B$ if and only if there exists some $N \in \mathbb{N}$ such that $|F_N(x)| \leq a$, so:

$$ B = \bigcup_{n=1}^{\infty} \{ x \in I : |F_n(x) | \leq a \}$$

Thus, couldn't the infimum be represented in (1) as the union of sets instead of the intersection?

Asaf Karagila
  • 405,794
MC2
  • 819
  • not quite what you are asking, but you might find useful https://math.stackexchange.com/questions/1456350/the-supremum-and-infimum-of-a-sequence-of-measurable-functions-is-measurable/1456367#1456367 – peter a g Oct 28 '24 at 01:29
  • 3
    For a sequence of constant functions with values $a+1/n$, all the sets ${|f_n|\le a}$ are empty although ${\limsup |f_n|\le a}=I$. However, $${\limsup |f_n|\le a}= \bigcap_{k\in\mathbb N}\bigcup_{n\in\mathbb N}\bigcap_{m\ge n} {|f_m|\le a+1/k}.$$ – Jochen Oct 28 '24 at 08:21

1 Answers1

1

As pointed out by @Jochen in the comments, the equality (1) in the question is not true.

To understand the order of the intersections and unions, suppose that we have a proposition $P_t(x)$ depending on a parameter $t\in T$ (a set). Then we have \begin{equation} \forall t\in T, P_t(x) \Longleftrightarrow x \in \bigcap_{t\in T}\{y: P_t(y)\} \end{equation} \begin{equation} \exists t\in T, P_t(x) \Longleftrightarrow x \in \bigcup_{t\in T}\{y: P_t(y)\} \end{equation} Hence the quantifier determines whether we have an intersection or a union.

Going back to the question we have \begin{equation} \limsup_{n\to \infty} |f_n(x)| \le a \Longleftrightarrow \forall k\ge 1, \exists j\ge 0, \forall n\ge j, |f_n(x)|\le a + \frac{1}{k} \end{equation} Hence \begin{equation} x\in A \Longleftrightarrow x\in \bigcap_{k\ge 1}\bigcup_{j\ge 0}\bigcap_{n\ge j} \{y: |f_n(y)|\le a + \frac{1}{k}\} \end{equation}

Gribouillis
  • 16,826